Quantitative Methods for Business and EconomicsJakub Kielbasa This book is written for students for free, so feel free to use, cite, steal example etc. It is a learning tool, but please cite it when you do use it. If you come across any mistakes, please contact
[email protected]. The wording of all theory, as well as all the content of all examples and exercises were the sole work of the author. ISBN: 5555000008952 𝑑𝑙𝑎𝑤𝑎𝑠 A huge thanks to my editor Alex Tharby, for helping me edit and publish this textbook. 1 Contents Chapter 1: Vital Mathematical Knowledge 1.1 Revision of Negative Numbers 1.2 Multiplying and Dividing Fractions 1.3 Adding and Subtracting Fractions 1.4 Notes on Fractions 1.5 Defining a Variable 1.6 Indices 1.7 BIMDAS 1.8 Equations 1.9 Factorisation 1.10 Inequalities and Absolute Values 1.11 The Number Zero Chapter One Summary Chapter One Questions p. 4 p. 6 p. 7 p. 8 p. 12 p. 13 p. 14 p. 18 p. 21 p. 25 p. 26 p. 28 p. 28 p. 29 Chapter 2: Linear Algebra 2.1 Linear Equations 2.2 Main Features of Linear Equations 2.3 Negative Gradients 2.4 Graphing Lines from Equations 2.5 Obtaining the Equation of a Line 2.6 Intersecting Lines 2.7 Microeconomic Applications 2.8 Elasticity 2.9 Interpreting Elasticity Chapter Two Summary Chapter Two Questions p. 30 p. 31 p. 32 p. 33 p. 34 p. 36 p. 38 p. 40 p. 42 p. 44 p. 46 p. 47 Chapter 3: Simultaneous Equations and Matrices 3.1 Simultaneous Equations 3.2 Two Simultaneous Equations 3.3 Three Simultaneous Equations 3.4 The Matrix 3.5 Solving Two Equation Matrices 3.6 Solving Three Equation Matrices 3.7 Notes on Solutions to Matrices 3.8 Applications 3.9 The Determinant of a 2 × 2 Matrix 3.10 The Determinant of a 3 × 3 Matrix 3.11 Using the Jacobian Determinant Chapter Three Summary Chapter Three Questions p.49 p. 50 p. 50 p. 52 p. 54 p. 55 p. 59 p. 61 p. 62 p. 65 p. 65 p. 68 p. 70 p. 71 Chapter 4: Non-Linear Functions 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 2 Defining Non-Linear Functions Defining a Quadratic Function Quadratic Graphs Sketching a Quadratic Function The Cubic Function The Exponential Function The Logarithmic Function Logarithmic Graphs p. 73 p. 74 p. 75 p. 76 p. 77 p. 80 p. 82 p. 84 p. 88 4.9 The Natural Number 𝑒 4.10 The Hyperbolic Function 4.11 Economic Applications Chapter Four Summary Chapter Four Questions p. 89 p. 90 p. 91 p. 93 p. 94 Chapter 5: Single Variable Differentiation 5.1 What is Differentiation? 5.2 Differentiation by First Principles 5.3 Differentiation Rules: Power Rule 5.4 Differentiation Rules: Chain Rule 5.5 Differentiation Rules: Product Rule 5.6 Differentiation Rules: Quotient Rule 5.7 Differentiation Rules: 𝑒 Rule 5.8 Differentiation Rules: ln Rule 5.9 The Second Derivative 5.10 The Gradient Function 5.11 Simple Applications Chapter Five Summary Chapter Five Questions p. 96 p. 97 p. 99 p. 103 p. 104 p. 105 p. 108 p. 109 p. 110 p. 111 p. 112 p. 113 p. 114 p. 115 Chapter 6: Applications of Differentiation 6.1 Graphical Optimisation 6.2 Mathematical Optimisation 6.3 The Nature of an Optimal Point 6.4 Inflection Points 6.5 Combining all Theory 6.6 Applications – Profit 6.7 Applications – Break-Even 6.8 Applications: Marginal and Average Values 6.9 Differentiation and Elasticity 6.10 Elasticity and Total Revenue Chapter Six Summary Chapter Six Questions p. 117 p. 118 p. 118 p. 120 p. 123 p. 124 p. 126 p. 129 p. 130 p. 133 p. 135 p. 137 p. 138 Chapter 7: Multiple Variable Differentiation 7.1 Additional Variables 7.2 Simple Partial Differentiation 7.3 Complex Partial Differentiation 7.4 Second Order Partial Derivatives 7.5 Application of Partial Differentiation 7.6 Total Differentiation 7.7 Optimisation with Many Variables 7.8 Economic Applications Chapter Seven Summary Chapter Seven Questions p. 140 p. 141 p. 142 p. 144 p. 146 p. 147 p. 148 p. 151 p. 155 p. 158 p. 158 3 170 p. 168 p. 164 p.4 Compound Interest 8.1 Index Numbers and Averages 8.5 Annual Interest Rates 8. 161 p. 174 p.6 Net Present Value 8. 160 p. 177 Solutions p.2 Series and Sums 8.7 Internal Rate of Return Chapter Eight Summary Chapter Eight Questions p.Chapter 8: Financial Mathematics 8. 177 p. 163 p. 180 4 . 172 p.3 Simple Interest 8. 9 1.11 Revision of Negative Numbers Multiplying and Dividing Fractions Adding and Subtracting Fractions Notes on Fractions Defining a Variable Indices BIMDAS Equations Factorisation Inequalities and Absolute Values The Number Zero 6 7 8 12 13 14 18 21 25 26 27 28 29 Chapter One Summary Chapter One Questions 5 .6 1.1 1.7 1.8 1.2 1.10 1.4 1.3 1.Chapter 1 Vital Mathematical Knowledge Things you must know 1.5 1. find the number 4 on the number-line. and − −9 becomes +9.1. E. because it cannot be changed using the rules. E. Then the −(−5) can be changed to +5 to get: = −4 − 3 + 5 = −2 Example 5: solve − −5 − 6 + −8 − −9 Solution: simplify each part separately. take 4. Similarly with negative numbers. After you get the hang of it. you will never need to use the number-line again.g. +(−8) becomes −8. then add 6. Do one section at a time. The answer is 1. Example 2: solve: −5 − 6 − 7 − 2 + 15 Solution: do it without the number-line = −5 Theory: Adding a negative number is the same as subtracting the positive value of that same number.g. 6 . but the +(−3) can be changed to −3. find the number −1 on the number-line. the −4 remains as it is. −(−5) becomes +5. then go back three places to get 1. Example 6: solve −6 × (−5) Solution: multiplying two negative numbers equals a positive number. −6 remains as it is. Working on the number line (a line with all negative and positive numbers) will help you revise the methods of adding and subtracting negative numbers. The multiplication of a positive and a negative number equals a negative number. = +5 − 6 − 8 + 9 = 0 A similar theory applies to multiplying negative numbers. −4 −3 −2 −1 0 1 2 3 4 Solution: this is a bit more complex.1 revision of negative numbers Example 3: solve 3 − −4 + 2 Solution: the two negatives (one after the other) can be changed into a positive sign: =3+4+2 Then solve: =3+4+2=9 Example 4: solve −4 + −3 − (−5) Adding and subtracting negative numbers is very similar to adding and subtracting positive numbers. Example 1: solve −4 + 5 − 2 − 4 + 6 Solution: find the number −4 on the number-line. then go back two places to get −3. Theory: The multiplication of two negative numbers equals a positive number. 5 + −3 = 5 − 3 Subtracting a negative number is the same as adding the positive value of that number. This is best explained with examples. for the problem −1 − 2. Since 6 × 5 = 30 then −6 × −5 = 30 For the problem 4 − 3. then add 5. 4 − −6 = 4 + 6 The two rules above are very important. take 2. Theory: to multiply two fractions. if two terms are in brackets without anything between the brackets. 5 ∙ 7 = 35 so −5 −7 = 35. e. then multiply the two bottoms (denominators) to give a new bottom. $100. Example 1: solve 2 4 ∙ 3 5 Solution: multiply the two tops: 2 ∙ 4 = 8.g. $4million is 1000 of a billion dollars.g. 6 4 = 6 × 4 Example 8: solve −5 (−7) Solution: two negative numbers are multiplied together which must equal a positive number. this is taken as a multiplication. Dividing with negative numbers also follows the rule of two negatives equals a positive. Simplify the following: 𝑎) −2 + 3 − 4 𝑏) −2 + −4 − −1 𝑐) −4 − −4 − 4 𝑑) 5 + 1 − 4 + (−3) 𝑒) −6 − 7 − 3 − −21 + 1 2.2 multiplying and dividing fractions Most students hate fractions. and these form the new numerator and the new denominator: = 2∙4 8 = 3 ∙ 5 15 4 7 . you will see the consequence of not using brackets appropriately. 50𝑐 is half a dollar. Simplify the following: 𝑎) 3 ∙ −2 𝑏) −6 −6 𝑐) −36 ÷ −4 𝑑) 81/ −9 𝑒) −20 −3 1.000 is one tenth of a million dollars. 24 =8 3 so −24 = −8 3 Note: there is a reason brackets were used so often in this section. but they are part of the real world. Later in this chapter.Example 7: solve (−6) × 4 Solution: a negative number multiplied by a positive number equals a negative number. 15 −15 = 5 then =5 3 −3 Example 10: solve −24 3 Solution: division with both a negative number and a positive number gives a negative number. Since 6 × 4 = 24. so you must be able to manipulate them. For example. E. 6 ∙ 4 is the same as 6 × 4 Similarly. The division of one negative and one positive number equals a negative number. Most things can have fractions. Example 9: solve −15 −3 Solution: dividing two negative numbers equals a positive number. Exercises: 1. then −6 × 4 = −24 Theory: mathematicians use the dot “∙” to mean multiplication. multiply the two tops (numerators) to give a new top. then multiply the two bottoms: 3 ∙ 5 = 15. Theory: The division of two negative numbers equals a positive number. You purchased a pie. you are hungry again and take a quarter of the original pie. making the bottom the new top. you eat half the pie. invert the second fraction then proceed just like multiplication. you ate three quarters. and change the division sign to a multiplication sign: = 2 4 ∙ 5 3 2∙4 8 = 5 ∙ 3 15 is the Common Denominator approach. Multiplying and dividing fractions is easier. 𝑒) 7 3 / − 2 10 Dividing fractions is very similar except it has one difference. The question is how much of the pie have you eaten? Write out the fractions: 1 1 + 2 4 That is. 𝑎 Simplify the fractions into a single fraction 14 −5 𝑎) −3 12 14 5 𝑏) / −3 −12 −3 23 𝑐) / 3 17 −5 −3 𝑑) / 9 −7 17 −1 𝑒) 23 12 1. but it works well. 𝑎 Inverting 𝑏 𝑏 it .3 adding and subtracting fractions The hardest aspect of fractions is adding and subtracting fractions. The denominator is the bottom of a fraction. You cut the pie into quarters. Apologies for the patronising subject. so that only one quarter is left. If we do that. an intuition of the concept is needed. then proceed as for multiplication. You might be tempted to simply add the top numbers and the bottom numbers. Theory: when dividing fractions. Before the theory is given. 2 16 32 2 ∙ = = 10 3 1 3 3 Exercises: 1. then you ate a quarter. we get: 1+1 2 1 = = 2+4 6 3 Have you really eaten only one third of the pie? No. The theory of common denominator helps find the true answer. first you ate half the pie. Inverting a fraction is flipping the fraction. Proceed with the multiplication: = Example 4: solve 2 1 / 3 16 Solution: invert the second fraction. After two hours. Simplify the following fractions: 3 2 𝑎) 4 3 5 12 𝑏) 7 5 3 3 𝑐) −8 1 5 2 𝑑) ÷ 4 3 8 . and making the top the new bottom. The concept used for adding/subtracting fractions makes Example 3: solve 2 3 ÷ 5 4 Solution: invert the second fraction.Example 2: solve 6 7 ∙ 5 3 Solution: = 6 ∙ 7 42 = 5 ∙ 3 15 2. and since you are hungry. and you take it home. while keeping the value of the fraction unchanged. Theory: the Common Denominator approach changes the form of all fractions so that the denominators are all the same. 7 needs to be multiplied by some number to get 21 on the bottom. change the form of each fraction without changing its value. Once all fractions have the same denominator. however if the 2 is erased and replaced with a 4. Theory: a Common Denominator is found by multiplying the denominators (bottoms) of all the fractions to be added or subtracted. but leaves the values of all fractions unchanged. 4 is used. solve 1 1 + 2 4 The common denominator process is finding the same denominator for all fractions. the value will no longer be one half. Another way of writing 2 is 4 or 6 or 8 but since 4 needs to be on the bottom.Intro example 1: from the pie example. solving: 1 1 + 3 7 2 1 2 3 4 1 What is the common denominator here? How can 1 3 be changed into something with 7 as a 1 denominator? Or alternatively. For example. The problem becomes: 2 1 + 4 4 The first part still has the value of a half. while leaving the value of the fraction unchanged. In this example. Multiply the 3 by 7: 1 7 ∙ 3 7 The number has value 1. What is wanted is to change the form of the fraction 2 so the bottom number is 4. 9 3 1 7 7 1 3 1 7 1 . so that all the denominators are the same. all the top numbers (numerators) can be written over the Common Denominator: 2 1 2+1 + = 4 4 4 Then simplify the top: 2+1 3 = 4 4 Exactly as is intuitive. The form of the first fraction has changed but the value is the same. the 1/2 must be changed to a fraction with a 4 on the bottom. having this over the common denominator. how can 7 be changed to have 3 as a denominator? There is a simple solution. Sometimes it is not so obvious how the fractions need to be changed. That number is 3: 1 3 3 ∙ = 7 3 21 Rewrite the original problem with the equivalent fractions: 7 3 7 + 3 10 + = = 21 21 21 21 Theory: to add/subtract fractions. Intro example 2: solve 1 1 + 3 7 A Common Denominator is the multiplication of the two denominators 3 and 7: 3 ∙ 7 = 21 The form of 3 needs to be changed to have 21 on the bottom but the value must be left unchanged. so the value of is unchanged however the form is changed: 1 7 7 ∙ = 3 7 21 Similarly. Then add/subtract the numerators (top numbers). and the second part is still one quarter. 1 5 5 ∙ = 9 5 45 Similarly for 5. Example 2: solve 3 4 + 11 5 Solution: a common denominator is 11 ∙ 5 = 55. Then. To change the 11 to a 44. The first fraction needs to be multiplied by to have 45 on the bottom. to change the form of each fraction without changing its value. Example 4: solve 3 3 1 + − 11 4 5 Solution: a Common Denominator is the 4 4 (where the 5 comes from the denominator of the other fraction): 3 5 15 ∙ = 11 5 55 Similarly for the other fraction. but a very similar theory applies for adding and subtracting more than two fractions. Theory: when adding or subtracting many fractions. to change 5 to have a 55 on the bottom. by multiplying it by 5 5 3 1 5 1 9 9 1 5 1 9 5 5 other fraction): 4 11 44 ∙ = 5 11 55 Rewrite the original problem: 15 44 15 + 44 59 + = = 55 55 55 55 Subtracting fractions uses the same process. we have only been using two fractions at a time. but leaving the value of the fraction unchanged. Similarly for 9 which was multiplied by 5 where the 5 came from the denominator of the other fraction. Example 3: solve 7 1 − 11 4 Solution: a Common Denominator is 11 ∙ 4 = 44. multiply by 4 (as this does not change the value of the fraction): 7 4 28 ∙ = 11 4 44 Then change the form (but not the value) of the second fraction: 1 11 11 ∙ = 4 11 44 Rewrite the original problem: 28 11 28 − 11 17 − = = 44 44 44 44 So far.(where the 11 comes from the denominator of the Example 1: solve 1 1 + 9 5 Solution: a Common Denominator is 9 ∙ 5 = 45. it needs to be multiplied by 11 11 multiplication of all the denominators. The fraction 3 11 needs to be 10 . except it has a negative sign in between. The 5 in 5 comes from the denominator of the other fraction. it needs to be multiplied by 9 where the 9 comes from the denominator of the other fraction: 1 9 9 ∙ = 5 9 45 Rewrite the original problem and solving: 5 9 5 + 9 14 + = = 45 45 45 45 Notice that the 5 was multiplied by 9 where the 9 came from the denominator of the other fraction. which is 11 ∙ 4 ∙ 5 = 220. it is multiplied by 𝑥/𝑥 where 𝑥 is the multiplication of all the other denominators. change the 11 to have 55 as a denominator without changing its value. to change the form of this fraction to get 45 on the bottom. She is ready to retire. Jane. 130 210 39 130 − 210 − 39 − − = 390 390 390 390 Simplify the numerator (top): =− 119 390 39 which is obtained by the multiplication of all the other denominators (i. the CEO of a company. Simplify the following fractions: 1 1 𝑎) + 4 3 2 1 𝑏) − 5 8 7 12 1 𝑐) + + 3 2 6 21 1 𝑑) − 13 11 9 15 2 𝑒) + − 13 39 1 2.e. The problem is rewritten with all the relevant signs added in after the form of each fraction has been changed to have a Common Denominator. owns 9/11 of the company. she sells 5/12 of the shares she owns. it must be multiplied by 130 . as so far. This number is 20 20 The third fraction must be multiplied by 39 which comes from 3 ∙ 13 = 39: 1 39 39 ∙ = 10 39 390 Do not use the negative signs yet.e. The 4 needs to be multiplied by 55 .multiplied by some number to get 220 as a denominator (bottom). The number 55 is obtained by multiplying the denominators of all the other fractions (i. 130 3 55 where the 130 comes from the multiplication of the other denominators (i.e. how much would she have left then? 11 . 4 ∙ 5 = 20): 3 20 60 ∙ = 11 20 220 Similar theory applies for the other two fractions. To change the form of the first fraction without changing its value. Example 5: Solve 1 7 1 − − 3 13 10 Solution: a Common Denominator is 3 ∙ 13 ∙ 10 = 390. Simplify the fractions into a single fraction: 1 1 1 1 𝑎) + + + 2 3 4 5 1 1 1 1 𝑏) − + − 2 3 4 5 21 3 2 𝑐) + − 11 17 23 99 3 12 7 𝑑) + − + 12 2 9 8 10 15 17 12 𝑒) − + + 3 4 9 7 3. If a shareholder owns 3/11 of a company. and his brother owns 9/83 of that same company. What proportion of the company does she still own after the sale. 5 ∙ 11 = 55): 3 55 165 ∙ = 4 55 220 Lastly: 1 44 44 ∙ = 5 44 220 Rewrite the original fraction: 60 165 44 60 + 165 − 44 + − = 220 220 220 220 181 = 220 Go over this last example to make sure you know how the form of each fraction was changed without changing its value. what proportion of the company do the brothers own combined? 4. 3 ∙ 10 = 30): 7 30 210 ∙ = 13 30 390 30 Exercises: 1. we are just changing the form of the fraction.e. What if Jane had sold 5/12 of the total company stock from her portfolio. 13 ∙ 10 = 130): 1 130 130 ∙ = 3 130 390 The second fraction must be multiplied by 30 where the 30 is from the multiplication of all the other denominators (i. so to have cash. so this is written as a fraction: 43 1 =7 6 6 12 . 2 denominator. The short-cut is shown below. then add that result to the top. Theory: to simplify a fraction involving whole numbers. a more common kind of fraction is one with a number out front. and any remaining numbers are written over the original This means “two wholes plus one third”.1. Theory: for a negative fraction. but the process is slightly different. the negative sign can be in three different places without changing the value of the fraction: 𝑎 −𝑎 𝑎 − = = 𝑏 𝑏 −𝑏 For example: 2 −2 2 − = = 5 5 −5 All three are the same. then add the result to the numerator (top): (44 + 6 = 50) 4 6 50 = 11 11 The reverse theory is the same. Then remove that number from the numerator. Theory: to find how many “whole” numbers there are in a fraction. Applying this to a larger context: 2 5 2 −5 2 5 − = + = + 3 3 3 3 3 −3 All three coloured parts are equal. Now. so you must be familiar with all of them. such as: 2 1 3 Solution: multiply the “whole” number with the denominator (4 × 3 = 12). This can also be rewritten as: 1 2+ 3 Write the 2 as 1: 2 1 + 1 3 Then find a Common Denominator: 6 1 6+1 7 + = = 3 3 3 3 This is the long way which shows the theory. so the number out front in 7. work out how many denominators will fit into the numerator without exceeding the numerator. It may be convenient to write it one of these three ways when you are doing a question with fractions. multiply the number out front with the denominator (bottom). Example 3: find how many “wholes” and a remaining fraction results from 43 6 Solution: 6 goes into 43 seven times (to give 42) as any more would exceed 43. then adding this to the numerator 12 + 1 = 13 giving: 1 12 + 1 13 4 = = 3 3 3 Example 2: find the single fraction equivalent of 4 6 11 Solution: multiply the “whole” number out front with the denominator (4 × 11 = 44).4 notes on fractions Example 1: change the following to a single fraction 4 1 3 Fractions can be written in many different ways. and the remainder is 1 (as 43 − 42 = 1). only like terms can be added. Also. 5 apples and 8 peaches. Change the following into whole numbers and a remaining fraction: 12 27 𝑎) 𝑑) 3 13 15 91 𝑏) 𝑒) 4 12 21 103 𝑐) 𝑓) 4 13 3. then it would exceed the numerator (97). therefore it is a variable. but the extent to which it changes ‘depends’ on another variable (i. This gives 7 as a remainder. To simplify the variables. Different letters are used to distinguish different variables.5 defining a variable profit will be made if 𝑄 units are sold – where 𝑄 is a variable number of computers. In business. Temperature is a variable. it is not always that simple. 6 apples and 12 peaches. so variables are used that allow those companies to forecast how much Where 𝑏 =bananas. Profit is also a variable but the amount of profit depends on how many computers are sold. To find out the total number of each fruit. Add Box 1 to Box 2: 𝑇𝑜𝑡𝑎𝑙 = 7𝑏 + 5𝑎 + 8𝑝 + 3𝑏 + 6𝑎 + 12𝑝 Since only apples can be added to apples.e. This is the same for bananas and peaches. 𝑇𝑜𝑡𝑎𝑙 = 10𝑏 + 11𝑎 + 20𝑝 13 . Things that are different cannot be grouped together. it can change). the number of computers sold). However. we need to learn the basics of manipulating variables and constants. Intro example 1: There are two boxes of fruit: Box 1: 7 bananas. a variable is often something like quantity sold (usually 𝑄).e. Simplify the following into a single fraction: 12 3 1 𝑎) − − 5 7 3 13 1 2 3 𝑏) 2 1 2 12 3 7 12 −2 −3 𝑐) + − 3 7 −10 1. but they cannot be added to the apples or the peaches (this is logical). Variables are found everywhere in life. Theory: “collecting like terms” is the process of bring together things that are the same. written over the original denominator: 97 7 =9 10 10 Exercises: 1. Give the following fractions as a single fraction: 1 1 𝑎) 2 𝑑) 3 2 3 2 3 𝑏) 12 𝑒) 9 3 9 7 13 𝑐) 9 𝑓) 11 8 27 2. the two boxes can be rewritten as: Box 1: 7𝑏 + 5𝑎 + 8𝑝 Box 2: 3𝑏 + 6𝑎 + 12𝑝 Theory: A variable is something that can change. That is. So profit is called a ‘dependent variable’ because it is a variable (i. then 5𝑎 + 6𝑎 = 11𝑎. because it can change. For now. 𝑎 = apples and 𝑝 = peaches. Companies do not know how many of a certain product they are going to sell (say computers).Example 4: change the following fraction into “whole” numbers and a remaining fraction: 97 10 Solution: 10 can go into 97 nine times as if we had ten times. The Woolworths share price can change. Box 2: 3 bananas. the bananas can be added to each other. the number of computers the company sells is called an ‘independent variable’ because it is not determined by other variables. and just do what you would do if they were numbers. 9𝑥 − 8𝑦 − 6 Note: lone numbers are only like other lone numbers. Simplify the following fractions into a single fraction: 12𝑥 −2𝑦 2𝑥 4𝑦 𝑎) − + + −5 3 30 10 5 5 3 𝑏) − − +2 3 𝑦 𝑧 3 3 𝑐) + 2𝑥 5𝑦 1. all the 𝑦’s together. the first fraction needs to be multiplied 𝑦 𝑦 by 𝑦 1 𝑦 𝑦 ∙ = 𝑥 𝑦 𝑥𝑦 It is easiest to ignore the fact that they are letters.6 indices 𝑥 𝑥 When mathematicians talk about an index. finding a Common Denominator of fractions with variables is easiest when you think of them as numbers. so must be multiplied by : 2 𝑥 2𝑥 ∙ = 𝑦 𝑥 𝑥𝑦 Rewrite the original problem and solve: 𝑦 2𝑥 𝑦 + 2𝑥 + = 𝑥𝑦 𝑥𝑦 𝑥𝑦 The terms on top cannot be added as they are not like terms. they are talking about numbers or variables to the power of other numbers or variables. so in this case will be 𝑥𝑦. Theory: an index (plural indices) is when a number (called a base) is put to the power of another number (called an index). Collect like terms using all the theory learnt so far: 1 1 2 𝑥 𝑥 2𝑥 𝑎) 𝑥 + 𝑥 − 𝑥 𝑑) + − 2 3 5 2 3 5 12𝑥 13𝑥 12𝑥 −2𝑥 2𝑦 13 𝑏) + 𝑒) − + − 5 6 3 4 3 12 3 4 5 5 𝑐) + 𝑓) 2+ + 𝑥 𝑦 𝑥 𝑦 3. Exercises: 1.Example 1: collect all like terms in the following 3𝑥 + 5𝑥 + 𝑦 − 𝑥 + 5𝑦 + 𝑧 Solution: collect all the 𝑥’s together. all the 𝑦′s together and all the 𝑧’s together: 7𝑥 + 6𝑦 + 𝑧 The 𝑥’s cannot be added to the 𝑦’s or 𝑧’s as they are not like terms. Then. Collect the like terms in the following 𝑎) 5𝑥 + 4𝑦 − 3𝑥 − 9𝑥 + 4𝑦 − 3 𝑏) 13𝑥 − 12𝑥 − 11𝑥 − 10 − 9𝑦 𝑐) − 8𝑥 − 4𝑦 − −3𝑥 − 4𝑦 − −1 𝑑) 8𝑥 + 8𝑥 + 8𝑦 + 8𝑧 − 8 + 12𝑥 − 13𝑦 2. 𝐵𝐴𝑆𝐸 𝑖𝑛𝑑𝑒𝑥 to get 𝑥𝑦 on the bottom (the 𝑦 in 𝑦 comes from the denominator of the other fraction): 14 . Example 4: find the Common Denominator and put the following two fractions under a single denominator: 1 2 + 𝑥 𝑦 Solution: a Common Denominator is the multiplication of all the denominators. and all the lone numbers together. Example 2: collect the like terms in 5𝑥 − 𝑦 + 4𝑥 + 3 − 7𝑦 − 9 Solution: collect all the 𝑥’s together. and are not like anything with a variable in it. Example 3: collect all like terms 5𝑥 + 3𝑦 − −4𝑥 + 7𝑦 − 13 Solution: get rid of the brackets: 5𝑥 + 3𝑦 + 4𝑥 + 7𝑦 − 13 Collect like terms: 9𝑥 + 10𝑦 − 13 Applying variables to fractions. work with them as if they were numbers. That is. The other fractions also needs 𝑥𝑦 on the bottom. Theory: RULE 2: when two numbers are divided and have the same base. For the index 36 The 3 is multiplied by itself 6 times: 3 ∙ 3 ∙ 3 ∙ 3 ∙ 3 ∙ 3 = 36 For 𝑥 3 . it is: 𝑥 ∙ 𝑥 ∙ 𝑥 = 𝑥 3 There are seven rules you need to learn before you can successfully manipulate indices: Theory: RULE 1: when two numbers with the same base are multiplied together. 𝑥 0 = 1 This is something you simply have to know. the indices can be added to one another. An index means that the base number is multiplied by itself the number of times of the index. Theory: RULE 3: any number with an index of zero equals one. 𝑥 2+𝑦 . Theory: RULE 4: the 𝑛𝑡 root of a number is the same as having the number to the power of 1/𝑛. 𝑥 ∙ 𝑥 = 𝑥 𝑎 𝑏 𝑎+𝑏 Three of the 𝑥’s on top cancel with three 𝑥’s on the bottom. the index on the bottom of the fraction is subtracted from the index on the top of the fraction.Examples of indices include 22 . 𝑥 𝑎 = 𝑥 𝑎−𝑏 𝑥 𝑏 Example 2: simplify the indices in: 𝑥 𝑥 3 Solution: use the above theory 𝑥 7 = 𝑥 7−3 = 𝑥 4 𝑥 3 Rewrite in extended form: 𝒙 ∙ 𝒙 ∙ 𝒙 ∙ 𝒙 ∙ 𝒙 ∙ 𝒙 ∙ 𝒙 = 𝒙𝟒 𝒙 ∙ 𝒙 ∙ 𝒙 7 𝑥 = 𝑥 1/2 2 Note: the square root sign is often written . 53 . 𝑥 2 . This is a bit complex in words. 2 Example 1: simplify the indices in 𝑥 3 ∙ 𝑥 5 = 𝑥 3+5 = 𝑥 8 Solution: write this out in extended form: 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 = 𝑥 8 The brackets do not matter as everything is multiplied. but simple in practice. 𝑥 𝑎 𝑏 = 𝑥 𝑎∙𝑏 Example 5: simplify the following to a single index 𝑥 3 𝑥 3 2 Solution: multiply the indices together: 2 = 𝑥 3∙2 = 𝑥 6 This can also be written out in extended form so you see where it comes from: 𝑥 3 𝑥 3 = 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 = 𝑥 6 15 . 𝑛 𝑥 = 𝑥 1/𝑛 Example 3: rewrite the following as an index 2 𝑥 Solution: the square root of 𝑥 is simply 𝑥 to the power of half. the two indices are multiplied together. with the number 2 omitted: 4 Example 4: write the following as an index 𝑥 𝑥 = 𝑥 1/4 Solution: it is 𝑥 to the power of one on four: 4 Theory: RULE 5: when a number to the power of one index is put to the power of another index. 1 + 𝑦 9 . then change the sign of the index: 1 = 𝑥 5 𝑥 −5 Theory: RULE 7: when the product of two or more bracketed terms are put to an index. The brackets can be rewritten as: −1 ∙ 𝑥𝑦 4 Then it can be seen that every term in the brackets is put to the index: −1 4 𝑥 4 𝑦 4 Simplifying −1 𝑥 4 𝑦 4 All these rules can work forwards and backwards.As the two brackets represent 𝑥 3 squared. so to have a positive power. and change the sign of the index: 𝑥 4 1 = −4 1 𝑥 Example 8: change the following to have a positive index 1 𝑥 −5 Solution: invert the fraction. all the terms are still multiplied together. Example 11: simplify the following to only have positive indices 𝑥 0.5 = 𝑦1 𝑦 4 = 1 gives: Example 12: simplify the following leaving all indices positive = 𝑥 𝑎 𝑦 𝑎 16 . The following examples involve the use of multiple rules. so using RULE 1 on the 𝑥’s and 𝑦’s respectively: = 𝑥 0. 𝑥 −𝑎 1 = 𝑎 𝑥 Example 9: expand 2𝑥𝑦 3 Solution: since all terms in the brackets are multiplied together.5 𝑦 −1 To have all indices positive. that index applies to each term within the brackets. the 𝑦 −1 must be brought to the bottom using RULE 6. Example 7: change the following to have a negative index 𝑥 4 Solution: invert the fraction. every term is raised to the index: 23 𝑥 3 𝑦 3 Simplify the 23 : 8𝑥 3 𝑦 3 Example 10: expand −𝑥𝑦 4 Example 6: rearrange the following to have a positive index: 𝑥 −3 Solution: invert the fraction. then the sign of the index is changed: = 𝑥 2. 𝑥𝑦 𝑎 Solution: despite the negative sign inside the brackets.5 𝑦 3 𝑥 2 𝑦 −4 Solution: only the like terms can be simplified. and change the sign on the index: 𝑥 −3 1 = 3 1 𝑥 The 𝑥 −3 is really a fraction with 𝑥 −3 “on one”.5+2 𝑦 3−4 = 𝑥 2. the fraction is inverted and the sign of the index is changed.5 𝑥 2. Theory: RULE 6: a number with a negative index can be made to have a positive index by inverting the fraction. 243 BASE 4: 4. 256 BASE 5: 5. to the power of a whole number: BASE 2: 2. 27. which can then be removed from the bottom: = 2−0. as it is of the same 𝑦 Go over this last example. 128 . as all the bases are different. and the 4/3 comes from 10/3 − 2. the bases must be the same! = 2−1 𝑥 3 𝑦 −0.5 𝑥 3 𝑦 0. For example. 625 Example 14: simplify the following as much as possible 3𝑥 272 𝑥 3 The 11/3 comes from 5/3 + 2. 125. Then using RULE 2 for both 𝑥 and 𝑦: = 81/3 𝑥 4/3 𝑦19/6 Where the 19/6 comes from 11/3 − 1/2.5 3−0.5 3 0. to make all the indices positive.5 Get rid of the brackets using RULE 5: = 𝑦 −6 9𝑥 𝑦 2 17 . numbers may need to be simplified to contain indices.5 30.5 20.5 Then use RULE 1 to bring the like terms together. Theory: be aware that each of the following numbers can be written as the indicated base. 81.5 𝑦 −1 3−0. 8. 64.5 𝑥 0 𝑦 30. In many questions.5 𝑥 0 𝑦 30.5 𝑥 0 𝑦 3 Solution: use RULE 4 to change all the root signs: = 20.5 𝑦 30. bring any bases with negative indices to the bottom while changing the sign of the index: = 𝑥 3 2𝑦 0. 64.5 𝑥 𝑦 20.5 2 −0. Remember. 4. Finally.5 20.5 𝑥 3 𝑦 0. 25. Example 13: collect all like terms and make all indices positive for 2 −1 3 0. change the 8 value: = 2𝑥 4/3 19/6 1/3 into 2. 16.5 2 𝑥 𝑦 0. 32.5 Solution: get rid of the root signs first (remember that if there is no number outside the root sign. 16.5 Get rid of the brackets using RULE 5 with RULE 7: = 81/3 𝑦 5/3 𝑥 3 𝑦 2 𝑥1/3 𝑦1/2 𝑥 2 Then use RULE 1 on the top for the 𝑥’s (as well as simplifying the addition of fractions): = 81/3 𝑦 5/3 𝑥10/3 𝑦 2 𝑦1/2 𝑥 2 And again use RULE 1 for the 𝑦’s on the top (a Common Denominator needs to be used): = 8 1/3 10/3 11/3 𝑥 𝑦 𝑦1/2 𝑥 2 This is as far as simplification can go. 9. use RULE 2 to bring everything on the bottom up top while changing the sign of each of the indices: = 2−0.256 BASE 3: 3.5 Now.5 2−0.5 −1 𝑥 3 𝑦 0. it is assumed to be a square root): = 8𝑦 5 𝑥 𝑦 𝑥 1/2 𝑥 2 𝑦 1/3 3 2 1/3 Doing this the long way to begin with. the number 81 may need to be simplified to either 92 which you should be familiar with. as the fractions were difficult. or maybe even 34 .3 8𝑦 5 𝑥 3 𝑦 2 𝑥 𝑦𝑥 2 3 Use RULE 3 to change anything to the power of zero to 1. and neither does the pair of addition/subtraction. 𝑎) 𝑥 2 𝑥 −3 𝑥 −6 𝑥 15 𝑓) 𝑥 2 𝑦 3 𝑥 4 𝑦 5 𝑥 6 𝑧 7 Subtraction The multiplication/division have no precedence over one another.5 𝑥 −6 𝑥 0. Similarly. Example 15: simplify 36 Solution: 36 = ±6 The ± sign means that there are two solutions. 6 6 = 36. It is pronounced “plus or minus”.5 𝑥 𝑥 3. one of the exercises was −6 −6 which you should have worked out to be 36. When solving: 24 + 5 to be able to add the 5.) is very important. Intro example: for a number to an index. one positive and one negative. as that does not follow the order of BIMDAS.1 𝑦 0.2 Change the 27 to 33 and 9 to 32 : 3𝑥 33 2 𝑥𝑦 −2 32 𝑥 𝑦 2 Use RULE 5 again to get rid of the brackets: 3𝑥 36 𝑥𝑦 −2 32𝑥 𝑦 2 Then use RULES 1 and 2 to collect all the indices with base 3: 3𝑥+6−2𝑥 𝑥𝑦 −2 𝑦 2 Simplify the index with base 3 and using RULE 6 to get the 𝑦 index: 3 −𝑥+6 −2 25 𝑥 4 625𝑥 2 𝑦 −5 625𝑥 2 𝑥 −1 4 𝑗) 1. BIMDAS is a method that gives the order in which operations must be done.Solution: get rid of the root sign using RULE 4: 3𝑥 272 𝑥 𝑦 −6 9𝑥 𝑦 2 3 27 𝑥𝑦 9𝑥 𝑦 2 𝑥 2 −2 1/3 𝑏) 𝑐) 𝑑) 𝑒) 𝑥 3 𝑥 5 𝑥 𝑥 −2 3 2 𝑔) ) 𝑖) 𝑥 −4 𝑥 2 𝑦 3 𝑥 −1 𝑦𝑥 2 𝑦1. BIMDAS stands for: Brackets Indices Multiplication Division Addition to the bottom and with a positive 𝑥 𝑦 4 Theory: the square root of a positive number will always have two solutions: one positive and one negative. it is sometimes called BIDMAS. the index must be simplified first: 16 + 5 = 2 If you were thinking that the 2 and 5 can be added to one another.4 𝑥 0.2 81𝑥 3 𝑥 −1/2 𝑦 −1/6 𝑥 0. Simplify the following to have all positive indices and fraction answers. the index takes precedence (as Index takes precedence over Addition in BIMDAS). and then put that to the power of 4 you would be incorrect.25 𝑥 −2 𝑦 −3 𝑥𝑦 3 3 5 𝑥 2 𝑦 3 8 𝑥 −2 𝑦 −4 4 −1 3 Then use RULE 5 to get rid of the brackets: 5 𝑥 4 10𝑥 2 𝑦 4 2 𝑦 2 𝑥 −2 3𝑥 3 27𝑥 3 𝑥 −5 𝑦 3 𝑥𝑦𝑥 −0. 18 . In the first section in this chapter. which is then added to another number. Example 16: simplify 144 Solution: 144 = ±12 Exercises: 1. multiplication etc.7 BIMDAS Theory: the order in which you conduct mathematical operation (such as addition. Because of this. it is called the crab-claw method.Example 1: using BIMDAS. informally. 𝑥 multiplies the 𝑦 (to give 𝑥𝑦). simplify the Index: = −8 ÷ 4 + 9 × 4 ÷ 2 + 7/2 Get rid of the Multiplication/Division: = −2 + 18 + 3. Last).5 Finally. For the multiplication of the following brackets: 𝑥 + 2 𝑦 + 4 Draw in lines to match the letters of FOIL or to look like a “crab-claw” (the reason is explained below). It will make sense soon. get rid of the Brackets first by multiplying the 7 into the −2 : = 5 ∙ 34 − −14 Since two negative signs equals a positive sign: = 5 ∙ 34 + 14 Expand the Indices: = 5 ∙ 81 + 14 Multiplication/Division: = 405 + 14 Addition/Subtraction: = 419 BIMDAS must be followed to simplify correctly. 𝑦 = 𝑥 + 2 𝑦 + 4 Theory: to expand the multiplication of two brackets. it is easiest to rewrite the problem with the “÷” changed: = −3 3×2 + −1 5 32 2 2 2 3 12 =− + −1 5 9 This is now a Common Denominator problem: =− 3 9 12 5 1 45 + − 5 9 9 5 1 45 27 60 45 =− + − 45 45 45 −27 + 60 − 45 12 = =− 45 45 Moving on to a harder concept that is common throughout this book. Example 2: using BIMDAS. Add/Subtract the numbers: = 19. First. That is. Outer. and it is used to simplify the multiplication of two (or more) brackets. simplify the following to a single fraction −3 ÷ 5 + 3 × 2 /3 − 1 Solution: before using BIMDAS. The 2 multiplies the 𝑦 to give 2𝑦 and the 2 also multiplies the 4 to give 8. each term in the first bracket is multiplying every term in the second bracket. simplify the following 5 ∙ 34 − 7 ∙ (−2) Solution: following BIMDAS. Inner. It is also called the FOIL rule (First. Outer. Inner. then that same 𝑥 also multiplies the 4 (to give 4𝑥). Last – FOIL) Looking back at the crab-claw that was drawn.5 Example 3: using BIMDAS. simplify the following −8 ÷ 4 + 32 × 4 ÷ 2 + 7/2 Solution: since there are no brackets.e. Adding these together: 𝑥 + 2 𝑦 + 4 = 𝑥𝑦 + 4𝑥 + 2𝑦 + 8 Example 4: expand the brackets for 𝑦 = 2𝑥 + 1 −𝑥 − 1 Solution: draw in the crab claw: 𝑦 = 2𝑥 + 1 (−𝑥 − 1) Get rid of the brackets: 3 3 × 22 =− + −1 5 32 Get rid of the indices: 3 3×4 =− + −1 5 9 Then the multiplication: 19 . every term in one bracket must be multiplied with every term in the other bracket (i. If given: 𝑥 + 3 𝑥 + 3 2 Example 6: expand the bracket −2𝑥 + 4 2 Solution: write out the two brackets: = −2𝑥 + 4 −2𝑥 + 4 Draw in the crab-claw: = −2𝑥 + 4 (−2𝑥 + 4) Follow the lines: = 4𝑥 2 − 8𝑥 − 8𝑥 + 16 Collect like terms: = 4𝑥 2 − 16𝑥 + 16 Example 7: using BIMDAS. take the −3𝑥 + 1 2 and expand this first (ignoring 2 = 𝑥 2 + 32 everything else): −3𝑥 + 1 = −3𝑥 + 1 −3𝑥 + 1 This is incorrect as there is an addition (or subtraction) sign separating two terms inside the brackets. However. Example 5: expand 𝑥 − 7 2 2 Using the crab-claw: −3𝑥 + 1 −3𝑥 + 1 Follow the lines gives: = 9𝑥 2 − 3𝑥 − 3𝑥 + 1 = 9𝑥 2 − 6𝑥 + 1 Replace this in the original statement (in brackets): 9𝑥 2 − 6𝑥 + 1 + 2 − 𝑥 + 1 Move on to the other bracket: 𝑥 + 1 2 2 Solution: write out the brackets twice: = 𝑥 − 7 𝑥 − 7 Draw in the crab-claw: = 𝑥 − 7 𝑥 − 7 Follow the lines: = 𝑥 2 − 7𝑥 − 7𝑥 + 49 Collect like terms: = 𝑥 2 − 14𝑥 + 49 + 52 = 𝑥 + 1 𝑥 + 1 Use the crab-claw method again: 𝑥 + 1 𝑥 + 1 Follow the lines: = 𝑥 2 + 𝑥 + 𝑥 + 1 = 𝑥 2 + 2𝑥 + 1 20 . This is a VERY common mistake. To get rid of the Brackets. expand and collect like terms for −3𝑥 + 1 2 Write the bracket out twice: 2 = 𝑥 + 3 𝑥 + 3 + 2 − 𝑥 + 1 2 + 52 This is index theory. write out the bracket that many times. The correct answer to this problem is 𝑥 + 6𝑥 + 9. Find it for yourself using the crab-claw method. so don’t be one of them! Theory: when a bracket containing two terms separated by an addition/subtraction sign is put to an index. many students think they can simply put the index into each of the two terms inside the bracket: 𝑥 + 3 2 Solution: This seems complex.Follow the lines: 𝑦 = 2𝑥 −1 + 2𝑥 −𝑥 + 1 −𝑥 + 1 −1 Simplify: 𝑦 = −2𝑥 − 2𝑥 2 − 𝑥 − 1 Collect like terms: 𝑦 = −3𝑥 − 2𝑥 2 − 1 The following is where many students make mistakes. but do it one small step at a time. = 𝑥 2 + 3𝑥 + 5𝑥 + 15 − (15𝑥 + 4) The other brackets need to have the negative sign introduced into each term: = 𝑥 2 + 3𝑥 + 5𝑥 + 15 − 15𝑥 − 4 Since there are no Indices (that can be simplified). 3𝑏 2 3 + 4𝑏 𝑏 2 𝑎) 𝑏 3 2𝑎 + 3 𝑏) 5𝑎 + 𝑎2 9 1 𝑐) 7𝑏 3 + 2𝑏 2 − 𝑏 2 𝑏 2 𝑏 2 𝑑) 3𝑏 +1 𝑏 0. 3. 21 . the easier it will become. Exercises: 1.Substitute back into the original statement: 9𝑥 2 − 6𝑥 + 1 + 2 − 𝑥 2 + 2𝑥 + 1 + 52 The brackets must still be there as that negative sign must go into every term in the brackets: 9𝑥 2 − 6𝑥 + 1 + 2 − 𝑥 2 − 2𝑥 − 1 + 52 All the brackets have been removed. so move on to Indices: 9𝑥 2 − 6𝑥 + 1 + 2 − 𝑥 2 − 2𝑥 − 1 + 25 The index of 𝑥 2 cannot be simplified as 𝑥 is a variable so it is left alone. The more practice you have with BIMDAS. by using the crab-claw: 𝑥 + 3 𝑥 + 5 − 15𝑥 + 4 Which gives: 1. 𝑥 + 3 𝑥 + 5 − 15𝑥 + 4 = 𝑥 + 3𝑥 + 15 − 15𝑥 + 4 Collecting like terms: = −11𝑥 + 19 For statement 2: use BIMDAS to get rid of the Brackets first. the only thing left is to collect like terms: = 𝑥 2 − 7𝑥 + 11 Despite the two statements having the same numbers. so only half a crab-claw is required. the two statements are not the same. Move on to Multiplication/Division (of which there is none). Simplify the following using BIMDAS 𝑎) 2 5 + 7 − 12 − 23 𝑏) 4 − 1 5 + 32 − 16 + 16 ÷ 8 𝑐) 12 ÷ 4 + 3 ÷ 3 × 2 + 1 𝑑) 7 3 − 22 ÷ −1 + 5𝑥 − 4 + 22 Simplify the following using the crab-claw 𝑎) 𝑥 + 1 𝑥 + 1 𝑏) 𝑥 + 3 𝑥 − 4 𝑐) 𝑥 + 2 2 + 1 𝑑) 2𝑥 + 2 𝑥 − 1 𝑒) −3𝑥 − 1 −𝑥 + 2 Simplify the following using the crab-claw and BIMDAS 𝑎) 3 − 42 − 𝑥 2 + 𝑥 + 13 2 𝑏) 𝑥 + 1 2 − 5 + 22 2 ÷ 3 + 1 𝑐) 5 − 1 − 13 2 + 𝑥 + −1 2 2 + 1 − 𝑥 𝑑) 𝑥 − 1 2 + 2𝑥 + 1 2 + 12 𝑒) 2 𝑥 − 1 2 + 3𝑥 − 22 − 4 3 − 𝑥 Simplify the following using all the theory from this chapter. The reason is that 𝑥 + 3 is not in brackets so only the 3 needs to be crabclawed: 4.5 𝑏 Solution: For statement 1: use BIMDAS to get rid of the Brackets: 𝑥 + 3 𝑥 + 5 − 15𝑥 + 4 The only thing multiplying 𝑥 + 5 is 3. nor multiplication/division. the placement of brackets is very important as obviously.8 equations You have already been introduced to equations without it being said explicitly. and then onto Addition/Subtraction by collecting like terms: = 8𝑥 2 − 8𝑥 + 27 Example 8: show that the expanded forms of the two statements below are not the same: Statement 1: 𝑥 + 3 𝑥 + 5 − 15𝑥 + 4 Statement 2: 𝑥 + 3 𝑥 + 5 − 15𝑥 + 4 2. 5 − 5 = 0 and on the right side. 𝑥 was isolated to give the number of computers to be ordered. The retailer has only 13 in stock. Example 2: isolate 𝑥 in 𝑥 = 10 4 Solution: to isolate 𝑥. and the right side (45) is what the customer wants. Examples of equations include: 5+1=6 3𝑥 − 4 = 7 𝑦 = 𝑥 2 + 6𝑥 − 3 All these have equals signs so all are equations. whatever is done to one side must be done to the other. but whatever is done to one side has to be done to the other side as well. Equations are used widely to find solutions to problems. and when 13 is taken from 45. If one side of an equation is changed without changing the other. rearranging an equation is to isolate a variable. Most of the time. But once again. Example 1: isolate 𝑥 in 𝑥 + 5 = 9 Solution: to isolate 𝑥. This sounds obvious. it will no longer be equal. the 4 needs to be removed. go back to the original equation and substitute 𝑥 = 4. so both sides must be multiplied by 4: 𝑥 4 ∙ = 4 ∙ 10 4 On the left side. Write the problem as an equation: 13 + 𝑥 = 45 The left side is what is in stock (13) plus what needs to be ordered (𝑥). Intro example 1: a customer wants to buy 45 laptop computers from a retail company. the more natural it becomes. Theory: RULE 1: whatever is done to one side of the equation must also be done to the other side. Does 4 + 5 = 9? Yes. In the above example. A “+5” is “reversed” by subtracting 5 from the left side. There are a few rules and a few hints which will make rearranging equations very simple. the number of computers that need to be ordered (𝑥 = 45 − 13 = 32). 22 .Theory: an equation is anything with an equals sign. but it is often forgotten. but the more you practice. This is a very simple example. the result will be 𝑥. Theory: make sure there is a reason for rearranging an equation. the 4’s cancel to give 1𝑥 or simply 𝑥 on the left side: 𝑥 = 40 Substitute this into the original equation: 40 = 10 ? 4 Yes Theory: RULE 2: when isolating a variable. So 5 is taken from both sides: 𝑥 + 5 − 5 = 9 − 5 This can then be simplified to: 𝑥 = 4 As on the left side. Manipulating equations is difficult to begin with. To make sure. This simply means work backwards to BIMDAS when deciding what to get rid of on the side you are trying to isolate 𝑥. 9 − 5 = 4. The problem is to find how many computers need to be ordered. The “reverse” of a “÷ 4” is to multiply the left side by 4. use the reverse order of BIMDAS to get rid of the elements around the variable to be isolated. the 5 needs to be removed. To reverse this +𝑦. but they must be done one at a time. the 3 and the 5 need to be “reversed”. Solution: to isolate 𝑥. the −3 is removed first by adding 3 to both sides. or down bottom (they are the same. get rid of the 5 as it is an addition. “reverse” the ÷ 7 by multiplying both sides by 7: 𝑥 = 7 ∙ 12 − 𝑦 Brackets are needed around the 12 − 𝑦 as RULE 1 above is really: Theory: RULE 1: whatever is done to the whole of one side of an equation must be done to the whole of the other side. and then the −5.Example 3: isolate 𝑥 in the following: 𝑥 +5=7 3 Solution: to isolate 𝑥. but the 𝑦 could have been removed first without changing the solution: 𝑥 + 𝑦 − 3 + 3 = 9 + 3 7 𝑥 + 𝑦 = 12 7 Get rid of the +𝑦 (treat this as you would any other number). work backwards through BIMDAS. Also. Example 5: isolate 𝑥 in the following 𝑥 + 𝑦 − 3 = 9 7 Solution: using the reverse of BIMDAS. as BIMDAS does not give precedence to Division/Multiplication. the +𝑦 or −3 must be removed. Get rid of the +7 by subtracting 7 from both sides: 3𝑥 + 7 − 7 = −4 − 7 5 3𝑥 − = −11 5 − The negative sign on the left can be put either up top. Example 4: isolate 𝑥 in the following: − 3𝑥 + 7 = −4 5 3 1 Divide both sides by 3 to get rid of the × 3 on the left side: 3𝑥 55 = 3 3 55 1 𝑥 = = 18 3 3 Note: the negative sign could have been moved to give −3𝑥 5 without changing the answer. Using the reverse of BIMDAS. subtract 𝑦 from both sides: 𝑥 + 𝑦 − 𝑦 = 12 − 𝑦 7 𝑥 = 12 − 𝑦 7 Finally. the 3 could have been “reversed” first. subtract 5 from both sides: 𝑥 +5−5=7−5 3 𝑥 =2 3 To “reverse” the ÷ 3 multiply both sides by 3 (remember that 3 = ): 3 𝑥 ∙ =3∙2 1 3 𝑥 = 6 If 𝑥 = 6 is substituted into the original equation: 6 +5=2+5=7 3 Which is correct. To “reverse” the +5. multiply both sides by −5 (remember −5 = −5 ): 1 −5 3𝑥 ∙ = −11 ∙ −5 1 −5 3𝑥 = 55 23 . and both will give the same answer): 3𝑥 = −11 −5 To get rid of the ÷ −5 . Below. put everything not with an 𝑥 onto the right side. which would give: 𝑥 = 30 + 10𝑦 39 On the left side. so 3𝑥/1 needs to be changed to have 5 on the bottom. however here. only the 12 is being multiplied by 7. by multiplying by 5/5: 2𝑥 3𝑥 5 2 2𝑦 − + = + 5 1 5 1 3 2𝑥 15𝑥 2 2𝑦 − + = + 5 5 1 3 Change the left side under a single division: −2𝑥 + 15𝑥 2 2𝑦 = + 5 1 3 Simplify the top of the left side: 13𝑥 2 2𝑦 = + 5 1 3 A common application of equations is finding exchange rates.You might have thought to write: 𝑥 = 7 ∙ 12 − 𝑦 Find a Common Denominator for the right side (1 × 3 = 3): 13𝑥 6 2𝑦 = + 5 3 3 13𝑥 6 + 2𝑦 = 5 3 The reason why a Common Denominator is found for both sides is that it makes it easier to “reverse” the ÷ 5 making it easier to simplify the right side: 13𝑥 6 + 2𝑦 ∙5= ∙5 5 3 Remember that ∙ 5 is really like multiplication by 1: 13𝑥 5 6 + 2𝑦 5 ∙ = ∙ 5 1 3 1 13𝑥 = 5 6 + 2𝑦 3 5 This is incorrect as the whole of the right side must be multiplied by 7. find how many Euros you will get per American dollar. This 24 . Half a crab claw could be used to put that 5 into the brackets. by adding 3 𝑦 to both sides and subtracting 5 from both sides: 2 2 2 2 − 𝑥 + 3𝑥 − 𝑦 + 𝑦 + 5 − 5 = 7 + 𝑦 − 5 5 3 3 3 2 2 − 𝑥 + 3𝑥 = 2 + 𝑦 5 3 Change each side to have a Common Denominator (the left side Common Denominator does not have to be the same as the Common Denominator on the right): − 2𝑥 3𝑥 2 2𝑦 + = + 5 1 1 3 2 Finally. Example 7: if the currency conversion rate from American dollars (𝑈𝑆$) to Euros (€) is: 𝑈𝑆$1. Given the correct form: 𝑥 = 7 ∙ 12 − 𝑦 It can be expanded using half a crab-claw: 𝑥 = 84 − 7𝑦 Example 6: isolate 𝑥 in 2 2 − 𝑥 + 3𝑥 − 𝑦 + 5 = 7 5 3 Solution: firstly. divide the whole of both sides by 13: 13𝑥 5 6 + 2𝑦 13 = ÷ 13 3 1 To get rid of the ÷ sign on the right.50 = €1 How many Euros would be obtained from converting 𝑈𝑆$173? Solution: to solve this sort of question. the Common Denominator is 5. to get rid of the × 13. invert the second fraction and multiply: 5 6 + 2𝑦 1 ∙ 3 13 5 6 + 2𝑦 𝑥 = 39 𝑥 = That is the answer. 33 1. then bring 𝑥 out front of brackets: 𝑦 = 𝑥 4 + 3𝑥 Example 2: factorise anything common in the equation 𝑦 = 4𝑥 2 + 8𝑥 3 Solution: each part on the right has an 𝑥 2 (the 𝑥 3 is the same as 𝑥 2 𝑥). then having that common aspect being put outside of the brackets. so as to get a single 𝑈𝑆$ on the left side: 𝑈𝑆$1 = € 1 1. leaving answers as a single fraction (where applicable): 𝑎) 3𝑦 − 2𝑥 = 𝑥 + 15 𝑏) 𝑦 = 3𝑥 − 15 − 𝑦 𝑐) 𝑦 + 1 + 𝑥 = 𝑦 − 1 3𝑥 5 𝑑) + 𝑦 − 4 = 0 4 3 𝑒) 17𝑥 − 13𝑦 − 11 = 7 4𝑦 − 1 𝑓) =5 𝑥 + 2 3.04 𝑐) ¥1801 𝑈𝑆$1.23 = 𝐶𝐴𝑁$0.5 change the equation. The application of factorising will be seen in subsequent chapters. Example 3: factorise 𝑦 = 3𝑥 3 − 12𝑥 + 6𝑥 2 Since 𝑈𝑆$173 is being converted. but rather is a method of manipulating an equation. so this means that 4𝑥 2 can be factorised out from each part on the right side: 𝑦 = 4𝑥 2 1 + 2𝑥 Theory: for a statement which is separated by additions/subtractions. Factorising does not 25 .5 Theory: for currency conversions. then each separate part of the statement can be “divided” by the common aspect. and each part of the statement has a certain common aspect. but then bringing 2 out front of brackets: 𝑦 = 2 𝑥 + 1 This may be confusing but if half a crab-claw was used to get rid of the brackets. Factorising out 2 from every part of the right side is analogous to “dividing” each term by the number 2. multiply both sides by 173: 1 𝑈𝑆$173 = € 173 ≈ €115. and given the exchange rate. the original equation will be obtained. 𝑎) 𝐴$181 𝐴𝑈𝑆$0. obtain the exchange rate for a single unit of the currency you have.94 = 𝐶𝐴𝑁$1. and also a 4 (as 8 = 4 ∙ 2).5𝑥 19 𝑑) 𝑥 − 15 = 17 − 5 3 −2𝑥 1 𝑒) + 5𝑥 − 𝑥 − 3 = 15 − 𝑥 3 3 𝑥 𝑥 1 𝑓) 𝑥 + − + = 5 2 3 4 2.80 = 𝑁𝑍$1. Example 1: factorise 𝑥 out of the following 𝑦 = 4𝑥 + 3𝑥 2 Solution: “divide” each of the parts on the right side by 𝑥. Given the initial amount of currency.1 𝑑) 𝐴𝑈𝑆$94 𝐴𝑈𝑆$1. Solve for 𝑥 in the following equations: 𝑎) 2𝑥 − 3 = 4𝑥 + 7 10 𝑏) 12𝑥 − 15 + = 2𝑥 2 𝑐) 17𝑥 − 3𝑥 = 15𝑥 − 2𝑥 + 13.03 = ¥96.00 𝑏) 𝑈𝑆$47 𝑈𝑆$0.9 factorisation Factorisation is the opposite of expansion of brackets.means both sides must be divided by 1. Isolate 𝑥 in the following equations. Exercises: 1. Intro example 1: for the equation 𝑦 = 2𝑥 + 2 Factorise out the number 2.50. then multiply both sides of that conversion equation by the number of units of currency initially had. determine the equivalent value in the other currency.92 1. Example 1: put in the inequality sign for 6∎8 Solution: since 6 is less than 8.g. The two “parts” are colour coded below: 0 = 3𝑥 𝑥 + 2 For the whole right side to equal zero.10 inequalities and absolute values Theory: inequalities are noted with the signs “<” and “>”. Solve for 𝑥 using factorisation: 𝑎) 18𝑥 2 − 9𝑥 𝑏) 8𝑥 − 4𝑥 2 𝑐) 19𝑥 2 − 38𝑥 𝑑) 17𝑥 − 18𝑥 2 − 𝑥 𝑒) 12𝑥 4 − 37𝑥 3 1. the inequality sign has to have the open part to the right: 6<8 Manipulating inequalities is a little different from equations. The two solutions are 𝑥 = 0 and 𝑥 = −2. there are at least two big parts being multiplied together. is that the side of the open part of the signs is greater than the pointed end. Example 2: take 5 from both sides of 11 > 7 Solution: 11 − 5 > 7 − 5 6>2 Theory: multiplication and division of inequalities is identical to equalities only when multiplied or divided by a positive number. either 3𝑥 = 0 making 𝑥 = 0 or 𝑥 + 2 = 0 making 𝑥 = −2. Factorise 𝑎) 2𝑥 + 4𝑥𝑦 + 8𝑥 2 𝑏) 3𝑥 2 + 9𝑥 − 12𝑦 𝑐) 3𝑥 3 + 2𝑥 2 𝑑) 13𝑥 2 + 𝑥 3𝑥 𝑥 2 𝑒) − + 4 4 2 𝑓) 18𝑥 𝑦 − 91𝑥𝑦 2 𝑔) 12𝑥 2 𝑦 2 − 48𝑥𝑦 2. one of the statements must be equal to zero. Example 3: divide both sides of the inequality by 5 50 > −40 Solution: 50 −40 > 5 5 10 > −8 An equality is when two things are equal. the sign ≥ means “greater than or equal to”. Exercises: 1.Solution: the common aspect in each part is 3𝑥. However. Inequalities are when two things are not equal. The easiest way of remembering which side is greater than and which side is less than. so put 3𝑥 out front of brackets and then “divide” each part by 3𝑥: 𝑦 = 3𝑥 𝑥 2 − 4 + 2𝑥 A common use of factorisation is finding the solution(s) to complex statements being equal to zero. 5 > 3 as 5 is on the side of the open end meaning that 5 is greater than 3. and the sign < means the left side is less than the right side. Similarly. Theory: addition and subtraction of inequalities is identical to that of equalities. Theory: when two statements are multiplied together and they equal zero. and the converse sign ≤ means “less than or equal to”. Example 4: solve 0 = 3𝑥 2 + 6𝑥 Solution: factorise a common 3𝑥 out of each part: 0 = 3𝑥 𝑥 + 2 Now there are two parts being multiplied to give zero. The sign > means the left side is greater than the right side. 26 . E. after the statements are factorised. 𝑎) | − 4| 𝑏) |79| 𝑐) |5 − 4| 𝑑) |32 − 42 | 𝑒) |4 − 6 − 1| 1. 5 ∙ 0 = 0 Zero divided by any number (other than another zero) is zero. and further investigation into the original function is required.g. Example 4: multiply both sides by −3 8>5 Solution: 8 −3 > 5 −3 −24 > −15 This is not true. This above example does not mean that 𝑥 cannot be simplified. Isolate 𝑥 in the following inequalities: 𝑎) 2𝑥 > 15 − 𝑥 1 𝑏) 13𝑥 + 𝑥 < 0 − 15 3 𝑐) 11𝑥 − 12𝑥 < 12 − 11 15 𝑥 𝑑) 2𝑥 − 3𝑥 + 𝑥 − > 21 − 3𝑥 2 3 3. 𝑥 is equal to 1.11 the number zero Zero is a special number. Theory: an absolute value makes any statement which is negative into a positive statement. Find the absolute values of the following (Hint: simplify before finding the absolute value). What it means is that when 𝑥 = 0. E. so the inequality sign must be reversed to give the correct answer: −24 < −15 Moving on to absolute values. 0 ≠ 0 Example 1: solve the following problem when 𝑥 = 0: 𝑥 𝑥 Solution: this is impossible. whenever 𝑥 is not equal to zero.Theory: when multiplying or dividing an inequality by a negative number. However. 15 = 0 Zero divided by zero does not exist. Exercises: 1. but it is what you will need for understanding this book. Divide both sides of the inequality by the number in brackets 𝑎) 15 > 4 2 𝑏) 13 < 15 −2 𝑐) − 12 < −6 −1 1 1 𝑑) > −5 5 8 2. but is not necessarily equal to zero. Absolute values are denoted with straight lines around the statement: 𝑥 means “the absolute value of 𝑥”. the inequality sign must be reversed. Example 5: find the absolute value of −5 Solution: −5 = 5 Example 6: Find the absolute value of 7 Solution: 7 = 7 This is not the extent of absolute values. 𝑥 𝑥 𝑥 0 0 27 . 𝑥 is undefined. when two statements are multiplied together and they equal zero. the value of 𝑥 is irrelevant: 15𝑥 2 0 = 0 Theory: as mentioned in the previous section. with special properties. one of the statements must be zero. Example 2: solve 15𝑥 2 0 Solution: because there is a multiplication by zero. Theory: Any number multiplied by zero has a result equal to zero. 256 BASE 5: 5. When multiplying or dividing an inequality by a negative number.81. obtain the exchange rate for a single unit of the currency you have. but leaves the values of all fractions unchanged. EQUATION RULE 2: when isolating a variable. work out how many denominators will fit into the numerator without exceeding the numerator.e. and each part of the statement has a certain common aspect. Indices. 4. Absolute values are denoted with straight lines around the statement: 𝑥 means “the absolute value of 𝑥”. “Collecting like terms” is the process of bring together things that are the same. then each separate part of the statement can be “divided” by the common aspect. To factorise a statement which is separated by additions/subtractions. 28 . BIMDAS is Brackets.27.Example 3: solve for: 0 = 𝑥 + 3 𝑥 − 1 Solution: the two parts (bracketed statements) are colour coded below: 0 = 𝑥 + 3 𝑥 − 1 Either 𝑥 + 3 = 0 which gives 𝑥 = −3 or 𝑥 − 1 = 0 giving 𝑥 = 1. Outer.243 𝑥 𝑎 BASE 4: 4. So the solutions are: chapter one summary Arithmetic with negative numbers: 𝑎 + −𝑏 = 𝑎 − 𝑏 𝑎 − −𝑏 = 𝑎 + 𝑏 −𝑎 −𝑏 = 𝑎 ∙ 𝑏 −𝑎 𝑎 = −𝑏 𝑏 𝑎 −𝑏 = −𝑎 ∙ 𝑏 −𝑎 𝑎 𝑎 = =− 𝑏 −𝑏 𝑏 𝑥 = −3 Exercises: 𝑎𝑛𝑑 𝑥 = 1 1. Multiplication. Inner. 32. and any remaining numbers are left over the original denominator. This approach changes the form of all fractions so that the denominators are all the same. Division. When two statements are multiplied together and they equal zero. 128. Inequalities are noted with the signs “<” and “>”. Addition. Addition and subtraction of inequalities is identical to that of equalities. Solve for 𝑥 in the following: 𝑎) 3 − 5𝑥 + 12 0 = 3 − 16𝑥 0 𝑏) 15𝑥 + 0 − 15 = 0 𝑐) 30 𝑥 + 0 2 + 10 𝑥 = 30𝑥 2 − 20 0 𝑑) + 1 = 𝑥 0 Different ways of denoting a multiplication: 𝑎 ∙ 𝑏 = 𝑎 × 𝑏 = 𝑎 𝑏 Multiplication of fractions: 𝑎 𝑐 𝑎 ∙ 𝑐 ∙ = 𝑏 𝑑 𝑏 ∙ 𝑑 Division of fractions: 𝑎 𝑐 𝑎 𝑑 𝑎 ∙ 𝑑 ÷ = × = 𝑏 𝑑 𝑏 𝑐 𝑏 ∙ 𝑐 Addition and subtraction of fractions uses the Common Denominator approach. then add that result to the top. 256 BASE 3: 3.9. Multiplication and division of inequalities is identical to equalities only when multiplied or divided by a positive number. BIMDAS is a method that gives the order in which operations must be done. To find how many “whole” numbers there are in a fraction.16. 16. then having that common aspect being put outside of the brackets. The sign > means the left side is greater than the right side. An absolute value makes any statement which is negative into a positive statement. INDEX RULE 1: 𝑥 𝑎 ∙ 𝑥 𝑏 = 𝑥 𝑎 +𝑏 INDEX RULE 2: = 𝑥 𝑎−𝑏 𝑥 𝑏 INDEX RULE 3: 𝑥 0 = 1 𝑛 INDEX RULE 4: 𝑥 = 𝑥 1/𝑛 INDEX RULE 5: 𝑥 𝑎 𝑏 = 𝑥 𝑎∙𝑏 1 INDEX RULE 6: 𝑥 −𝑎 = 𝑎 𝑥 INDEX RULE 7: 𝑥𝑦 𝑎 = 𝑥 𝑎 𝑦 𝑎 Common numbers and their base: BASE 2: 2.625 The square root of a positive number will always have two solutions: one positive and one negative. A Common Denominator is found by multiplying the denominators (bottoms) of all the fractions to be added or subtracted. and the sign < means the left side is less than the right side. For currency conversions.25. To expand the multiplication of two brackets. 8. use the reverse order of BIMDAS to get rid of the elements around the variable to be isolated. The multiplication/division have no precedence over one another. then multiply both sides of that conversion equation by the number of units of currency initially had. The form of each individual fraction is changed by multiplying by 𝑥/𝑥 where 𝑥 is the multiplication of all the other denominators. The numerators (tops) are then put over the common denominator. First. Last – FOIL) EQUATION RULE 1: whatever is done to the whole of one side of the equation must also be done to the whole of the other side. the inequality sign must be reversed. one of the statements must be equal to zero. and neither does the pair of addition/subtraction. every term in one bracket must be multiplied with every term in the other bracket (i.64. Simplifying a fraction involving whole numbers: multiply the whole number out front by the denominator (bottom). 64. Subtraction. Then remove that many denominators from the numerator.125. Isolate 𝑥 in the following equations: 𝑎) 𝑥 − 4 = 1 𝑏) 3𝑥 + 1 = 7 𝑐) 2𝑥 − 1 = 4𝑥 − 3 𝑑) 3𝑥 − 7 = 4 2𝑥 + 2 2 𝑒) =5 𝑓) 7𝑥 + = 1 3 3 𝑔) 12𝑥 + 3 = −6 + 2𝑥 ) 3𝑥 + 𝑦 − 7 = 2 4𝑥 − 2 𝑖) =2 𝑗) 𝑥 − 7 = 5𝑥 − 3𝑦 + 2 𝑦 9.1 𝑓) |2 − 3| 2 29 . ≠ 0 0 chapter one questions 1.5 𝑥 8 𝑥 −3 𝑦 −4 1 216𝑥 6 3 4𝑥 0.25 𝑦 3𝑥𝑦 2 0.125 8𝑦 −2 𝑥 −1 𝑔) ) 6.Any number multiplied by zero has a result equal to zero. 3. leaving positive indices: 𝑎) 𝑥𝑦 2 3 𝑏) 𝑐) 𝑑) 2𝑥 4 15 2. Solve for 𝑥 in the following equations: 𝑎) 𝑥 − 3 𝑥 − 2 = 0 𝑏) 𝑥 + 7 𝑥 − 4 = 0 𝑐) 2𝑥 − 4 𝑥 − 3 = 0 𝑑) −3𝑥 − 1 6𝑥 + 15 = 0 3𝑥 5 − 6𝑥 4 𝑒) =0 𝑥 3 3 27𝑥 6 − 𝑥 2 2 + 𝑓) 4 − 𝑥 − 𝑥 2 = 0 13. Find the absolute value of the following: 𝑎) −3 𝑏) 4 𝑐) | − 1| 𝑑) −15 𝑒) −0.5 𝑥 −1. Zero divided by any number (other than another zero) is 0 zero. Simplify and solve for 𝑥 in the following: 𝑎) 𝑥 2 − 𝑥 = 0 𝑏) 𝑥 2 + 3𝑥 = 0 2 𝑐) 2𝑥 − 6𝑥 = 0 𝑑) 4𝑥 2 + 4𝑥 = 0 𝑒) 𝑥 3 − 𝑥 2 = 0 𝑓) 3𝑥 4 − 5𝑥 3 = 0 11. 216𝑥 −3 𝑦 8 36𝑥𝑦𝑥𝑦𝑥 1.5 25𝑥 4 𝑦 −0. the following: 𝑎) 2𝑥 + 3 2 + 3 𝑏) 𝑐) 9 − 𝑥 2 9𝑦𝑥 4 −2 3 − 𝑥 2 + 3 − 3 + 𝑥 2 + 3 + 𝑥 −2 3𝑥 2 + 6𝑥 3 12𝑥 𝑥(𝑥 −3 −2 𝑥 4.5 −𝑥𝑦 3 3 𝑑) 𝑒) −3 − 2 𝑥 − 32 + 1 − 3 2 3 𝑥𝑦 125𝑦 3 Simplify the following using BIMDAS: 𝑎) 3 + 5 ∙ 7 − (−3 + 5) ∙ 2 𝑏) 4 −2 + 4 ∙ 32 − 2 ∙ 2 𝑐) 3 − 2 ∙ 2 − 3 2 + 23 𝑑) 14𝑥 − 𝑥 3 + 4 2 + 3 − 1 𝑥 𝑒) 2 − 3 ∙ 32 − 1 2 + 4 𝑓) 2 − 6 3 + 33 + −2 − 1 + 2 4𝑥 −125𝑥 9 + 3𝑥 3 𝑥 −2 𝑓) − 5𝑥 2 + 3𝑥 1 𝑥 2 12. Simplify.5 81𝑥 9 3𝑥𝑦 2 3 𝑒) 27𝑥𝑦 −2 𝑓) 2𝑥𝑦 4 𝑦 0. 5. 𝑔) −2 −3 − 43 + 2 ∙ −3 ) 2𝑥 − 3 𝑥 + 2𝑦 − 1 + 2 3 𝑦 − 𝑥 + 3 7. 4.5 𝑔) 3 𝑥 − 1 2 + 1 ) − 2 − 𝑥 + 1 2 − 14 8. then factorise.25 −0. Simplify the following negative numbers: 𝑎) −3 −5 𝑏) −2 2 𝑐) 5 ∙ −5 ∙ −2 𝑑) 3 ∙ −2 Simplify the following fractions: 1 1 2 1 𝑎) ∙ 𝑏) ∙ 2 3 3 8 1 1 4 2 𝑐) / 𝑑) − 2 3 3 5 9 −3 18 1 𝑒) ∙ 𝑓) ÷ −7 7 2 3 27 1 3 5 4 3 2 1 𝑔) × ÷ ) × ÷ × ÷ 13 2 8 3 3 3 3 3 Simplify the following fractions: 1 1 2 1 𝑎) + 𝑏) + 2 3 3 8 3 4 2 1 𝑐) + 𝑑) − 4 3 11 2 2 1 1 1 1 𝑒) + 𝑓) − + 13 5 2 3 4 6 5 4 2 3 𝑔) + − ) 3 + − 7 6 5 3 7 Simplify the following fractions: 1 2 1 2 1 1 𝑎) ∙ − 𝑏) ∙ + 3 −7 2 2 6 5 1 1 1 −2 1 𝑐) + 𝑑) − − 𝑥 3 −2 3 2 4 3 1 1 1 1 1 𝑒) / − 𝑓) 1 + + + + 5 4 6 2 3 4 𝑥 𝑥 3 5 𝑥 1 1 2 𝑔) 2 / ∙ + ) + 2 4 1 3 𝑥 𝑦 3 Simplify using index rules. Factorise out anything possible in the following: 𝑎) 2𝑥 + 2𝑥𝑦 + 2 𝑏) 2𝑥 2 + 6𝑥𝑦 − 8𝑦 2 𝑥 2 𝑐) 13𝑦𝑥 − 12𝑦 + 𝑦 𝑑) 3𝑥 2 + 9𝑥 3 + 12𝑥 5 1 2 1 4 1 𝑒) 4𝑄𝑃 + 𝑟𝑃𝑄 − 𝑥𝑃𝑦𝑄 𝑓) 𝑥 + 𝑥 + 3 3 6 1 1 1 3 4 𝑔) 15𝑥 + 25𝑥 − 50𝑥 ) + − 𝑥 𝑥 2 𝑥 3 10. = 0 Zero divided by zero does not exist. Expand and simplify the following: 𝑎) 𝑥 + 2 2 𝑏) 𝑥 − 3 2 2 𝑐) 2𝑥 + 1 𝑑) 1 − 𝑥 2 𝑒) 𝑥 + 3 2 + 2𝑥 − 4 𝑓) 2 − 2𝑥 − 1 2 + 3𝑥 − 40. but is not necessarily 0 equal to zero.5 𝑦 −0. 2 2.5 2.9 Linear Equations Main Feature of Linear Equations Negative Gradients Graphing Lines from Equations Obtaining the Equation of a Line Intersecting Lines Microeconomic Applications Elasticity Interpreting Elasticity 31 32 33 34 36 38 40 42 44 46 47 Chapter Two Summary Chapter Two Questions 30 .1 2.4 2.Chapter 2 Linear Algebra How to describe and draw lines in maths 2.3 2.7 2.6 2.8 2. 𝑦 𝐴 𝐵 The graph below shows how this definition is used: 𝑦 𝐺 𝑟𝑖𝑠𝑒 𝐹 𝑟𝑢𝑛 𝑥 Going from point 𝐹 to point 𝐺. because its value depends on 𝑥. how much does the line rise for every unit it goes across (runs)? Example 1: determine the gradient of the following line: 𝑦 𝑥 Line 𝐴 is steeper than line 𝐵. 𝑦 𝐷 Plan: use the equation 𝑚 = Solution: 𝑚 = 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 3 10 𝑥 𝑟𝑖𝑠𝑒 3 = = 0. 𝑐 is a constant.3 𝑟𝑢𝑛 10 Theory: the gradient of a line is the same at all points along it.1 linear equations These lines have the same slope (𝑚). which is the independent variable (see chapter 1). Graphs of linear equations are straight lines. The value of 𝑐 is a constant that determines how much the graph of the line is shifted up or down (line 𝐷 has a larger value of 𝑐 than line 𝐸). 𝐸 𝑥 The values of 𝑚 and 𝑐 are actual numbers. ask. 𝑦 is the dependent variable. 𝑚 is the gradient of the line – which is another way of saying the slope of the line. In simple terms.2. For example 𝑦 = 2𝑥 + 3. 𝑐 = 3). The 𝑥 and the 𝑦 are variables. Theory: a mathematical definition of gradient is: 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝑚 = 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 Theory: linear equations have the general form: 𝑦 = 𝑚𝑥 + 𝑐 Where 𝑚 is the gradient. The graph below shows how the value of 𝑐 affects the position of a line. and 𝑥 and 𝑦 are variables. Below is a graph of two lines which might help you understand this theory. Steeper lines have larger values of 𝑚. so it has a larger gradient (slope=gradient). Thus. line 𝐴 has a larger value of 𝑚 than line 𝐵. such as: 𝑦 = 4𝑥 + 5 𝑦 = 81𝑥 − 7 𝑚 = 4 𝑚 = 81 𝑐 = 5 𝑐 = −7 31 . but the values of 𝑐 differ. The 𝑐 is a constant which is added to everything else (it is just a number). Both 𝑚 and 𝑐 are definite numbers (𝑚 = 2. 2 main features of linear equations The graph below shows a couple of lines. lines cutting the 𝑦 − 𝑎𝑥𝑖𝑠 higher up have larger 𝑐 value. Exercises: 1. Solution: multiply both sides by (𝑥 + 1): 3𝑦 − 4 (𝑥 + 1) = 6(𝑥 + 1) 𝑥 + 1 3𝑦 − 4 = 6𝑥 + 6 Add 4 to both sides: 3𝑦 − 4 + 4 = 6𝑥 + 6 + 4 3𝑦 = 6𝑥 + 10 Divide both sides by 3: 3𝑦 6𝑥 + 10 = 3 3 10 𝑦 = 2𝑥 + 3 The gradient (𝑚) and the constant (𝑐) of each line is unknown until it is rearranged into the general form. but there are many other forms. Solution: add 7𝑥 to both sides. 2. −7𝑥 + 7𝑥 + 4𝑦 = 𝑦 − 3 + 7𝑥 Take 𝑦 from both sides (to have all the 𝑦’s on the left side): +4𝑦 − 𝑦 = 𝑦 − 𝑦 − 3 + 7𝑥 3𝑦 = 7𝑥 − 3 Then divide the whole of both sides by 3. There are two aspects to note when graphing lines: 1. the 𝑐 value shifts the line up or down. steeper lines have larger gradients (𝑚 values). Similarly. Determine the exact gradient and 𝑐 − 𝑣𝑎𝑙𝑢𝑒 of the following equations: 𝑎) 8𝑦 + 3𝑥 − 7 = 0 𝑏) − 3𝑥 − 5𝑦 = 7 − 𝑥 𝑥 4 𝑐) + 𝑦 = 7 −3 −1 −5 6 𝑑) − 𝑥 + 𝑦 = 𝑦 − 1 4 7 𝑒) 𝑦 = 9 2𝑦 − 1 𝑓) =5 𝑥 + 1 2. it is best practice to manipulate equations of lines to get them into this easier. you should be able to rearrange equations easily: Example 2: rearrange to get into the general form −7𝑥 + 4𝑦 = 𝑦 − 3 Plan: use the reverse of BIMDAS to isolate 𝑦. 𝑦 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑥 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 Theory: the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is where the line cuts the 𝑥 − 𝑎𝑥𝑖𝑠. the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is where the line cuts the 𝑦 − 𝑎𝑥𝑖𝑠. From Chapter 1. 32 . fractions or decimals. and can be whole numbers. general form. Despite all these other forms. The 𝑦 = 𝑚𝑥 + 𝑐 equation is the general form of a line. Get the following equations in general form: 𝑎) − 𝑥 − 𝑦 = 7 𝑏) 2𝑥 − 2𝑦 − 4 = 0 𝑐) 3𝑦 + 4𝑥 = 17 𝑑) − 3 𝑥 − 2𝑦 = 𝑦 − 2 𝑒) 𝑦 = 2𝑥 − 0.5𝑦 + 8 𝑦 + 3 𝑓) =3 𝑥 + 3 2.1 3 𝑦 = − 𝑥 + 4 4 𝑚 = − 1 4 𝑐 = 3 4 The constants 𝑚 and 𝑐 can be positive or negative. 3𝑦 7𝑥 − 3 = 3 3 7 3 𝑦 = 𝑥 − 3 3 7 𝑦 = 𝑥 − 1 3 Example 3: rearrange to get into the general form 3𝑦 − 4 =6 𝑥 + 1 Plan: apply the reverse of BIMDAS to isolate 𝑦. Order the lines from the one with the largest gradient to the one with the lowest gradient: 𝐴 is steeper than line 𝐵. look at where the lines cut the 𝑦 − 𝑎𝑥𝑖𝑠. but in absolute terms (ignoring the negative signs) 3 > 1. 𝐵 𝑦 𝑦 𝐷 𝐵 𝐴 𝑥 2. line 𝐴 has gradient −3 and line 𝐵 has gradient −1. 𝐷. Order the above lines from the one with the largest 𝑐 − 𝑣𝑎𝑙𝑢𝑒 to the one with the lowest. so the order of 𝑐 values is 𝐴. and lines with a 33 . Example 1: for the three lines graphed below. Theory: the steeper a line with a negative gradient. and at the same time has a more negative 𝑚 value than lines less steep. all linear equations have had a positive gradient. Theory: lines with a positive gradient go from the bottom left to the top right. The graph below shows lines with negative gradients: 𝑦 𝐵: 𝑦 = −𝑥 + 3 𝐷 𝑥 𝑥 𝐴: 𝑦 = −3𝑥 − 1 Both lines have a negative gradient. The line with the higher 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 has a larger 𝑐 value.Example 1: from the following graph. That is. determine 1) which has a larger 𝑚 value.3 negative gradients Up to this point. Solution: Looking only at the slopes of the lines. Then order the lines from the largest 𝑐 value to the smallest 𝑐 value. Exercises: 1. the larger the absolute value of 𝑚. Remember from Chapter 1 that −1 > −3. Now. So the order of gradients is 𝐵. 𝐵. So the gradient 𝑚 is bigger for 𝐴 than for 𝐵 in absolute terms. 2. 𝐵 is steeper than 𝐷 which is steeper than 𝐴. 𝑦 𝐴 𝐵 negative gradient go from the top left to the bottom right. 𝐴 cuts it higher than 𝐵 which cuts it higher than 𝐷. 2) which has a larger absolute value of 𝑚. order the lines from the largest 𝑚 value to the lowest 𝑚 value. 𝐷 𝑥 𝐴 Plan: the line which is steeper has a higher absolute 𝑚 value. the smaller is the gradient 𝑚 (as it is more negative) and at the same time. 𝐷. and 3) which has a larger 𝑐 value. and 2) lines cutting the 𝑦 − 𝑎𝑥𝑖𝑠 higher have larger 𝑐 values. however line Plan: use the fact that 1) steeper lines have larger gradients. 𝐴. Solution: From the graph, 𝐴 is steeper than 𝐵 which is steeper than 𝐷. So the absolute values of the gradient 𝑚 is 𝐴, 𝐵, 𝐷. Because 𝐴 is steeper than the other two, it must have a more negative gradient; similarly 𝐵 is steeper than 𝐷 so it must have a more negative 𝑚 value than 𝐷. Thus, the normal 𝑚 value order is the opposite of the one above: 𝐷, 𝐵, 𝐴. Finally, 𝐵 cuts the 𝑦 − 𝑎𝑥𝑖𝑠 higher than 𝐷, which cuts it higher than 𝐴, so the order for the size of 𝑐 is 𝐵, 𝐷, 𝐴. Exercises: For the following lines: 𝐴 𝐵 𝐷 Theory: to find: 1. the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 (where the line crosses the 𝑥 − 𝑎𝑥𝑖𝑠), replace all values of 𝑦 with zero, then solve for 𝑥. 2. the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, replace all values of 𝑥 with zero, then solve for 𝑦. This may be a little bit confusing at first. Example 1: plot the line 𝑦 = 2𝑥 + 6 Plan: set all 𝑦′𝑠 equal to zero then solve for 𝑥 to find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. Then set all the 𝑥's equal to zero, and solve for 𝑦 to find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. Solution: for 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑦 = 0: 0 = 2𝑥 + 6 2𝑥 = −6 𝑥 = −3 This gives the coordinate −3,0 . Remember that at the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, the value of 𝑦 = 0, and coordinates are written with 𝑥 first then 𝑦: (𝑥, 𝑦). To find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑥 = 0: 𝑦 = 2 0 + 6 𝑦 = 6 This gives the coordinate (0,6), as at the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, the value of 𝑥 = 0. These two points can be plotted on a set of axes, then joined with a ruler. (0,6) 𝑦 𝑦 𝑥 a) Order the lines from the one with the lowest gradient to the one with the largest gradient: b) Order the lines from the shallowest to the steepest (in absolute terms). Compare a) and b). c) Order the lines from the one with the largest 𝑐 − 𝑣𝑎𝑙𝑢𝑒 to the one with the lowest. 2.4 graphing lines from equations Lines can be easily plotted on a set of axes. Theory: to plot a line, points are needed which are called coordinates. Coordinates are written as (𝑥, 𝑦), where 𝑥 is always written first, and 𝑦 second. For the coordinate (3,7), 𝑥 = 3 and 𝑦 = 7. To be able to draw a line, you need at least two points (sets of coordinates). It is good practice to have equations of lines in the general form, but the following method allows plotting of lines in any initial form. (−3,0) 𝑥 Theory: the 𝑥 − and 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 are usually the easiest points to find to be able to plot a line. 34 Example 2: Plot 𝑦 = 3𝑥 − 7 Plan: find: 1) the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑦 = 0 and solving for 𝑥 2) the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑥 = 0 and solving for 𝑦. Plot these two points then join them with a ruler. Solution: for the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑦 = 0: 0 = 3𝑥 − 7 3𝑥 = 7 𝑥 = 7 3 7 ,0 3 For the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑥 = 0: 𝑦 = 3 0 − 7 𝑦 = −7 Which gives the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 0, −7 . Plot these two points and join with a ruler: 𝑦 7 ,0 3 𝑥 𝑥= −2.5 The coordinate of the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is (−2.5,0). To find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑥 = 0: 3𝑦 − 2 0 = 5 3𝑦 = 5 𝑦 = 5 3 5 3 The coordinates of the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is 0, . Plot these two points, then join with a ruler: 5 0, 3 𝑦 Thus the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is: (−2.5,0) 𝑥 Note, that if asked for the gradient of this last equation (3𝑦 − 2𝑥 = 5), you would have to rearrange the equation into the general form. Try rearranging it, and plotting the rearranged equation. Example 4: find the gradient and plot the line 3𝑦 + 4 = −1 2𝑥 − 1 Plan: rearrange into the general form to find the gradient. Then find two points: (0, −7) 1) the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑦 = 0 and solving for 𝑥 2) the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑥 = 0 and solving for 𝑦. Plot these two points then join them with a ruler. Solution: rearrange into the general form: 3𝑦 + 4 (2𝑥 − 1) = −1(2𝑥 − 1) 2𝑥 − 1 3𝑦 + 4 − 4 = −2𝑥 + 1 − 4 3𝑦 −2𝑥 − 3 = 3 3 2 3 𝑦 = − 𝑥 − 3 3 2 𝑦 = − 𝑥 − 1 3 35 Example 3: plot the equation 3𝑦 − 2𝑥 = 5 Plan: find: 1) the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑦 = 0 and solving for 𝑥 2) the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑥 = 0 and solving for 𝑦. Plot these two points then join them with a ruler. Solution: to find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑦 = 0: 3 0 − 2𝑥 = 5 −2𝑥 = 5 Now that it is in the general form, the gradient is 𝑚 = − 2 3 ) 8𝑦 + 𝑥 =4 𝑥 − 𝑦 To plot the line, use either the original equation, or the equation in the general form. The following uses the original equation: 1. To find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑦 = 0: 3 0 +4 = −1 2𝑥 − 1 4 = −1 2𝑥 − 1 4 = −2𝑥 + 1 3 = −2𝑥 𝑥 = −1.5 Giving 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 coordinates (−1.5,0). 2. To find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑥 = 0: 3𝑦 + 4 = −1 2 0 −1 3𝑦 + 4 = −1 −1 3𝑦 + 4 = 1 3𝑦 = −3 𝑦 = −1 This gives coordinates of (0, −1). Plot these two points: 𝑦 −1.5,0 𝑥 2.5 obtaining the equation of a line In many cases you will need to find the equation of a line without being given a graph. Theory: when given two points, approximately plot the points on a set of axes (if not given the graph), then find how much the line “runs” along the 𝑥 − 𝑎𝑥𝑖𝑠, and then how much it “rises” along the 𝑦 − 𝑎𝑥𝑖𝑠 between the two points. Applying the 𝑟𝑖𝑠𝑒/𝑟𝑢𝑛 formula, the gradient can be found. Using the general form of a line, substitute in the gradient and any one of the points given, and solve for 𝑐. Rewrite the equation of the line. You will always be given either two coordinates, or a coordinate and a gradient. The second case is a subset of the first case, so if you know how to do the first case, you will be able to do the second. Intro example 1: find the equation of the line passing through the points (1,3) and (3,7). Solution: approximately plot the points: 𝑦 (3,7) 4 (1,3) (0, −1) 2 intersection Exercises: 1. Find the gradient of each of the equations, then plot the function: 𝑎) 𝑦 = 2𝑥 − 3 𝑏) 3𝑦 = 6𝑥 − 9 𝑐) 2𝑦 + 3𝑥 = 7 𝑑) 12𝑥 − 𝑦 = 3 𝑒) 7𝑦 − 3 = −2 + 𝑥 𝑓) 𝑦 − 𝑥 = 4 − 𝑥 3𝑦 − 1 𝑔) =3 2𝑥 + 1 𝑥Then draw a horizontal line from the point that is most left, and a vertical line at the point most right. Ask yourself: how much does the horizontal line “run” before it reaches the intersection? The answer is two units along the 𝑥 − 𝑎𝑥𝑖𝑠. Then ask: 36 4) will give the same answer. for every 7 that the line runs. how much does it rise? If the line falls.2) 𝑥 The line “runs” between −16 and −9 .from this intersection. This formula allows you to find the gradient between two points without plotting the points. The point (−9. 𝑦1 ) and 𝑥2 . and “rises” between 4 and 2 which is 2. substitute either one of the two point in for 𝑥 and 𝑦.4) and (−9. substitute in either of the original points.4) (−9. Try it! Theory: a gradient can be found using the 𝑟𝑖𝑠𝑒/𝑟𝑢𝑛 formula.2) is used below. which is 7. 𝑦2 .7) will give the same answer. Use the 𝑟𝑖𝑠𝑒/𝑟𝑢𝑛 formula to find the gradient: 𝑟𝑖𝑠𝑒 4 = =2 𝑟𝑢𝑛 2 Rewrite the general form of a line: 𝑚 = 𝑦 = 𝑚𝑥 + 𝑐 Then replace the gradient with the value found: 𝑦 = 2𝑥 + 𝑐 To find 𝑐. substitute 1 for 𝑥 and 3 for 𝑦 in the general form of the line: 3 = 2(1) + 𝑐 3 = 2 + 𝑐 1 = 𝑐 So the equation of the line is: 𝑦 = 2𝑥 + 1 Using the other point (3. Substitute this gradient into the general form: where the subscripts refer to the two points (𝑥1 . 2) Plan: use the 𝑟𝑖𝑠𝑒/𝑟𝑢𝑛 formula to find the gradient. then substitute that gradient and one of the points into 𝑦 = 𝑚𝑥 + 𝑐 to solve for 𝑐. as in this case.3). how much does the vertical line “rise”? The answer is four units on the 𝑦 − 𝑎𝑥𝑖𝑠. Substitute 𝑥 = −9. Solution: approximately plot the points: 37 . we ask. Now. but the other point (−16. Example 1: find the equation of the line going through the points: (−16. Using the point (1. 𝑦 = 2: 2 −9 + 𝑐 7 18 2= + 𝑐 7 14 18 − = 𝑐 7 7 4 𝑐 = − (leave this as a fraction) 7 2=− So the equation of this line is: 2 4 𝑦 = − 𝑥 − 7 7 You must know how to manipulate fractions to be able to find equations of lines (see Chapter 1). the rise is negative! So: 𝑚 = 𝑟𝑖𝑠𝑒 −2 = (leave it as a fraction) 𝑟𝑢𝑛 7 2 This could also have been done using the mathematical 𝑟𝑖𝑠𝑒/𝑟𝑢𝑛 formula: 𝑚 = 𝑦2 − 𝑦1 4−2 2 = = 𝑥2 − 𝑥1 −16 − −9 −7 2 𝑦 = − 𝑥 + 𝑐 7 Next. which can be written in mathematical terms as: 𝑚 = 𝑦2 − 𝑦1 𝑟𝑖𝑠𝑒 = 𝑥2 − 𝑥1 𝑟𝑢𝑛 7 𝑦 (−16. −3) 𝑒) (−2. At the intersection. as no other point will satisfy this condition. 𝑚 = −3 3 𝑐) −1.1)(3. similarly.7) 𝑑) (1. This example did not require you to sketch the line. this gives the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 at the 2.6 intersecting lines All this mathematical theory is leading to real world applications. as it involved only substitution. −1 𝑚 = −1 3 𝑑) 15. So at the intersection: 𝑦1 = 𝑦2 𝑎𝑛𝑑 𝑥1 = 𝑥2 Theory: when linear equations with different gradients intersect.6). Theory: lines with different gradients will intersect only once.6)(3. having the gradient shown: 𝑎) 2. Set the two 𝑦's equal to each other and solve for 𝑥. This only applies to the intersection. Exercises: 1.13) 𝑏) (1. there is only one set of coordinates that is similar on both lines. Plan: use the general form of a line: 𝑦 = 𝑚𝑥 + 𝑐 Substitute the given facts (𝑚 = −3. Another way of thinking about this is that if you look along the blue line.16) 𝑐) (−1. 38 . Determine the equation of a line passing through one point.10)(2.3)(1.0) 21 63 𝑓) −1. the only point that it touches the red line is at the intersection. Example 1: find the intersection of the following lines 𝑦 = 2𝑥 + 5 𝑦 = −𝑥 − 3 Plan: the only point where the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 of both lines are equal. −7. If you draw two lines on a set of axes.7 𝑚 = 4 2 𝑏) 6. the two lines are labelled with subscripts on the 𝑦 and the 𝑥 to distinguish between the two lines. the only time it touches the blue line is at the intersection.Example 2: find the equation of a line which has a gradient of −3 and passes through the point (−1. the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 of both lines are the same and at the same time the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 of both lines are the same. 15 15 2.8 𝑚 = 7 𝑦 𝑦1 = 5𝑥1 + 1 𝑦2 = −4𝑥2 + 11 𝑥 The two lines above cross only once. and the point (−1. if you look along the red line.6)) and solve for 𝑐. and at the same time. the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 of both lines are equal is at the intersection. On this graph. chances are they will cross. The only point on the graph where the lines do not need to be distinguished is at the intersection.1)(−5. Determine the equations of the lines passing through each of the two points (respectively): 𝑎) (1. Solution: substitute everything into the general form of a line: 6 = −3 −1 + 𝑐 6 = 3 + 𝑐 𝑐 = 3 So the equation of the line is 𝑦 = −3𝑥 + 3. as it is common to both lines. Solution: rearrange the first equation: 4𝑦 − 3𝑥 = 12 4𝑦 = 3𝑥 + 12 3 𝑦 = 𝑥 + 3 4 And the other equation: 𝑦 = −2𝑥 + 4 Set 𝑦1 = 𝑦2 and substitute the equations: 𝑦1 = 𝑦2 3 𝑥 + 3 = −2𝑥 + 4 4 11 𝑥 = 1 4 4 𝑥 = 11 This is the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the intersection. and solve for 𝑥.5𝑥 + 15 3𝑦 = −𝑥 + 15 𝑓) 5𝑦 − 3𝑥 = 31 8𝑦 − 2𝑦 = 16 𝑔) 2𝑦 = −3𝑥 − 19 9𝑦 = −3𝑥 − 12 39 . Using 𝑦1 : 𝑦1 = 2 − 1 𝑦 = − 3 So the intersection is at − 3 . Substitute 𝑥 into one of the equations to find 𝑦. 8 + 5 3 If you are struggling with fractions. The 𝑦 − 𝑣𝑎𝑙𝑢𝑒 comes from one of the original equations (it doesn’t matter which one. then set the 𝑦's equal to each other. as at the intersection. Subscripts were used to distinguish between the two functions. 11 11 . Try using the other equation to see if you get the same solution. Solution: rewrite the above equations with subscripts to distinguish between the two equations: 𝑦1 = 2𝑥1 + 5 𝑦2 = −𝑥2 − 3 Set the two 𝑦’s equal to each other: 𝑦1 = 𝑦2 Then substitute the 𝑦's for their respective equations: 2𝑥1 + 5 = −𝑥2 − 3 At the intersection.intersection. they both have the same 𝑥 − and 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠). go back to Chapter 1. however. this is not necessary. and now for the 𝑦 − 𝑣𝑎𝑙𝑢𝑒: 3 𝑦 = 𝑥 + 3 4 3 4 𝑦 = +3 4 11 36 𝑦 = 11 Thus. Substitute with the 8 1 equations. but 𝑥 = − the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 is still required to get the coordinates.5𝑥 − 7 𝑒) 𝑦 = 0. Determine where the following pairs of equations intersect: 𝑎) 𝑦 = −4𝑥 + 3 𝑦 = −6𝑥 + 7 𝑏) 𝑦 = 3𝑥 + 8 𝑦 = 7𝑥 + 4 𝑐) 𝑦 = −7𝑥 + 12 𝑦 = 2𝑥 + 21 𝑑) 𝑦 = −4𝑥 − 12 𝑦 = −1. Example 2: find the intersection of: 4𝑦 − 3𝑥 − 11 = 1 𝑦 − 4 = −2𝑥 Plan: rearrange into the general form. the 𝑥’s are the same so the subscripts can be removed and 𝑥 solved: 2𝑥 + 5 = −𝑥 − 3 3𝑥 = −8 8 3 This gives the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the intersection. − 3 . the two lines intersect at 4 36 . Substitute this into either of the original equations to give the 𝑦 − 𝑣𝑎𝑙𝑢𝑒. Exercises: 1. shops would want to sell more so supply more. Despite there not being any 𝑥's or 𝑦's. Solution: at equilibrium. making them the supplier. but the quantity of guitars to be made is still unknown. meaning how many guitars you would make at certain prices. most probably. 𝑄𝑠 = 𝑄𝑑 Then substitute the functions of 𝑃: 4𝑃 − 500 = −2𝑃 + 700 Rearrange to find 𝑃: 6𝑃 = 1200 𝑃 = 200 This is the price the guitars should be sold for. there is no excess supply or excess demand (don’t worry too much if you don’t understand excess supply/demand). you would buy less. You demand goods and services. would you buy many? Probably not.) 7𝑦 + 12𝑥 = 72 4𝑦 = −3𝑥 + 36 Intro example: you are thinking of starting a business selling guitars so you do a little research into the market for guitars. the shop supplies them. when the supply and demand curves intersect). the demand curve would be: 𝑄𝑑 = −2𝑃 + 700 Now looking at the supply side of the market. Looking at the demand side of the market. the price of a good (𝑃) is on the 40 . This explains the downward slope of the demand function: 1000 estimate that if you started your business. 𝑃 1000 𝑆 2 𝑄1 𝑄2 set of axes. in 𝑄 economics. you are demanding those clothes. If you are asked to do this. to make a big profit. then the shop wouldn’t make much profit. So at low prices. Would you buy many? Yes. You may also be asked to draw the two lines on a 𝑃 2 𝑄1 𝑄2 𝑄 SUPPLY: when you go to purchase board-shorts. and at higher prices. This quantity comes from either 𝑄𝑠 or 𝑄𝑑 (either one as it is at equilibrium): 𝑄𝑠 𝑃 = 200 = 4 200 − 500 𝑄𝑠 𝑃 = 200 = 300 The interpretation is that 300 guitars should be produced and sold for $200 each.7 microeconomic applications Economic theory: DEMAND: this is what you do as a consumer. simply apply the same theory of lines. At higher prices. you would buy more. as it’s too expensive. If board shorts cost $1000 each. If the price of board-shorts was $1000 (totally ignoring demand) the shop would supply a lot. This is why the supply curve slopes upwards. What about if they cost $2 each. If the price went down to $2. when you go to a surf shop to purchase clothes. For example. set the 𝑄𝑠 = 𝑄𝑑 (similar 𝐷 to 𝑦1 = 𝑦2 in the previous section).e. you 2. the supply function is given by: 𝑄𝑠 = 4𝑃 − 500 You would like to know the equilibrium price and quantity (i. so they wouldn’t supply very many. Solution: set 𝑄𝑑 = 𝑄𝑠 1 −0.5𝑃 + 25 = 𝑃 − 2 4 Rearrange to isolate 𝑃: 1 27 = 𝑃 + 0.5𝑃 4 Use fraction theory from Chapter 1 to find a Common Denominator: 𝑃 2𝑃 + 4 4 3𝑃 27 = 4 27 = 108 = 3𝑃 𝑃 = 108 = 36 3 41 . remembering that 𝑃 is on the 𝑦 − 𝑎𝑥𝑖𝑠 and 𝑄 on the 𝑥 − 𝑎𝑥𝑖𝑠. This is a standard which is used by economists. in economics.5𝑄𝑑 + 350 0. You cannot have a negative quantity.125) (700. To find the 𝑃 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. For the supply equation. or a negative price (hence the dotted extensions).5𝑃 + 25 1 𝑄𝑠 = 𝑃 − 2 4 Then sketch the two lines. Example 1: find the equilibrium quantity and price for the demand and supply equations 𝑄𝑑 = −0. set 𝑃 = 0: 0 = −0. Plan: set 𝑄𝑑 = 𝑄𝑠 then solve for 𝑄 and 𝑃. For the demand equation: 𝑄𝑑 = −2𝑃 + 700 𝑄𝑑 − 700 = −2𝑃 𝑄𝑑 − 700 = 𝑃 −2 𝑃 = −0.0) 𝐷 𝑄 Theory: demand and supply can only exist when quantity and price are positive or zero. and draw in the equilibrium. For the demand equation: 1. 2. Then plot both demand and supply on the same axes: (−500.350 . To find the 𝑄 intercept.25𝑄𝑠 + 125 Plot both equations individually using the method earlier in the chapter. and the quantity of that good (𝑄) is on the 𝑥 − 𝑎𝑥𝑖𝑠.0) (0.𝑦 − 𝑎𝑥𝑖𝑠.0). Also. you should get 𝑄𝑠 = −500 and 𝑃 = 125. Sketch each of the two equations by finding the four axis intercepts. 𝑃 is on the 𝑦 − 𝑎𝑥𝑖𝑠 and 𝑄 is on the 𝑥 − 𝑎𝑥𝑖𝑠.5 0 + 350 𝑃 = 350 Giving a coordinate of 0.5𝑄𝑑 = 350 𝑄𝑑 = 700 Giving a coordinate of (700.350) 𝑃 𝑆 (0. you will have to rearrange both the supply and demand equation to have the price (𝑃) by itself. set 𝑄 = 0: 𝑃 = −0. so to be consistent. Make sure you get this.5𝑄𝑑 + 350 For the supply equation: 𝑄𝑠 = 4𝑃 − 500 𝑄𝑠 + 500 = 4𝑃 𝑃 = 0. But if the original price was $50. the percentage change in price is 𝐷 −6%. 𝑎) 𝑃𝑠 = 0. and in the second instance. To find the 𝑄 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. If the original price of the car was $6.000.5𝑄𝑠 + 26 𝑃𝑑 = −2𝑄𝑑 + 158 𝑐) 𝑃𝑠 = 1.8𝑄𝑠 + 15 𝑃𝑑 = −1. In the first instance.2𝑄𝑠 + 8 𝑑) 𝑃𝑠 = 0. right? Well. Let’s say that you own a business selling cars.5𝑃 = 25 𝑃 = 50 This gives coordinates of (50.0) 25. and 𝑄 is on the horizontal axis). set 𝑃 = 0: 𝑄𝑑 = −0.3𝑄𝑠 + 24 𝑃𝑑 = −1.000 then the reduction is not very big at all.3𝑄𝑑 = 182 𝑑) 𝑃𝑠 = 1. find the two intercepts (remember that 𝑃 is on the vertical axis.50 (0.7𝑄𝑠 + 𝑃𝑠 = 6 7𝑄𝑑 + 𝑃𝑑 = 61 2. set 𝑄 = 0: 𝑄𝑑 = −0. then yes.7𝑄𝑠 + 21 𝑃𝑑 + 2. That’s a large reduction. Find the equilibrium price and quantity given the following demand and supply equations: 𝑎) 𝑃𝑑 = −7𝑄𝑑 + 50 𝑃𝑠 = 0.0 42 .5𝑃 + 25 0 = −0.9𝑄𝑠 + 12 𝑃𝑑 + 0. so you decide to reduce the price by $3.9𝑄𝑠 + 𝑃𝑠 = 4 7𝑄𝑑 = −𝑃𝑑 + 61 2.0). Demand is used below: 𝑄𝑑 = −0.2𝑄𝑑 + 115 𝑒) −0. we don’t really know. set 𝑃 = 0: 𝑄𝑠 = 1 0 −2 4 Exercises: 1.5𝑃 + 25 𝑄𝑑 = 25 Supply: to find the 𝑃 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. To sketch the lines. Find the equilibrium price and quantity given the following demand and supply equations.5𝑄𝑑 + 148 𝑒) −0.8 elasticity The concept of elasticity is best explained with intuition. You have a car that is not selling well.9𝑄𝑑 = 117 𝑏) 𝑃𝑠 = 2. 𝑄 Theory: percentage changes are found like any other percentage is found: 𝑄𝑠 = −2 Plot demand and supply separately. Demand: to find the 𝑃 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. use either demand or supply (it does not matter). What we really want to know it the percentage 𝑆 change in price.8) (−2. the reduction is huge.36 because the price has decreased). set 𝑄 = 0: 1 𝑄𝑠 = 𝑃 − 2 4 1 0 = 𝑃 − 2 4 0. 𝑃 = 36.5𝑃 + 25 0.5 36 + 25 𝑄𝑑 = −18 + 25 = 7 The equilibrium is 𝑄 = 7.25𝑃 = 2 𝑃 = 8 To find the 𝑄 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡.To find the quantity.25𝑃𝑑 + 23 𝑃𝑠 = 0.000. the percentage change is price is −50% (it’s negative 7. then plot the functions. then join the respective dots: 𝑃 0.5𝑄𝑠 = 5 𝑃𝑑 77 𝑏) 𝑄𝑑 = + −9 9 𝑄𝑠 = 4𝑃𝑠 − 12 𝑐) 𝑄𝑑 − 0. Theory: elasticity of demand is the responsiveness of quantity demanded to a change in price.000 The form of elasticity used most often is: 𝜀𝑑 = 𝛥𝑄 𝑃 × 𝛥𝑃 𝑄 Back to the elasticity concept: will you sell more cars if you reduce the price by 6% or 50%? You probably said 50%. Now. If there is a $3.500) = 5 So 6 boats are sold initially.000 each. Yes that would be sort–of correct: you would sell more cars.25 . would the number of cartons of beer sold increase a little or increase a lot? This is what elasticity determines.500) + 10 𝑄𝑑 (7.%∆𝑃 = ∆𝑃 𝑐𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑖𝑐𝑒 × 100 = × 100 𝑃 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑝𝑟𝑖𝑐𝑒 2. Theory: the elasticity formula is: 𝜀𝑑 = 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒 𝑖𝑛 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 %𝛥𝑄 = 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑖𝑐𝑒 %𝛥𝑃 ∆𝑃 𝑃 This last equation is simpler than you think. and sell 4 cars when you decrease price by 50%. Substituting this into the demand equation: 𝑄𝑑 (12. but then decides to increase the price by 25%. And if the original price 𝑃 = 50. Intro example 1: a business selling motorboats to James Bond decides to hire an economist to estimate the demand curve. Another way of thinking about this is: how sensitive is the quantity demanded to changes in price. what is the change in 𝑄𝑑 ? ∆𝑄𝑑 = 5 − 6 = −1 What is the percentage change in 𝑄𝑑 ? %𝛥𝑄 = 𝛥𝑄 −1 × 100 = × 100 𝑄 6 𝛥𝑄 The %𝛥𝑄 and %𝛥𝑃 can be replaced with their respective definitions (%∆𝑃 = %∆𝑄 = ∆𝑄 𝑄 × 100 and × 100): 𝛥𝑄 𝛥𝑄 𝑃 𝑄 × 100 𝜀𝑑 = = × 𝛥𝑃 𝛥𝑃 𝑄 𝑃 × 100 NOTE: 1. respectively. The second part is a point along the line. the 100 cancels off 100 %𝛥𝑄 ≈ −16. The 𝛥𝑄 and 𝛥𝑃 mean change in 𝑄 and change in 𝑃. if the shop reduced the price of beer by $1. The economist comes back and gives the following function before he is shot: 𝑄𝑑 = −0.67 ≈ ≈ −0.67 %𝛥𝑃 25 43 . the business sells 6 boats per year at $10.0004𝑃 + 10 Initially. The question is. what is a better option? (You don’t need to answer this.500 = 10. and this is reduced to 5 after the 25% price increase.000 decrease. but by what percentage will your car sales increase? If you sell 3 cars when you decrease price by 6%. An intuitive example is.0004(12. The first part 𝛥𝑃 is simply the slope of the line (𝑚) when you have the equation in the general form (with 𝑄 by itself).000 × 1. It uses percentages to get rid of the error we encountered in the car example before.000. then: %∆𝑃 = −3000 × 100 = −6% 50. the 𝛥𝑄 and 𝛥𝑃 are symbols which are different from 𝑄 and 𝑃. how responsive are boat sales to changes in price? Solution: a 25% price increase means the new price will be $12. the ∆𝑃 = −3000.67% Putting all this information into the original elasticity equation: 𝜀𝑑 = %𝛥𝑄 −16. just think about the concept).500) = −0. 000 and original quantity sold is 6.0004 × 10.11 9 The elasticity of supply is approximately 1. Example 1: a producer of laptops has a supply curve approximated by 𝑄𝑠 = 4𝑃 − 400 The price the laptops are sold to a retailer is $1000 each.0004. however 𝑄 is still required.5𝑄𝑑 determine the elasticity at 𝑄𝑑 = 20 and also determine by what percentage and in what direction quantity will change when price is increased by 10%.0004𝑃 + 10 Is the 𝑄 by itself? Yes. How sensitive is the quantity supplied to changes in price? Plan: use the elasticity formula. then the quantity demanded will move in the opposite direction. Substitute this into the elasticity equation: 𝜀𝑑 = −0. Using the easier form of elasticity: 𝜀𝑑 = 𝛥𝑄 𝑃 × 𝛥𝑃 𝑄 𝑄𝑠 (1000) = 4(1000) − 400 = 3600 Substitute 𝑄 = 3600. Exercises: 1. Determine the elasticity of demand for the given information: 𝑎) 𝑃𝑑 = −0. the elasticity of demand will always be negative.Now the much easier way of doing this question. when 𝑄 is by itself is 4. Substitute this into the elasticity formula: 𝜀𝑠 = 4 × 𝑃 𝑄 2. The elasticity of supply will always be positive. The original price is $10.15𝑃𝑑 + 103 𝑎𝑡 𝑄 = 27 2.5𝑃𝑑 + 75 𝑎𝑡 𝑄 = 31 𝑐) 𝑄𝑑 = −4.2𝑃𝑠 − 37. which is −0. the meaning of this number will be discussed.4𝑃𝑠 − 44 𝑎𝑡 𝑄 = 17 𝑐) 𝑄𝑠 = 1. so remembering that the slope of the demand function.0004 × 𝑃 𝛥𝑄 𝛥𝑃 is = 10 ≈ 1. Plus this is a more precise answer. Plan: find the elasticity using 𝜀𝑑 = 𝛥𝑄 𝑃 × 𝛥𝑃 𝑄 The price of the laptops is $1000.5 𝑎𝑡 𝑄 = 21 𝑃 𝑄 The 𝑄 needs to be found and it is simply the original point.000 2 = − ≈ −0.11. The positive value of the elasticity of supply means that as prices change. Also. Determine the elasticity of supply for the given information: 𝑎) 𝑃𝑠 = 3 + 𝑄𝑠 𝑎𝑡 𝑃 = 15 𝑏) 𝑄𝑠 = 0.9 interpreting elasticity Theory: the negative sign of the elasticity of demand means that as prices changes in one direction (either a positive or negative change). so substitute these in: 𝜀𝑑 = −0. it is found using the supply curve: 44 . 𝑃 = 1000 into the elasticity formula: 𝜀𝑠 = 4 × 1000 3600 Look at the original demand equation: 𝑄𝑑 = −0.67 6 3 The same answer as before.5𝑄𝑑 + 50 𝑎𝑡 𝑃 = 20 𝑏) 𝑄𝑑 = −3. Example 1: given the demand equation 𝑃𝑑 = 100 − 0. the quantity will change in the same direction as the price changes. In the next section. but now supply instead of demand: 𝛥𝑄 𝑃 𝜀𝑠 = × 𝛥𝑃 𝑄 Solution: the gradient 𝑚 of the supply curve. leave it as a positive number). the percentage change in sales would not be as large (i. the elasticity of supply 𝜀𝑠 was approximately 1.67). the quantity will change by that same percentage. Specifically. and if it is positive. only 0. the quantity will change by a smaller percentage. if price of the motorboats changed by a certain percentage. if price changed by a certain percentage.11 times larger percentage change.e. if the price of a good is changed by a certain percentage. then apply the following rules: ε < 1 : INELASTIC – in this case. make it positive.5 20 = 90 𝜀𝑑 = −2 90 = −9 20 Now substitute the 10% increase in price and the elasticity into the other elasticity equation: %∆𝑄 −9 = 10 %∆𝑄 = −90 This means that as price is increased by 10%. That is.67 < 1. it means that this good is inelastic to price changes. take the absolute value of the elasticity (if it is negative.67. giving 0. Substitute this into the elasticity formula: 𝜀𝑑 = − 1 𝑃 × 40 𝑄 45 . the quantity demanded will decrease (i. Since 0. ε = 1 : UNITARY ELASTIC – if the price of a good is changed by a certain percentage.Then use the other definition of elasticity to determine the %∆𝑄 ε > 1 : ELASTIC – if the price of a good is changed by a certain percentage. and the demand function to find the elasticity. Theory: to interpret elasticity. The interpretation of the actual number of an elasticity tells you the degree to which quantity changes when price is changed.e. the producer would respond with a 1. move in the opposite direction due to the negative sign) by 90%. Solution: 𝑄 needs to be isolated in the demand equation so the gradient can be determined: 𝑃 + 400 = −40𝑄𝑑 𝑃 + 400 = 𝑄𝑑 −40 1 𝑄𝑑 = − 𝑃 + 100 40 The gradient observed is −1/40.11.67 as large) In the laptop example. 𝑃𝑑 is found using the demand equation: 𝑃𝑑 𝑄 = 20 = 100 − 0. Find and interpret the elasticity of demand for cigarettes.e. Example 2: the demand function for cigarettes is (with quantity in millions): 𝑃𝑑 = −40𝑄𝑑 − 400 and the price of cigarettes is $80/carton. which means that the supply of laptops is elastic. the quantity will change by a larger percentage. take the absolute value of this number (i. so to interpret this number. Plan: use the elasticity equation 𝜀𝑑 = 𝛥𝑄 𝑃 × 𝛥𝑃 𝑄 𝜀𝑑 = Solution: isolate 𝑄 in the demand equation: 𝑄 = 200 − 2𝑃 The gradient is −2 so substitute this into the first elasticity equation: 𝜀𝑑 = −2 × 𝑃 𝑄 %𝛥𝑄 %𝛥𝑃 Since 𝑄 = 20. The James Bond question from the previous section had an elasticity of demand 𝜀𝑑 ≈ −0. Now.2. 46 .19 𝑏) 𝜀𝑑 = −2. Use the original definition of elasticity to work out by what percentage quantity will change: chapter two summary Linear equations have the general form: 𝑦 = 𝑚𝑥 + 𝑐 𝑟𝑖𝑠𝑒 𝑦2 − 𝑦1 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝑚 = = 𝑟𝑢𝑛 𝑥2 − 𝑥1 The gradient of a line is the same at all points along it. 1. and determine the effect on quantity of a 10% price decrease.Given 𝑃 = 80. This means that if price changes by a certain percentage.7𝑃𝑠 − 104 𝑎𝑡 𝑃 = 20 2. using the original definition of elasticity. smoking is addictive so when price increases.2 = %𝛥𝑄 15% Rearrange to isolate %𝛥𝑄: −1. Exercises: 1. Example 3: a company selling beer conducted research and found that elasticity of demand was −1. people still need to smoke. quantity demanded will change by a larger amount.7𝑄𝑑 𝑎𝑡 𝑄 = 40 𝑐) 𝑄𝑑 = −2. Interpret this number. Interpret the following elasticities. 2. and since 1.02 which means it is (highly) inelastic. the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. To be able to draw a line. replace all values of 𝑥 with zero. but in the opposite direction (as it is the elasticity of demand). then solve for 𝑥. you need at least two points (sets of coordinates).0 𝑒) 𝜀𝑑 = 0.02 of the change in price). For any price changes. the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. In reality. 𝜀𝑑 = %𝛥𝑄 %𝛥𝑃 Solution: the absolute value of 𝜀𝑑 is 1. then beer is an elastic good.02 3920 49 𝜀𝑑 = − Taking the absolute value.97 Coordinates are written as (𝑥. 𝑎) 𝜀𝑠 = 1. 𝑦). to get 𝑄 substitute this into the rearranged demand function: 1 𝑄𝑑 (80) = − (80) + 100 40 𝑄𝑑 80 = 98 So 98million cartons are sold.3 𝑐) 𝜀𝑠 = 0. replace all values of 𝑦 with zero. but maybe won’t buy as many cartons. substitute into the original definition: −1. steeper lines have larger (absolute) gradients (𝑚 values). Determine and interpret the elasticity for the following functions: 𝑎) 𝑄𝑠 = 1. this is the case.7𝑃𝑑 + 111 𝑎𝑡 𝑄 = 100 𝑑) 𝑃𝑠 = 50 + 0. the 𝑐 value shifts the line up or down. |𝜀𝑑 | = 0. lines cutting the 𝑦 − 𝑎𝑥𝑖𝑠 higher up have larger 𝑐 value. A highly inelastic product like cigarettes means prices can be changed by a large percentage without the loss of many sales. then solve for 𝑦.2. The elasticity is: 1 80 × 40 98 80 1 𝜀𝑑 = − =− ≈ −0. If the company increased price by 15%.6𝑃𝑠 − 27 𝑎𝑡 𝑄 = 27 𝑏) 𝑃𝑑 = 49 − 0. and the fact that 𝜀𝑑 = −1.1𝑄𝑠 𝑎𝑡 𝑄 = 50 𝑒) 𝑄𝑠 = 1. how much would quantity demanded change? Plan: interpret the elasticity from the theory.43 𝑑) 𝜀𝑑 = 1.2 × 15% = %𝛥𝑄 %∆𝑄 = −18% So the percentage change in 𝑄 will be larger than the percentage change in 𝑃. 2. Linear equations with different gradient intersect only once and there is only one set of coordinates that is similar on both lines. To find: 1.2 and %𝛥𝑃 = 15%. the change in quantity will be significantly less (only 0.2 > 1. The negative sign of the elasticity of demand means that as prices changes in one direction (either a positive or negative change).0 𝑎𝑛𝑑 3.1 𝑚 = 2. You demand goods and services. %𝛥𝑄 𝛥𝑄 𝑃 𝜀𝑑 = = × %𝛥𝑃 𝛥𝑃 𝑄 The 𝛥𝑄 and 𝛥𝑃 are symbols which are different from 𝑄 and 𝑃. −9 5.0 𝑥 2.4 𝑑) −1. −1 𝑒) −5.4. b) The order of the steepness (the absolute value of the gradients) from shallowest to steepest.3 𝑚 = 2 𝑏) −1. respectively. −2 𝑎𝑛𝑑 3. Demand and supply can only exist when quantity and price are non-negative.3 𝑎𝑛𝑑 3.DEMAND: this is what you do as a consumer. The positive value of the elasticity of supply means that as prices change. A demand curve slopes downwards. Rearrange the following linear equations into the general form: 𝑎) 𝑦 − 4 = 𝑥 − 5 𝑏) 2𝑦 + 𝑥 = 3 𝑐) 2𝑦 − 𝑥 = 𝑥 + 3 − 1 𝑑) 2𝑥 − 𝑦 = 4 2𝑦 2𝑦 + 1 𝑒) 14𝑥 − = 𝑦 − 3 𝑓) = 𝑥 5 4 𝑦 𝑥 𝑦 − 𝑥 𝑔) + =4 ) =4 −5 3 𝑥 − 3 3𝑦 − 4𝑥 𝑖) = 14 − 5 3 − 𝑥 For the following graph: 𝑦 5. determine the equation of the line: 𝑎) chapter two questions 1. −6 𝑚 = −2 𝑒) 1. −6) (−2. c) The order of the values of 𝑐 from lowest to highest. −3. then apply the following rules: ε < 1 : INELASTIC ε = 1 : UNITARY ELASTIC ε > 1 : ELASTIC 4. To interpret elasticity: take the absolute value of the elasticity. Percentage changes are found: ∆𝑃 𝑐𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑖𝑐𝑒 %∆𝑃 = × 100 = × 100 𝑃 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑝𝑟𝑖𝑐𝑒 Elasticity of demand is the responsiveness of quantity demanded to a change in price.16 𝑐) 0. then plot them: 𝑎) 𝑦 = 4𝑥 + 1 𝑏) 2𝑦 = 6𝑥 − 4 𝑐) 3𝑦 − 1 = 4𝑥 + 1 𝑦 − 𝑥 𝑑) =2 3 3 𝑒) 2𝑥 − 3𝑦 = 2 𝑓) 12𝑦 − 18𝑥 − 16 = 4 + 2𝑥 𝑦 + 3𝑥 𝑔) =4 2𝑦 − 3 ) 𝑦 = 4 𝑦 𝑥 2. −1 𝑚 = 3 𝑐) 7. A supply curve slopes upwards.7 𝑎𝑛𝑑 7. SUPPLY: this is what a firm does.15 𝑏) 2.1 𝑎𝑛𝑑 5.3) 𝐴 𝑥 Determine: a) The order of the gradients 𝑚 from lowest to highest. For each of the following graphs. 6. Determine the gradients of each of the following linear functions. 3.7 𝑚 = 1 𝑑) −3. with the corresponding gradient: 𝑎) 2. −12 Determine the equation of the line passing through the point indicated.5 47 . Determine the equation of the line passing through each of the two points: 𝑎) 1. the quantity will change in the same direction as the price change. ? ? 𝐷 ?? 𝐹 𝑦 𝐶 𝐵 𝑏) (0. The 𝛥𝑄 and 𝛥𝑃 mean change in 𝑄 and change in 𝑃. then the quantity demanded will move in the opposite direction. 5𝑥 = 44. c) The elasticity of supply at the equilibrium. and its meaning.6𝑄𝑑 + 89 Determine the elasticity at 𝑄 = 45. 9.48 1 𝑃𝑠 = − 𝑄𝑠 + 21. Given the demand equation: 𝑃𝑑 = −0. and its meaning.75𝑄𝑠 + 4. 10.06 And its supply curve is estimated to be: 𝑃𝑠 = 3.5 Determine the equilibrium price and quantity.5𝑄𝑑 + 89 𝑃𝑠 = 2𝑄𝑠 + 44 Plot the two functions.1𝑄𝑠 + 15. c) The elasticity of supply at the equilibrium. Determine the coordinates where the following lines intersect: 𝑎) 𝑦 = 3𝑥 + 9 𝑦 = 2𝑥 + 3 𝑏) 𝑦 = 4𝑥 − 9 𝑦 = 3𝑥 − 3 𝑐) 𝑦 = −5𝑥 + 7 𝑦 = 2𝑥 + 14 𝑑) 2𝑦 + 3𝑥 − 26 = 0 5𝑦 + 4𝑥 = −51 𝑒) 𝑦 + 4. 13.5𝑥 + 2𝑦 = 58 8.7𝑄𝑑 + 26.4𝑄𝑑 + 24 And the supply equation: 𝑃𝑠 = 3. Given the following supply and demand functions. A car manufacturer estimates its demand curve to be (in hundreds of thousands): 𝑃𝑑 = −0. determine the equilibrium quantity and price. The market for steel has an estimated demand and supply curve (in millions of tonnes) to be: 𝑃𝑑 = −1. Determine the elasticity of supply at 𝑄 = 12 for the supply equation: 𝑃𝑠 = 1. For the demand equation: 𝑃𝑑 = −0.2𝑄𝑑 + 23. and its meaning.8 12. and plot the market on a set of axes.6 3 Determine: a) The equilibrium quantity and price. 𝑃𝑑 = −0.75 5.58 Determine: a) The equilibrium quantity and price. 48 . b) The elasticity of demand at the equilibrium.7.1𝑄𝑠 + 6. and interpret its meaning. and include the axis intercepts and the equilibrium. b) The elasticity of demand at the equilibrium. and its meaning. 11. 1 3.6 3.3 3.Chapter 3 Simultaneous Equations and Matrices Solving the intersection of lines using different methods 3.9 3.4 3.10 3.5 3.7 3.11 Simultaneous Equations Two Simultaneous Equations Three Simultaneous Equations The Matrix Solving Two Equation Matrices Solving Three Equation Matrices Notes on Solutions to Matrices Applications The Determinant of a 2 × 2 Matrix The Determinant of a 3 × 3 Matrix Using the Jacobian Determinant 50 50 52 54 55 58 61 62 65 66 68 70 71 Chapter Three Summary Chapter Three Questions 49 .8 3.2 3. 3. substitute the solutions of 𝑥 back into either of the original equations to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the solution. there needs to be at least as many equations as there are variables (this is not the only condition). 2. the intersection of two lines was found mathematically. In the following case. so there will be no unique solution. and substitute for the functions of 𝑦. Note: you could also set the 𝑥’s equal to each in step 1. and in such a case. Essentially. so there is no reason why more variables cannot be introduced. Exercises: 1. and x and y are the independent variables. simultaneous equations were used to do this. This is explained in detail in Section 3. Simultaneous equations are simply a certain number of equations to help solve for a certain number of variables. so there are an infinite number of solutions. then solve for 𝑥. z is the dependent variable. Remember that 𝑥 and 𝑦 are two different variables. Substitute back into one of the 50 . The and are notes to distinguish the two different equations. as according to the theory above. Example 1: solve for 𝑥 and 𝑦 in 𝑦1 = 2𝑥 + 1 𝑦2 = −𝑥 + 4 Here. sometimes a set of lines do not all cross at a single point. 𝑧 = 4𝑥 + 3𝑦 − 1 𝑧 = −2𝑥 + 2𝑦 + 4 Plan: set 𝑦1 = 𝑦2 and substitute the equations. It involves finding the common intersection of multiple lines. substitute the functions of 𝑥 in and solve for 𝑥. Can the following equations be solved? 𝑧 = 4𝑥 + 3𝑦 − 1 𝑧 = −2𝑥 + 2𝑦 + 4 The answer is no. there are more variables (3) than there are equations (2). Given the two equations: 𝑦 = 2𝑥 + 5 𝑦 = −3𝑥 + 6 𝑧 = 3𝑥 − 3𝑦 − 3 Thus a solution is possible (Note: see Section 3. set the 𝑦’s equal to each other. Similarly. and ) and three variables (𝑥. 3. Theory: simultaneous equations are used to solve for the variables when there are two or more equations.2 two simultaneous equations Theory: to solve simultaneous equations: 1. Theory: to be able to solve simultaneous equations. there is no unique solution. This will give the same solution. 𝑦 and 𝑧). sometimes sets of equations cross at an infinite number of points.7). Determine if the following sets of equations can be solved: 𝑎) 𝑦 = 2𝑥 + 3 𝑦 = −3𝑥 − 14 𝑏) 𝑦 = 𝑥 + 𝑧 − 3 𝑦 = −2 − 𝑥 − 𝑧 𝑦 − 𝑧 = 15 𝑐) 𝑦 = 𝑧 + 𝑥 − 1 𝑦 = 12𝑥 − 3 𝑦 = 𝑧 + 𝑥 + 𝑎 − 3 Simultaneous equations can be used to solve for 𝑥 and 𝑦.7. One such equation could be: 𝑧 = 4𝑥 + 3𝑦 − 1 3.1 simultaneous equations In Chapter 2. However. there are three equations (. Now. and try to isolate 𝑥 in . Substitute this solution into any of the modified or original equations to solve for the other variable. Using equation : 𝑦 = − 1 + 4 =3 So the solution to the two equation is 𝑥 = 1. Also. ’ is easiest as 𝑦 is already isolated: 𝑦 = −𝑥 + 4 𝑦 = − 3 + 4 𝑦 = 1 Thus the solution is 𝑥 = 3. or in coordinates (3. Go back to the beginning of this example. 4. even if you choose the harder one. It is also a good idea to substitute the solution you get back into both original equations. Exercises: 1. In the following. ’ → (this means the modified equation 1 (labelled ’) is substituted into equation ) 5 4 − 𝑥 − 3𝑥 + 4 = 0 Use the crab-claw from Chapter 1: 20 − 5𝑥 − 3𝑥 + 4 = 0 24 − 8𝑥 = 0 𝑥 = 3 To find the value of 𝑦. choose the variable easiest to isolate. ’ or .original equations to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 at the intersection.3). a 𝑦 − 𝑣𝑎𝑙𝑢𝑒 is still needed.1). Example 2: solve for 𝑥 and 𝑦 in 2𝑦 + 2𝑥 − 1 = 7 5𝑦 − 3𝑥 + 4 = 0 compared to . 𝑦 = 3 or in coordinate form: (1. All these should give the same solution. Theory: another method of solving simultaneous equations is: 1. Substitute (4 − 𝑥) in for 𝑦 in the other equation. try to isolate 𝑥 or 𝑦 in . substitute the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 back into either of the original equations (both will give the same answer). isolate one variable in one equation 2. Solution: 𝑦1 = 𝑦2 2𝑥 + 1 = −𝑥 + 4 3𝑥 = 3 𝑥 = 1 So 𝑥 = 1. substitute this solution back into either of the original equations to solve for the other variable. the answer will still be the same. Try substituting the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 into equation to see if the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 is the same (it should be). substitute it into the other equation. Solution: choose as the numbers in front of the variables are even and look easy to solve 51 . Solve the following sets of equations 𝑎) 7𝑦 = −5𝑥 + 9 𝐴𝑁𝐷 5𝑦 = −4𝑥 + 15 Plan: isolate one variable in one of the equations. to double check that you have the correct solution. 𝑦 is isolated however 𝑥 is just as easy. 𝑦 = 1. and substitute it into the other equation to solve for the other variable. However. but for a coordinate. 2𝑦 + 2𝑥 − 1 = 7 2𝑦 + 2𝑥 = 8 2𝑦 = 8 − 2𝑥 𝑦 = 4 − 𝑥 ’ This rearranged version is labelled ’. To find this 𝑦 − 𝑣𝑎𝑙𝑢𝑒. substitute 𝑥 = 3 into either . solve for the other variable. The hardest part using of this method is knowing which equation is easier to use and this will take practice. 3. substitute one of those equations into the other to solve for one of the unknowns. is no longer used.3 . 𝑦 and 𝑧 in 𝑧 − 3𝑥 + 8 = 4𝑦 4𝑧 = −𝑥 + 7𝑦 − 1 4𝑥 − 2𝑦 = 3𝑧 − 9 The two modified equations ’ and ’ are now two simultaneous equations with two variables (like in the last section). and substitute it into the other two equations (below.𝑏) 𝑐) 𝑑) 𝑒) 2𝑦 = −3𝑥 + 8 5𝑥 + 9𝑦 = 15 −6𝑥 + 7𝑦 = 15 2𝑥 = −7𝑦 + 13 𝐴𝑁𝐷 𝐴𝑁𝐷 𝐴𝑁𝐷 𝐴𝑁𝐷 4𝑥 + 𝑦 = 9 6𝑦 + 4𝑥 = 15 6𝑦 = 3𝑥 + 12 − 4𝑥 − 5𝑦 − 17 = 0 This is the solution to one variable. or in three dimensional coordinates 𝑥. 3. 4. This will make two new equations with only two variables. use these two solutions to solve for the third variable. Work backwards to solve for the other variables. . use either ’ or ’’ (’’ is easier as 𝑥 is already isolated): 𝑥 = −2𝑦 − 2 ’’ 3. lastly. then use those two modified equation to solve for one of the variables. 𝑧 = 0. 𝑦.e. 𝑦. Example 1: Solve for 𝑥. Solution: To solve for three equations. 𝑦 and 𝑧 in 𝑧 = 2𝑥 + 3𝑦 + 6 𝑧 = 𝑥 + 𝑦 + 4 𝑧 = −2𝑥 + 5𝑦 + 8 𝑥 = −2(−1) − 2 𝑥 = 0 With two variables solved. first. use this solution to solve for one of the other variables. the last variable can now be solved (i. Substitute this value of 𝑦 into an equation with only 𝑥 and 𝑦. Theory: to solve three simultaneous equations: 1. is put into both and ): →: 𝑧1 = 𝑧2 2𝑥 + 3𝑦 + 6 = 𝑥 + 𝑦 + 4 𝑥 + 2𝑦 = −2 →: 𝑧1 = 𝑧3 2𝑥 + 3𝑦 + 6 = −2𝑥 + 5𝑦 + 8 4𝑥 − 2𝑦 = 2 ’ ’ 2. 𝑧) using : 𝑧 = 2𝑥 + 3𝑦 + 6 𝑧 = 2(0) + 3(−1) + 6 𝑧 = 3 So the solution is 𝑥 = 0. Isolate 𝑥 in ’: 𝑥 + 2𝑦 = −2 𝑥 = −2𝑦 − 2 Substitute ’’→’: 4 −2𝑦 − 2 − 2𝑦 = 2 −8𝑦 − 8 − 2𝑦 = 2 −10𝑦 = 10 𝑦 = −1 52 ’ ’’ Plan: substitute one equation into the other two. 𝑧 = 3. take one of the equations. take one equation and substitute it into the other two equations. −1. 𝑦 = −1. Plan: use the substitution method to solve one variable at a time. 𝑧 . Example 2: Solve for 𝑥. Note: the hardest part is organising and manipulating the three equations.3 three simultaneous equations A similar theory applies to three equation systems. The solution is also written in coordinate form 𝑥. A simple example will show you the technique before the theory is introduced. this equation is labelled as ’ as it has been modified. so use either. Working with only ’ and ’ isolate any variable in either ’ or ’. Equation ’ has been used below: −5𝑥 − 14𝑦 = −33 −5𝑥 = −33 + 14𝑦 −33 + 14𝑦 −5 33 14 𝑥 = − 𝑦 5 5 𝑥 = ’ ’’ Again. use any of the equations with all three variables to solve for 𝑧. ’’→’: 33 14 9𝑦 + 13( − 𝑦) = 31 5 5 429 182 9𝑦 + − 𝑦 = 31 5 5 137 274 − 𝑦 = − 5 5 274 − 5 =2 𝑦 = 137 − 5 53 .3 . Substitute 𝑦 = 2 into either of the equations with only two variables (i. 𝑧 = 3. 𝑦 = 2→’ 𝑧 = 4𝑦 + 3𝑥 − 8 𝑧 = 4(2) + 3(1) − 8 𝑧 = 3 The solution is 𝑥 = 1.2. ’ ). Solutions will not always be nice round numbers. Substitute 𝑦 = 2 →’’: 33 14 − 2 5 5 33 28 𝑥 = − 5 5 5 𝑥 = = 1 5 𝑥 = Finally. ’→: 4𝑥 − 2𝑦 = 3(4𝑦 + 3𝑥 − 8) − 9 4𝑥 − 2𝑦 = 12𝑦 + 9𝑥 − 24 − 9 −5𝑥 − 14𝑦 = −33 Again. This equation (’’) is substituted into the other equation (i. ’’ or ’).Solution: 𝑧 in looks easiest to isolate 𝑧 − 3𝑥 + 8 = 4𝑦 𝑧 = 4𝑦 + 3𝑥 − 8 ’ This solution for 𝑦 is used to find the other two unknowns.e. 𝑦 = 2. Then put ’ into both and : ’→: 4(4𝑦 + 3𝑥 − 8) = −𝑥 + 7𝑦 − 1 16𝑦 + 12𝑥 − 32 = −𝑥 + 7𝑦 − 1 9𝑦 + 13𝑥 = 31 ’ Again.e. Exercises: 1. so it is a good habit to always work with fractions. as then you will always have an exact answer. The easiest is ’ as 𝑧 is already isolated. Solve the following sets of equations: 𝑎) 𝑧 = 4𝑥 + 𝑦 − 6 2𝑧 = −𝑥 + 2𝑦 + 6 𝑧 = −𝑥 + 2𝑦 + 2 𝑏) 5𝑥 − 𝑦 − 𝑧 = 13 2𝑥 − 2𝑦 + 4𝑧 = 8 𝑥 − 3𝑦 + 2𝑧 = 2 𝑐) −𝑧 + 2𝑦 = −3𝑥 + 12 2𝑥 = 5𝑦 + 2𝑧 + 7 −𝑥 + 2𝑦 − 5𝑧 − 20 = 0 𝑑) 2𝑦 + 5𝑧 = 4 + 2𝑥 4𝑥 + 4𝑦 + 6𝑧 = 10 4𝑦 = 2𝑥 + 2𝑧 + 8 𝑒) 5𝑧 = −7𝑦 − 6𝑥 + 8 5𝑦 = −2𝑥 − 8𝑧 + 12 2𝑥 + 2𝑦 + 2𝑧 = 16 Correctly label this as ’ as it has been slightly modified. or 1. Both are difficult. this equation is labelled appropriately. Substitute 𝑥 = 1. notice the labelling. g. rearrange it: −5𝑥 − 3𝑦 + 𝑧 = −4 Here. and any numbers without variables (constants) on the right. Theory: a matrix is a set of square brackets summarising equations in a simpler form. For the equation 𝑧 − 3𝑦 + 4 = 5𝑥 To get it into the general matrix form. as it is easy to remember.4 the matrix The substitution method for solving equations was quite easy for two equations. To understand what a matrix is and how it is created. Theory: every equation has to be in the general matrix form: 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑 Where 𝑎. The order of the variables is alphabetical order. take your left index finger and put it onto the −2 on the top left of the left matrix. alphabetical). 3. after writing an equals sign. 𝑦 and 𝑧 so that the variables are defined. The next matrix has the 𝑥. Example 2: write the following equations in matrix form −2𝑥 + 3𝑦 + 𝑧 = 9 −4𝑥 − 2𝑦 + 3𝑧 = 4 −𝑥 − 𝑦 + 𝑧 = −2 Solution: −2 3 1 −4 −2 3 −1 −1 1 𝑥 9 𝑦 = 4 𝑧 −2 ’ ’ ’ Plan: rearrange to get variables in alphabetical order on the left. 𝑐 = 1 and 𝑑 = −4.g. and any numbers without variables (i. with the 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 from the equations as another matrix.e. 𝑏 = −3. Example 1: rearrange the following equations into the general matrix form 𝑧 = 2𝑥 − 3𝑦 + 9 3𝑧 − 4 = −4𝑥 + 2𝑦 −𝑦 + 𝑧 = 𝑥 − 2 −𝑥 − 𝑦 + 𝑧 = −2 ’ Rewrite the original equations into a new set: −2𝑥 + 3𝑦 + 𝑧 = 9 −4𝑥 − 2𝑦 + 3𝑧 = 4 −𝑥 − 𝑦 + 𝑧 = −2 ’ ’ ’ It is from this new set that a matrix can be constructed.3. draw a new set of brackets and vertically write the order of the variables you have chosen (e. alphabetical). the constants on the right side are written vertically in square brackets. To understand how this system works. Solution: use alphabetical order : 𝑧 = 2𝑥 − 3𝑦 + 9 −2𝑥 + 3𝑦 + 𝑧 = 9 : 3𝑧 − 4 = 4𝑥 + 2𝑦 −4𝑥 − 2𝑦 + 3𝑧 = 4 : 54 Look at the left matrix. equations need to be in a particular order. Essentially. the middle column is the 𝑦 column and the right column is the 𝑧 column. There is a short–cut method for solving simultaneous equations and it involves using matrices. 𝑏. but was a lot harder for three equations. 2. 𝑐 and 𝑑 are constants. The first column is the 𝑥 column. Then there is the equals sign. solitary constants) on the right side. To construct a matrix: 1. 𝑎 = −5. this form has all the variables on the left in a consistent order (e. then put your right index ’ ’ −𝑦 + 𝑧 = 𝑥 − 2 . take the numbers in front of the variables and put them into square brackets. e. it is −2𝑥). Extract the equations from the following matrices. you will get equation ’. you will find there is an easier way of writing all the information in a matrix: −2 3 1 ⋮ 9 4 −2 3 ⋮ 4 −1 −1 1 ⋮ −2 In this form. the number 4). these equations must be rearranged into the general matrix form: : 2𝑦 = 3𝑥 + 5 −3𝑥 + 2𝑦 = 5 : 0 = −𝑥 + 10 − 𝑦 ’ 55 . Exercises: 1. 𝑦 and 𝑧 in an easier way than the substitution method.finger on the 𝑥 in the middle matrix. the first column is the 𝑥 column. Once you are comfortable with moving between equations and matrices. The 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 are separated from 𝑥. the second column the 𝑦 column and the third column First. which is the top entry of the last matrix. If you go back to the beginning and put your left index finger at the start of the second row (i. Solution: −𝑥 − 𝑦 + 𝑧 = 11 3𝑥 + 7𝑦 + 3𝑧 = 5 −3𝑥 + 2𝑧 = 2 3. and your right hand down one entry. You should have the last entry of 1𝑧 (or just 𝑧). Going across these three numbers gives the left side of ’. 𝑦 and 𝑧 by a dotted line. just like in ’). Now move your left hand across one entry. Example 3: find the original equations for the matrix −1 −1 1 3 7 3 −3 0 2 𝑥 11 𝑦 = 5 𝑧 2 the 𝑧 column.5 solving two equation matrices Having a set of equations in matrix form allows you to solve for 𝑥. and this is equal to 9. Intro example 1: solve the following equations using matrices 2𝑦 = 3𝑥 + 5 0 = −𝑥 + 10 − 𝑦 This is how matrices work. Add these together and make this equal to the entry in the respective row of the right matrix. You should be on 3 with your left hand. −2 1 −3 ⋮ 5 6 7 15 ⋮ −2 𝑎) 𝑐) 5 2 6 ⋮ 14 3 −2 0 ⋮ −3 4 7 −1 ⋮ 11 2 12 2 ⋮ 5 4 8 3⋮ 5 5 8 0 ⋮5 𝑏) 𝑑) −5 0 −4 ⋮ 2 −5 12 3 ⋮ −7 9 −4 0 ⋮ 0 1 2 1 ⋮2 Plan: go across each row of the left matrix and multiply by each entry of the variables column. Put the following sets of equations into a matrix: 𝑎) −3𝑥 + 5𝑦 − 12𝑧 − 12 = 4 5𝑥 + 2𝑦 − 3 = 15 −2𝑥 + 5𝑦 − 3𝑧 = 3𝑧 + 1 𝑏) 12𝑥 − 11𝑦 + 10𝑧 = 9 7𝑧 − 3𝑥 + 2𝑦 = 7𝑦 − 1 2𝑥 + 3𝑥 − 4𝑥 + 𝑦 = 𝑥 − 14 𝑐) −2𝑥 + 4𝑦 = 14 7𝑥 + 7𝑧 = −7 9𝑧 − 𝑦 + 5 = 0 𝑑) 𝑧 − 12𝑥 − 5𝑦 = 3 𝑥 = 3𝑥 − 2𝑦 + 4 13𝑥 = 12𝑦 − 4 2. A good habit is to write what the columns represent with a different colour: 𝑥 𝑦 𝑧 𝑐𝑜𝑛𝑠𝑡 −2 3 1 ⋮ 9 4 −2 3 ⋮ 4 −1 −1 1 ⋮ −2 You have to understand how a matrix relates to a set of equations before you can solve that system. What you are pointing to is the number in front of the 𝑥 in the first equation (look back up at ’. your right index finger back on 𝑥 and repeat the process. and 𝑦 with your right. Move your left finger across one more place and your right finger down one place. as this would change the equation. For the two equations below to be added to one another: −3𝑥 + 2𝑦 = 5 𝑥 + 𝑦 = 10 ’ ’ subtracting the COLUMNS of a matrix. to get a zero in the bottom left corner. What must be done is use something called row operations to manipulate that bottom row to get a zero in the bottom left corner. and the two right sides must be added. Theory: A row operation is a process where a multiple of one ROW is added/subtracted from another ROW. This ratio cannot be only applied to one column. Using this theory. this same ratio must be applied to all columns.𝑥 + 𝑦 = 10 ’ The results are in red. if the columns are added together: −3 2 ⋮ 5 1 1 ⋮ 10 −2 3 15 56 ⋮ 5 ⋮ 10 ⋮ 5 ⋮ 35 𝑅2’ = 3𝑅2 + 𝑅1 . and knowing row operations can be used without changing the equalities. This is similar to finding a common denominator is Chapter 1. the answer would be zero (as 3 × 1 + (−3) = 0). However. The reason is that an equation made from that row. In the above example. it is best to understand what is being done. Then 0𝑥 is just zero. leaving: 𝑏𝑦 = 𝑐 Which can then be solved for 𝑦. The easiest method of doing this is to come up with a short sentence once you have figured out the ratio. the sentence would be: “three lots of bottom plus top” −3 2 ⋮ 5 1 1 ⋮ 10 For the first column: “three lots of one plus negative three” is 3 × 1 + (−3) = 0 For the second column: “three lots of one plus two” is 3×1 +2=5 For the third column: “three lots of ten plus five” is 3 × 10 + 5 = 35 The way this is represented on a matrix is: −3 1 −3 0 2 1 2 5 The two left sides must be added. which is the same as adding the two equations together. If 3 times ROW2 were added to ROW1. how many ROW2’s will need to be added/subtracted to how many ROW1’s? Look only at the first column (the 𝑥 column). Theory: ROW operations involve adding/ Equations ’ and ’ allow construction of the matrix: −3 2 ⋮ 5 1 1 ⋮ 10 The idea now is to change the second row to get the number zero in the bottom left corner (circled in blue). Before row operations are done on a matrix. the circled number cannot simply be erased and replaced by 0. ’ + ’: −3𝑥 + 2𝑦 + 𝑥 + 𝑦 = 5 + 10 Simplifying this gives: −2𝑥 + 3𝑦 = 15 Back to the matrix. would be of the form: 0𝑥 + 𝑏𝑦 = 𝑐 Where 𝑏 and 𝑐 are constants. then write it out: 4 × −1 ∎ −4 = 0 −4∎ −4 = 0 If you have an addition. This is where many students stuff up. 𝑦 = 1. solve for the unknowns: 5𝑦 = 35 𝑦 = 7 Move up to the next equation. 57 . but if you are not sure whether to add or subtract. and substitute in 𝑦 = 7: −3𝑥 + 2(7) = 5 −3𝑥 + 14 = 5 −3𝑥 = −9 𝑥 = 3 The hardest part when using matrices is finding the ratio to be able to get zero in the bottom left corner. Apply this ratio to all columns. Example 1: solve for 𝑥 and 𝑦 in −1 6 ⋮ 7 −4 2 ⋮ −5 Plan: use row operations to get a zero in the bottom left corner. and solve for 𝑥 and 𝑦 from the bottom up.This is pronounced: “the new ROW2 is three times the old ROW2 plus ROW1”. Write out your saying: “four lots of top minus bottom” For the first column: “four lots of negative one subtract negative four” 4 × −1 − (−4) = 0 For the second column: “four lots of six subtract two” 4 × 6 − 2 = 22 For the third column: “four lots of seven subtract negative five” 4 × 7 − (−5) = 33 In matrix form. Once the zero is in the bottom left.1. Solve for the unknowns. Theory: use ROW operations to find the ratio of ROW 1 added/subtracted from ROW 2. Solution: how many lots of −1 's need to be added/subtracted to how many lots of −4 's to get zero in the bottom left? Four of the top are required to make the bottom. the equations can be extracted from this matrix: −3𝑥 + 2𝑦 = 5 0𝑥 + 5𝑦 = 35 Then working from the bottom equation up. so don’t be one of them. so the equations can be extracted: −𝑥 + 6𝑦 = 7 0𝑥 + 22𝑦 = 33 Solve from the bottom up: 22𝑦 = 33 𝑦 = 33 = 1.5 .5 22 Up to the next equation: −𝑥 + 6(1. then extract the equations from the modified matrix.5 or 2. this whole process would be written: −1 6 ⋮ 7 −4 2 ⋮ −5 𝑅2’ = 4𝑅1 − 𝑅2 −1 6 ⋮ 7 0 22 ⋮ 33 There is now a zero in the bottom left corner. working from the bottom equation up.5) = 7 −𝑥 + 9 = 7 𝑥 = 2 So the solution is 𝑥 = 2. −1 6 ⋮ 7 −4 2 ⋮ −5 Then extract the equations from this matrix. which will give a zero in the bottom left corner. it will make −8 so the black space must be a subtract sign. 5. then solve for the unknowns. Theory: to solve a three equation system. 35 𝑦 = − 35 52 Then check that the solution is correct: −4𝑦 = −3𝑥 + 17 52 129 = −3 + 17 35 35 208 208 = 35 35 −4 − Exercises: 1. then use row operations to get a zero in the bottom left corner. 𝑎) 2𝑥 − 𝑦 = −7 𝑐) 3𝑥 + 7𝑦 = 13 −4𝑥 + 4𝑦 = 8 5𝑥 + 9𝑦 = 11 𝑏) 𝑦 = −𝑥 + 8 𝑑) 9 − 3𝑦 = 5𝑥 4𝑥 = −2𝑦 − 3 −𝑥 = −3𝑦 + 13 3. This is the reason why zeros are needed in Then solve for 𝑥: 52 3𝑥 − 4 − = 17 35 208 3𝑥 + = 17 35 58 . the bottom equation will be 0𝑥 + 0𝑦 + 𝑎𝑧 = 𝑏 This will have some multiple of 𝑧 equal to a constant. Solution: put and into general matrix form: 3𝑥 − 4𝑦 = 17 2𝑥 + 9𝑦 = −6 Construct the matrix: 3 −4 ⋮ 17 2 9 ⋮ −6 Find the correct ratio from the first column “two lots of top minus three lots of bottom” Apply this to all columns: 3 −4 ⋮ 17 2 9 ⋮ −6 3 −4 ⋮ 17 0 −35 ⋮ 52 3𝑥 − 4𝑦 = 17 0𝑥 − 35𝑦 = 52 Working up. Find the equation of the demand function. zeros are needed in the bottom left triangular area: 𝑋 0 0 𝑋 𝑋 0 𝑋 ⋮ 𝑋 𝑋 ⋮ 𝑋 𝑋 ⋮ 𝑋 Write the equations from this matrix: From this matrix. and the top row will have all three variables. so 𝑦 can be solved as 𝑧 will already be known. one at a time. then determine the equilibrium price and quantity given that supply is 𝑄𝑠 = 0. so 𝑥 can be found. The middle row will have 𝑧 and 𝑦. solve for 𝑦: −35𝑦 = 52 𝑦 = − 52 35 𝑅2’ = 2𝑅1 − 3𝑅2 ’ ’ 3.Check if the solution is correct by substituting it into either of the original equations (the bottom row of the matrix. A company initially sets a price for a good at 𝑃 = 15. Solve the following matrices for the unknowns: 2 1 ⋮ 12 1 −1 ⋮ 9 𝑎) 𝑐) 3 4⋮ 8 2 1 ⋮9 3 −2 ⋮ −7 5 −1 ⋮ 9 𝑏) 𝑑) 4 2 ⋮ 0 −1 7 ⋮ 39 2. and sells a quantity of 𝑄 = 30.5𝑃𝑠 − 1. Solve the following equations using matrices. just for consistency): −𝑥 + 6𝑦 = 7 − 2 + 6 1.5 = 7 −2 + 9 = 7 This means the solution is correct. but there are some differences. When they increase price to 𝑃 = 20.6 solving 3 equation matrices Solving three equation matrices uses row operations. the company sells 25 units. Example 2: use matrices to solve for 𝑥 and 𝑦 −4𝑦 = −3𝑥 + 17 6 + 9𝑦 = −2𝑥 595 208 − 35 35 387 3𝑥 = 35 129 𝑥 = 35 3𝑥 = The solution is 𝑥 = 129 . Plan: construct a matrix from the equations. Extract the equations. so 𝑥. you may not be sure whether to add or subtract. To change the 6 into a zero. work with ROW1 and ROW2: Find the ratio by asking: for the new ROW2. How many ROW1’s will need to be added/subtracted from how many ROW3’s? The short sentence is: “two tops minus three bottoms” 3 0 2 3 0 0 −1 1 ⋮ 8 −4 −1 ⋮ 6 2 3 ⋮ 18 −1 1 ⋮ 8 −4 −1 ⋮ 6 −8 −7 ⋮ −38 ′ 𝑅3 = 2𝑅1 − 3𝑅3 Now that the circled entries are zeros. it would ruin the zero in the bottom left corner. Extract the equations from this matrix: 3𝑥 − 𝑦 + 𝑧 = 8 −4𝑦 − 𝑧 = 6 5𝑧 = 50 Work from the bottom equation up: 𝑧 = 50 = 10 5 Substitute into the next equation up: −4𝑦 − 10 = 6 59 . the bottom left corner). For now. how many ROW1’s will need to be added/subtracted from how many ROW2’s? The short sentence is: “two tops minus bottom” 3 −1 1 ⋮ 8 ′ 6 2 3 ⋮ 10 𝑅2 = 2𝑅1 − 𝑅2 2 2 3 ⋮ 18 3 −1 1 ⋮ 8 0 −4 −1 ⋮ 6 2 2 3 ⋮ 18 Remember that the “two tops minus bottom” applies to every entry in the second row. 𝑦 and 𝑧 can now be solved. then the blue zeros will remain unchanged (as any ratio of two zeros added or subtracted is still zero!). ignoring ROW2. Theory: the two bottom entries of the first column must be changed to zero first. as the number 3 from the first row would be introduced again. Intro example 1: solve for the unknowns in the matrix 3 −1 1 ⋮ 8 6 2 3 ⋮ 10 2 2 3 ⋮ 18 Firstly. ask: how many ROW2’s will you have to add/subtract from ROW3? Two lots of ROW2 are needed. If the top row is used. so: “two tops minus bottom” In matrix form: 3 0 0 3 0 0 −1 1 ⋮ 8 −4 −1 ⋮ 6 −8 −7 ⋮ −38 −1 1 ⋮ 8 −4 −1 ⋮ 6 0 5 ⋮ 50 ′ 𝑅3 = 2𝑅2 − 𝑅3 The matrix is reduced into the form that is required. these zeros need to be found in a particular order. To get a zero in place of the highlighted −8 . 𝑦 and 𝑧. ROW1 and ROW3 must be used. both the 6 and 2 in the blue box must be changed to zeros. However. To get zeros in the blue box. To change the number 2 to a zero in the third row (i. we can solve for 𝑥. but later you can do them at the same time. they will be done one at a time. if ROW3 is changed using ROW2. but because of the negative signs. working from the bottom up. So write it out: 2 −4 ∎ −8 = 0 −8∎ −8 = 0 The sign can only be a subtraction. the next requirement is to change the −8 (highlighted) into a zero. However. and then the bottom entry of the second column can be changed to zero. ROW1 must be used. The reason for this will soon emerge.the bottom left triangular area.e. to check if the solution is correct. get a zero in the bottom entry of the second column. −4. −1 −1 ⋮ 17 −1 4 ⋮ −11 −3 −5 ⋮ 9 −1 −1 ⋮ 17 −1 −11 ⋮ 73 1 5 ⋮ 67 Check this solution is correct: 𝑅2’ = 3𝑅1 − 2𝑅2 𝑅3′ = 5𝑅1 − 2𝑅3 5𝑥 − 3𝑦 − 5𝑧 = 9 5 266 551 70 −3 −5 − =9 3 3 3 . 3 .− 3 3 266 . Finally. 𝑧 3 =− 70 3 or . 𝑦 = −4 and 𝑧 = 10. row operations need to be used to get zeros in the bottom left triangle corner of the variables matrix. or −2.10 . substitute it back into the bottom row of the original matrix: 2𝑥 + 2𝑦 + 3𝑧 = 18 2 −2 + 2(−4) + 3(10) = 18 −4 − 8 + 30 = 18 Theory: to solve a system of three equations. 𝑦 3 = 551 . Example 1: solve for 𝑥. Then using the modified ROW2.−4𝑦 = 16 𝑦 = −4 Next equation up: 3𝑥 − −4 + (10) = 8 3𝑥 + 14 = 8 𝑥 = −2 The solution is 𝑥 = −2. 𝑦 and 𝑧. Rewrite the equations from the final matrix then work up to solve for 𝑥. 𝑦 and 𝑧 in the matrix 2 −1 −1 ⋮ 17 3 −1 4 ⋮ −11 5 −3 −5 ⋮ 9 Plan: use row operations to get zeros in the first column where 3 and 5 are located using ROW1. Solution: For ROW2: “three lots of top minus two lots of bottom” For ROW3: “five lots of top minus two lots of bottom” 2 3 5 2 0 0 60 To get a zero in the bottom entry of the second column: “one lot of top plus bottom” 2 0 0 2 0 0 −1 −1 ⋮ 17 −1 −11 ⋮ 73 1 5 ⋮ 67 𝑅3’ = 𝑅2 + 𝑅1 −1 −1 ⋮ 17 −1 −11 ⋮ 73 0 −6 ⋮ 140 Rewrite the equation from this matrix: 2𝑥 − 𝑦 − 𝑧 = 17 −𝑦 − 11𝑧 = 73 −6𝑧 = 140 Solve for 𝑧 using the bottom equation: 𝑧 = − 140 70 =− 6 3 Solve for 𝑦 using the next equation up: 70 = 73 3 770 −𝑦 = 73 − 3 219 770 −𝑦 = − 3 3 551 𝑦 = 3 −𝑦 − 11 − Solve for 𝑥 using the next equation up: 551 70 − − = 17 3 3 481 2𝑥 − = 17 3 51 481 2𝑥 = + 3 3 532 2𝑥 = 3 266 𝑥 = 3 2𝑥 − The solution is: 𝑥 = 266 551 70 . getting zeros in the left-most columns first. then there would be infinite intersections (every point on the two lines is an intersection).00/𝑆 At full capacity.00/𝑆 State 2: $7. The 𝑅1.00/𝐿. 3. These notes are also a good way of finding any mistakes. $21. and the solution is shown in the following matrix. b) The amount of 𝐿. Each factory uses labour 𝐿 . just the matrix the note is written next to. Solve the following matrices.7 notes on solutions to matrices When the intersection of two lines was found back in Chapter 2. Exercises: 1. a unique. if the final check is found to be incorrect. then there can be: A single solution (what we have been doing in the last two sections) No solutions (if there is no common intersection of all lines) Infinite solutions (if we do not have enough equations and/or if some of the equations are the same) Only after you have tried getting the zeros in the bottom left triangle can you determine if there is infinite. if two lines had the same gradient 𝑚 and the same constants 𝑐 (that is. Due to differences in union powers. if the two lines had the same gradient 𝑚 but different constants 𝑐. such as 𝑅3’ = 𝑅2 + 𝑅1 refer to the what is being done to that matrix.00 State 2: $3589.50/𝑀 and $31. Theory: since matrices are just a simpler form of writing out a number of equations.00 State 3: $3564. Solve the following sets of equations using matrices.00/𝑆 State 3: $9.50/𝑀 and $30.50/𝐿. $16. 𝑀 and 𝑆 in these three states are not the same: State 1: $7. taxation laws and property rights. 𝑎) −2𝑥 + 6𝑦 − 3𝑧 = 6 𝑥 + 3𝑦 − 𝑧 = 17 4𝑥 + 2𝑦 + 4𝑧 = 4 𝑏) 6𝑥 + 4𝑦 + 6𝑧 = 2 −𝑥 + 2𝑦 − 𝑧 = 3 2𝑥 − 3𝑦 + 3𝑧 = 6 𝑐) 3𝑦 − 6 = −𝑧 + 4𝑥 2𝑧 + 𝑥 = 2𝑦 + 𝑥 + 12 6𝑥 + 𝑦 + 𝑧 + 4 = 0 𝑑) 4𝑥 = −7𝑦 − 𝑧 + 5 4𝑦 = 𝑥 + 𝑧 + 15 −2𝑥 − 2𝑧 − 10 = −2𝑦 3. 𝑅2 etc. 61 . or no solutions. then there would be no intersection. two equations for the same line). metal 𝑀 and silicone chips 𝑆 . the total cost for each of the three factories is: State 1: $3373.00/𝑀 and $29.1330 1653 350 − + =9 3 3 3 27 = 9 3 Note: the notes next to any matrix.00 Determine: a) Three expressions equating units of each of the three inputs to total cost. do not refer to the original matrix. the costs of 𝐿. 𝑀 and 𝑆 used in each of the three identical factories.20/𝐿. A computing company makes computers at to capacity in three identical factories in three different states. 4 −1 −2 ⋮ 8 5 2 −9 ⋮ 9 𝑎) 𝑐) 7 3 4 ⋮ −10 2 2 3 ⋮ 12 3 3 −1 ⋮ 25 1 3 −4 ⋮ −13 2 −3 4 ⋮ 8 1 −5 1 ⋮ 15 𝑏) 𝑑) 2 5 2 1 7⋮1 6 ⋮ 12 2 −3 7 ⋮ 5 3 −3 14 ⋮ 2 2. Similarly. thus there are no solutions. The following is a matrix where there are no solutions (notice that there are zeros in the bottom left triangle): 2 0 0 −1 4 ⋮ −9 2 3 ⋮ 5 0 0 ⋮ 4 If the bottom row is written in equation form: 0𝑥 + 0𝑦 + 0𝑧 = 4 0=4 But it is impossible for 4 to be equal to zero. $18. 5𝑦 − 4. Exercises: 1. 𝐶. thus there are infinitely many solutions. other independent equations need to be found. 𝐼 is the sum of all investment expenditure in an economy. it comes from savings. and 𝐺 the government spending.8 applications An application you will surely come across is the macroeconomic model of an economy.5𝑧 − 52. 𝑎) −12𝑥 − 5𝑧 = −7𝑥 + 19 𝑧 = −8𝑦 − 4𝑥 + 8 −6𝑥 + 34. To be able to solve for 𝑌.5𝑦 = 24𝑧 − 10 3. 𝑆 = 𝑠𝑌𝑑 Where “𝑆” is the total savings of all consumers. so from the above equation: This matrix has the same number of equations as there are variables. that money needs to come from somewhere. infinite solutions and a unique solution. Determine if the following sets of equations can be solved.g. then solve them if possible.20). so that income is reduced.An example with an infinite amount of solutions (again. which is already known. but rather save a proportion of it. so it cannot be solved. above is one equation with four variables. so this cannot be solved for a unique solution. 𝑦 and 𝑧 will give 0 = 0. 𝐼 or 𝐺. Determine the number of solutions for the following matrices: 6 2 4⋮ 5 7 5 −3 ⋮ −17 𝑎) 6 6 4 ⋮ 3 𝑐) 21 3 7 ⋮ 5 3 4 2 ⋮ 20 14 34 −38 ⋮ −146 12 2 6 ⋮ 5 5 2 4⋮5 𝑏) 6 3 7 ⋮ 5 𝑑) 6 6 4 ⋮ 3 18 19 41 ⋮ 12 3 4 2⋮2 2. Find the value(s) of 𝑎 where there are zero solutions. and is written: 𝑌 = 𝐶 + 𝐼 + 𝐺 Where 𝑌 is the total income of an economy.5 = 0 𝑏) −8𝑥 − 3𝑦 = 14 − 𝑧 −5𝑧 − 12 = −6𝑥 − 4𝑦 4𝑥 + 5𝑦 + 2𝑧 = 2 𝑐) 12𝑧 + 4𝑥 = −9𝑦 + 16 2𝑥 + 9. the income they physically receive). for anyone in an economy to invest money (say in constructing an office block).e. If the other equations are extracted: −5𝑥 + 𝑦 + 5𝑥 = 9 𝑦 + 3𝑧 = 6 The result is two equations and three variables. Whatever is not saved. After earning income from working. Thus 𝑆 = 𝐼. 𝐶 is the total consumption expenditure in an economy. 62 . however one of these equations is all zeros: 0𝑥 + 0𝑦 + 0𝑧 = 0 This gives no information. the government takes a proportion of it through taxes. notice that the zeros in the bottom left triangle): −5 1 0 1 0 0 5 ⋮9 3⋮ 6 0 ⋮ 0 Economic theory: the macroeconomic model of an economy determines the income of all the people in an economy. The assumption is made that individuals save at a rate “𝑠” of their disposable income (i.5𝑦 + 3𝑧 = 15 −6𝑥 − 3. because any values of 𝑥. then the income actually obtained is called disposable income (denoted 𝑌𝑑 ): 𝑌𝑑 = 1 − 𝑡 𝑌 Most people do not spend everything they earn. That is. 𝑌𝑑 = 1 − 𝑡 𝑌 so this allows us to get rid of the 𝑌𝑑 in the above two equations: 𝑆 = 𝑠 1 − 𝑡 𝑌 𝐶 = 1 − 𝑠 1 − 𝑡 𝑌 Lastly. 20%: 𝑡 = 0. must be spent on consumption goods: 𝐶 = 1 − 𝑠 𝑌𝑑 From before. 6 4 −3 ⋮ 12 3 8 3 ⋮ 8 1 1 𝑎 ⋮ 3 3. If the government taxes at a rate 𝑡 (e. 44 ⋮ 300 0 1 −0.3𝑌 = 300 Work from the bottom up: 𝑌 = 300 = 1000 0. the savings rate 𝑠 takes a long time to change. then solve for 𝐶.44 ⋮ 300 ′ 𝑅3 = 𝑅2 + 𝑅3 0 1 −0. 𝐼 and 𝑌. and six variables: 𝑌 = 𝐶 + 𝐼 + 𝐺 𝐼 = 𝑠 1 − 𝑡 𝑌 𝐶 = 1 − 𝑠 1 − 𝑡 𝑌 The variables will be arranged in alphabetical order: 𝐶.2 1 − 0. with the citizens.14 ⋮ 0 −1 −1 1 ⋮ 300 0 −1 0. Working through the following example will show you how to solve this set of equations. the government sets the tax rate 𝑡 and is usually the same for at least a year.44 1000 = 300 𝐼 = 140 −𝐶 − 𝐼 + 𝑌 = 300 −𝐶 − 140 + 1000 = 300 𝐶 = 560 63 . Solution: 𝑌 = 𝐶 + 𝐼 + 300 𝐶 = 1 − 0.3 ⋮ 300 Extract the equations from this last matrix: −𝐶 − 𝐼 + 𝑌 = 300 −𝐼 + 0.56 ⋮ 0 0 1 −0. saving 20% of their income. what is the consumption. This gives three equations.44 ⋮ 300 0 0 0.2 1 − 0. Example 1: if the government of an economy spends $300billion and taxes at a rate of 30%.14 ⋮ 0 Get a zero in the bottom entry of the second column: −1 −1 1 ⋮ 300 0 −1 0. 𝐼 and 𝑌.56𝑌 = 0 𝐼 − 0. on average. use row operations to get the zeros in the bottom left triangle. investment and total income of the economy? Plan: use the model of a closed economy 𝑌 = 𝐶 + 𝐼 + 𝐺 𝐶 = 1 − 𝑠 1 − 𝑡 𝑌 𝐼 = 𝑠 1 − 𝑡 𝑌 Construct the matrix: −1 −1 1 ⋮ 300 1 0 −0.14 ⋮ 0 −1 −1 1 ⋮ 300 0 −1 0.14 ⋮ 0 Use row operations to get zeros in the bottom two entries of the first column: −1 −1 1 ⋮ 300 ′ 𝑅2 = 𝑅1 + 𝑅2 1 0 −0. the government spends approximately the same amount). then rearrange into the general matrix form.3 𝑌 Lastly: Substitute this into the next equation up: −𝐼 + 0. so is assumed to be constant.14𝑌 = 0 However. it is not changed much.56 ⋮ 0 0 1 −0.𝐼 = 𝑠 1 − 𝑡 𝑌 Where the right side is the equivalent of “𝑆”. Government spending (𝐺) is constant regardless of the level of income (as even in a recession. assumptions are made about 𝐺. 𝑠 and 𝑡.44𝑌 = 300 0.3 Substitute all the values that are known.3 𝑌 𝐼 = 0. and if it is changed. Rearranging the equations into the general matrix form: ’: ’: ’: −𝐶 − 𝐼 + 𝑌 = 300 𝐶 − 0. 85𝑠𝑌 = 0 ’ ’ ’ 64 . investment is $180billion and total income is $1. on average. investment and total income of the economy? Plan: use the model for a closed economy 𝑌 = 𝐶 + 𝐼 + 𝐺 𝐼 = 𝑠 1 − 𝑡 𝑌 𝐶 = 1 − 𝑠 1 − 𝑡 𝑌 0. If you think it would be easier to solve the model using the substitution method.12𝑌 = 0 ’ ’ ’ and assuming the tax rate is 15% and government spending is $600billion.85𝑠𝑌 Rearrange to get into the general matrix form −𝐶 − 𝐼 + 𝑌 = 600 𝐶 − (0. what is the solution to this system of equation in terms of the savings rate 𝑠? Plan: substitute the numbers that are known. and the people in the economy save. assuming 𝑠 is just a number. Example 2: If a government has a tax rate of 40% and spends $600billion.85𝑌 − 0. 𝐼 and 𝑌 in terms of 𝑠. Solve for 𝐶. what is the consumption.15 𝑌 𝐶 = 0. 𝑡 = 0. Solution: substitute in the known numbers 𝑌 = 𝐶 + 𝐼 + 600 𝐶 = 1 − 𝑠 1 − 0.4 ⋮ 600 Extract the equations from this last matrix: −𝐶 − 𝐼 + 𝑌 = 600 𝐼 − 0.12𝑌 = 0 𝐶 − 0.15 𝑌 = 0. then solve for 𝐶. then rearrange to get into the general matrix form.85 − 0. 𝐼 and 𝑌.12 1500 = 0 𝐼 = 180 Up one more equation: −𝐶 − 180 + 1500 = 600 𝐶 = 720 Thus consumption is $720billion.500billion.48 ⋮ 0 𝑅3′ = 𝑅1 + 𝑅3 −1 −1 1 ⋮ 600 0 1 −0.4 𝑌 = 0.4 𝑌 = 0. Unless you are not given enough information to solve for actual numbers. Use row operations to get zeros in the bottom left triangle.52 ⋮ 600 𝑅3′ = 𝑅2 + 𝑅3 −1 −1 1 ⋮ 600 0 1 −0. 𝐺 = 600).000billion (or $1trillion). 20% of their income.85𝑠𝑌 𝐼 = 𝑠 1 − 0. you would be correct. and then rearrange to get into the general matrix form.4 Then up one equation: 𝐼 − 0.4.12𝑌 𝐶 = 1 − 0.12 ⋮ 0 0 0 0. invests $140billion and has a total income of $1.48𝑌 = 0 Construct the matrix: −1 −1 1 ⋮ 600 0 1 −0. Example 3: Given the following closed economy 𝑌 = 𝐶 + 𝐼 + 𝐺 𝐶 = 1 − 𝑠 1 − 𝑡 𝑌 𝐼 = 𝑠(1 − 𝑡)𝑌 Substitute in the known information (𝑠 = 0.12 ⋮ 0 0 −1 0.48𝑌 Rearrange into the general matrix form: −𝐶 − 𝐼 + 𝑌 = 600 𝐼 − 0.2.85𝑠)𝑌 = 0 𝐼 − 0.4𝑌 = 600 Work upwards: 𝑌 = 600 = 1500 0.12 ⋮ 0 1 0 −0.This economy has consumption of $560billion.2 1 − 0.2 1 − 0. Solution: substitute in all the known information: 𝑌 = 𝐶 + 𝐼 + 600 𝐼 = 0. Government spending = 400. solve for 𝐶. using matrices seems pointless. Government spending = 700. Given an economy with the following facts.85𝑠 ⋮ 0 1 −0. 2. 65 . and then subtracting the multiplication of the numbers in the left hand diagonal. 𝐼 and 𝑌 using matrices. solve for 𝐶. This is why you should know a few different ways of solving matrices.85𝑠 ⋮ 600 0 0 0. Government spending = 500.15𝑌 = 600 Work from the bottom up: 𝑌 = 4000 Up one more equation: −𝐼 + 0.85𝑠 𝑌 = 600 0. straight lines are drawn either side of the matrix: 3 −4 5 −8 Example 1: find the determinant of the matrix 3 5 −4 −8 Matrices make easy work of a large number of equations.Remember that 𝑠 is the savings rate and is a constant.15 ⋮ 600 Rewrite the equations: −𝐶 − 𝐼 + 𝑌 = 600 −𝐼 + 0.9 the determinant of a 𝟐 × 𝟐 matrix For a small number of equations. the determinant is written with two vertical lines around either the whole matrix. tax rate = 10% and savings rate = 25%. 𝐼 and 𝑌 using matrices in terms of the tax rate 𝑡. Given an economy with the following facts. Theory: the determinant of a 2 × 2 matrix is found by multiplying the numbers in the right hand diagonal. 600 ′ 0 𝑅2 = 𝑅1 + 𝑅2 0 3. but with a large number of equations.15 + 0. Construct the relevant matrix: −1 −1 1 ⋮ 1 0 −0. To show that the determinant is being found.85𝑠 4000 = 600 −𝐼 + 600 + 3400𝑠 = 600 𝐼 = 3400𝑠 Then: −𝐶 − 3400𝑠 + 4000 = 600 𝐶 = 3400 − 3400𝑠 The answer in terms of the savings rate 𝑠 is: Consumption is $(3400 − 3400𝑠)billion.15 + 0. solve for 𝐶.85𝑠 ⋮ 600 ′ 0 1 −0. 𝐼 and 𝑌 using matrices. For the matrix: 𝑎 𝐹 = 𝑏 𝑐 𝑑 The determinant of matrix 𝐹 is: 𝐹 = 𝑎𝑑 − 𝑏𝑐 Not to be confused with absolute values. Manipulate this to get the zero in the middle entry 3. or the letter defining a matrix. tax rate = 20% and savings rate = 15%.85𝑠 ⋮ of the first column: −1 −1 1 ⋮ 600 0 −1 0.85 + 0.85𝑠 ⋮ 0 𝑅3 = 𝑅2 + 𝑅3 Then get a zero in the bottom entry of the second column: −1 −1 1 ⋮ 600 0 −1 0.15 + 0. Exercises: 1. Investment is $(3400𝑠)billion.15 + 0. savings rate = 18%. For the 2 × 2 matrix: 2 4 3 7 The determinant is: 2 × 7 − (4 × 3) = 14 − 12 = 2 An easy way of remembering this (especially if you’re right handed) is right hand diagonal take away left hand diagonal (actually do the karate chops!). and Total income of the economy is $4000billion. Given an economy with the following facts. matrices simplify the mathematics. 66 . −. This pattern works for the top row. so make sure you are very comfortable with all the material in Section 3. The determinant of the original 3 × 3 matrix is: 𝐷𝑒𝑡 = + 2 −1 − 5 7 + −3 8 = −61 The +. sometimes you will not want to use the top row.9. + pattern does NOT come from the sign of the numbers in the original matrix. 𝐷𝑒𝑡 = −1 −2 − −3 −1 = −1 Go on to the next number along in the original 3 × 3 matrix.10 the determinant of a 𝟑 × 𝟑 matrix This is a harder section. Notice also the highlighted signs and how they alternate (+. draw a matrix: −1 −1 −3 −2 Find the determinant of this matrix using the right hand/left hand method. +) which we will come back to soon. Find the determinant of the following matrices: 2 −1 4 4 −1 −2 𝑎) 𝑑) 𝑔) 2 3 7 5 2 −6 −3 2 2 4 8 −3 𝑏) 𝑒) ) −29 5 3 6 −16 6 6 12 −2 4 −3 5 𝑐) 𝑓) 𝑖) 8 11 7 0 −5 −3 Draw the matrix from the remaining numbers. then cross out all the numbers in that row and column. Many students stuff up here. Intro example 1: find the determinant of the 3 × 3 matrix: 2 5 −3 −1 −1 −1 5 −3 −2 Solution: Circle the first entry (2). −. but leave the circled number alone: 2 5 −3 −1 −1 −1 5 −3 −2 With the remaining numbers. but as you will soon see. Theory: The determinant of a 3 × 3 matrix is: 𝐷𝑒𝑡 = + 1𝑠𝑡 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 − 2𝑛𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 3. circle it and cross out the vertical and horizontal numbers: 2 5 −3 −1 −1 −1 5 −3 −2 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 + 3𝑟𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × 𝑑𝑒𝑡 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 And the “circled numbers” are the numbers in the top row.Solution: the determinant is the multiplication of the right diagonal minus the multiplication of the left diagonal which gives: 3 × −8 — 4 × 5 = −24 − (−20) = −24 + 20 = −4 Exercises: 1. and find the determinant: −1 −1 5 −2 𝐷𝑒𝑡 = −1 −2 − 5 −1 = 7 Go on to the last number: 2 5 −3 −1 −1 −1 5 −3 −2 Finding the determinant of the remaining numbers: −1 −1 5 −3 𝐷𝑒𝑡 = −1 −3 − 5 −1 = 8 Then bring it all together. so don’t be one of them. but it is a pattern unrelated to the original matrix. Example 1: find the determinant of the matrix −2 3 1 𝐴 = −1 0 4 4 −5 −3 Plan: use the middle column (because it has a zero): circle the top number and cross out the other numbers in that row and column. but any row or any column can be used. the signs in the determinant equation will be −. because the determinant of a circled zero does not need to be found. the pattern for the determinant equation is given by the matrix: + − + − + − + − + An easy way of remembering this matrix is that the top left is an addition sign. −: 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 = − 1𝑠𝑡 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 + 2𝑛𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 − 3𝑟𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 Solution: Circle the first number: −2 3 1 −1 0 4 4 −5 −3 Find the determinant of the remaining matrix: 𝐷𝑒𝑡 = −1 4 = −13 4 −3 Circle the next number: −2 3 1 −1 0 4 4 −5 −3 𝐷𝑒𝑡 = −2 1 =2 4 −3 For the last number: −2 3 1 −1 0 4 4 −5 −3 𝐷𝑒𝑡 = 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 = − 1𝑠𝑡 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 + 2𝑛𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 − 3𝑟𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × 𝑑𝑒𝑡 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 −2 1 = −7 −1 4 𝐷𝑒𝑡 𝐴 = 𝐴 = − 3 −13 + 0 2 — 5 −7 =4 But the middle number is (0)(2) = 0. Try the above example using the middle row. +. Use the “sign” matrix to get the correct pattern: + − + − + − + − + Since the middle column is being used. this does NOT mean that the numbers in the original 3 × 3 matrix are positive or negative. Then find the determinant of the remaining numbers. as the following example demonstrates. Repeat for the other two numbers in the middle column. and then the signs alternate across every row and down every column. There is a good reason why the +. this is the pattern of signs you put in front of the “𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡” in the formula. so the 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 × 𝑧𝑒𝑟𝑜 will always be zero. 67 . the top row was used.Theory: when finding the determinant using any row or column. −. Theory: when asked to find the determinant of a matrix and one of the rows or columns has one (or more) zeros in it. Remember. This is why columns/rows with lots of zeros are chosen. You should get the same answer. It makes life a lot easier. + matrix should be remembered. use that row or column. but rather that when you come to find the determinant. In the introductory example. Then replace the appropriate column (for the variable to be solved) with the constants column. firstly find the determinant of the variables matrix (|𝐴|). The following is a different way of finding the solution to a set of equations. 𝑦 and 𝑧 were found by finding the zeros in the bottom left triangle. Divide this determinant by the determinant of the original variables determinant to solve for the unknown variable. Exercises: 1. Next. The solution is this second determinant divided by the determinant of the variables matrix: 𝑏 = 𝐴𝑏 𝐴 3.e. The determinant was −61. and find the determinant of this (|𝐴𝑏 |). Find the determinants of the following 3 × 3 matrices: 𝑎) 𝑏) 𝑐) 2 5 10 5 6 3 1 −2 8 3 7 −1 5 8 4 1 9 9 11 12 15 −11 19 18 3 23 27 𝑑) 𝑒) 𝑓) 𝐴𝑥 = − 0 19 −3 5 −3 + −1 −10 −2 −3 −2 19 5 − (−1) −10 −3 𝐴𝑥 = −1 −68 − −1 −7 = 61 Then to find the actual value of 𝑥. but it requires a lot of work.e. the column of the variable to be solved is REPLACED by the column of constants. the easier it will be.Finding determinants of 3 × 3 matrices is not too difficult. This is denoted by |𝐴| and it is the determinant of the original 3 × 3 “variables” matrix. find the determinant of the variables matrix.11 using the Jacobian determinant Previously. Intro example 1: to solve the matrix 2 5 −3 −1 −1 −1 5 −3 −2 ⋮ 19 0 ⋮ ⋮ −10 Example 1: solve for 𝑦 and 𝑧 using the Jacobian Method in the matrix that was started above 2 5 −3 −1 −1 −1 5 −3 −2 find the determinant of the 3 × 3 “variables” matrix (left of the dotted line) which was done in the previous section. Then replace the constants column into the column of the variable being solved. the numbers on the right of the dotted line): 19 5 −3 0 −1 −1 = 𝐴𝑥 −10 −3 −2 Find the determinant of this new matrix using the method from the last section. and find the determinant of this new 3 × 3 matrix. to solve for 𝑥. replace the 𝑦 column with the constants column 2 19 −3 −1 0 −1 = 𝐴𝑦 5 −10 −2 Find the determinant of this matrix using the middle row: 68 . the values of 𝑥. That is. Use the middle row as it has a zero (remembering the “signs” matrix): ⋮ 19 0 ⋮ ⋮ −10 Plan: use the Jacobian method. replace the 𝑥 column (i. first column) in the matrix with the constants column (i. The more you practice. the following fraction is used: 𝑥 = 𝐴𝑥 61 = = −1 𝐴 −61 −1 4 −1 4 5 4 −1 4 −1 3 2 −8 0 7 5 1 2 1 0 5 3 11 −6 2 5 −6 0 Theory: to solve for a variable using the Jacobian Method. Solution: to solve for 𝑦. Example 2: solve for 𝑥. 3. To make sure this answer is correct. Call this 𝐴 . which then gives: 𝑧 = 𝐴𝑧 122 = = −2 𝐴 −61 Use the bottom row: 𝐴 = + 0 𝐴 = −1 −2 + −3 −8 = 26 To find 𝐴𝑥 : 𝐴𝑥 = 0 5 10 4 4 −2 1 1 −3 Thus 𝑥 = −1. replace the middle column with the constants. Find the determinant of this matrix. but the more you practice the easier it becomes. and use the top row: 𝐴𝑧 = 𝐴𝑧 2 4 0 1 −2 5 0 1 10 −2 5 1 5 =+ 2 − (4) 1 10 0 10 ⋮ 0 ⋮ 5 ⋮ 10 Plan: find the determinant of the variables matrix. 𝑦 and 𝑧 in the following matrix using the Jacobian Method 2 4 4 1 −2 1 0 1 −3 Use the top row (leaving the zero out from now on): 𝐴𝑥 = − 4 5 1 5 + (4) 10 −3 10 −2 1 𝐴𝑥 = −4 −25 + 4 25 = 200 To find 𝐴𝑦 replace the middle column with the constants. when finding the determinant. 𝑦 = 3. −2 . substitute it back into the bottom equation of the original matrix: 5𝑥 − 3𝑦 − 2𝑧 = −10 5 −1 − 3 3 − 2 −2 = −10 −10 = −10 This is a complicated method. or −1. use it. 𝑧 = −2. using the 3 × 3 method. and call it 𝐴𝑥 . for a solution: 𝑥 = 𝐴𝑥 200 100 = = 𝐴 26 13 69 .𝐴𝑦 = − −1 19 −3 + 0 −10 −2 2 19 − (−1) 5 −10 2 5 −3 −2 Repeat for 𝑦 and 𝑧 with the appropriate columns. Replace the 𝑥 column with the constants column. When you have a zero in a matrix. remember the “signs” matrix: + − + − + − + − + Solution: determinant of the variables matrix: 𝐴 = 2 1 0 4 4 −2 1 1 −3 4 4 2 4 − 1 −2 1 1 1 2 4 + (−3) 1 −2 𝐴𝑦 = 1 −68 − −1 −115 = −183 Solve for 𝑦: 𝑦 = 𝐴𝑦 −183 = =3 𝐴 −61 To solve for 𝑧: 2 5 19 −1 −1 0 = 𝐴𝑧 5 −3 −10 Find the determinant (yourself!) to make sure it is 𝐴𝑧 = 122. Solve for 𝑥 by: 𝑥 = 𝐴𝑥 𝐴 𝐴𝑧 = 2 −25 − 4 10 = −90 Finally. Also. and use the top row: 𝐴𝑦 = 𝐴𝑦 0 4 5 1 10 −3 5 1 1 =+ 2 + (4) 10 −3 0 2 1 0 5 10 𝐴𝑦 = 2 −25 + 4 10 = −10 To find 𝐴𝑧 . and remember that you do not need to find the determinant of the smaller matrix when that zero is circled. plastic costs $8/𝑢𝑛𝑖𝑡. The solution is 𝑥 = 100/13. 8 −1 2 ⋮ 1 𝑎) −2 −2 −7 ⋮ 14 𝑐) 4 1 3 ⋮ 5 3 6 8 ⋮ 12 𝑏) 𝑑) −1 1 2 ⋮ 13 −3 −2 1 ⋮ 11 3. draw a new set of brackets and vertically write the order of the variables you have chosen (e. 𝑎) 6𝑥 + 3𝑦 + 3𝑧 = 18 2𝑥 − 𝑦 − 𝑧 = 24 −2𝑥 + 2𝑦 − 4𝑧 = 15 𝑏) 4𝑥 + 5𝑦 = 2𝑧 + 6 5𝑥 = 𝑦 + 5𝑧 + 2 −3𝑥 − 𝑧 = −5𝑦 + 6 𝑐) 6𝑥 + 4𝑦 + 6𝑧 = 3 𝑧 − 𝑦 + 4𝑥 − 9 = 0 2𝑦 − 2𝑥 − 3𝑧 = 8 𝑑) 𝑥 − 9𝑦 + 3𝑧 = 2 −𝑥 − 2𝑦 + 2𝑧 = 6 4𝑥 − 2𝑦 + 2𝑧 = 8 A toy manufacturing company uses metal (𝑚) plastic (𝑝) and labour (𝑙). this is the process of adding/subtracting the COLUMNS of a matrix. labour costs $3/𝑢𝑛𝑖𝑡. The company owns three identical factories in a country where the costs of each of the three inputs is different: City 1: metal costs $2/𝑢𝑛𝑖𝑡.− . If all factories are at full capacity.𝑦 = 𝑧 = 𝐴𝑦 −10 −5 = = 𝐴 26 13 𝐴𝑧 −90 −45 = = 𝐴 26 13 100 5 45 . take one equation and substitute it into the other two equations. set the 𝑦’s equal to each other. For three equation systems: zeros are needed in the bottom two entries of the first column (this must be 3. substitute it into the other equation. 3. 2. use these two solutions to solve for the third variable. Note: setting the 𝑥’s equal to each other initially will give the same answer. take the numbers in front of the variables and put them into square brackets. plastic and metal used in each of the three identical factories. the costs for each of the factories is as follows: City 1 factory costs are $342 City 2 factory costs are $255 City 3 factory costs are $375 Determine: a) The equation in terms of 𝑚. so leave them as fractions! Exercises: 1. 3.g. To solve three simultaneous equations: 1. To be consistent. 70 . Check if the answer is correct by substituting it back into one of the rows of the original matrix. then substitute this solution back into any one of the original equations to solve for the other variable. Method 2 to solve two equation systems: 1. 𝑦. This will make two new equations with only two variables. a zero is needed in the bottom left corner. alphabetical). Manipulating matrices involve ROW operations. 𝑦 = −5/13. the constants on the right side are written vertically in square brackets. 4. after writing an equals sign. City 1: metal costs $7/𝑢𝑛𝑖𝑡. 4. labour costs $9/𝑢𝑛𝑖𝑡. 1 −1 −3 ⋮ 7 2 −1 −4 ⋮ 7 −8 5 −3 ⋮ 7 4 −6 2 ⋮ 10 −2 −1 −5 ⋮ 6 1 −7 −1 ⋮ 5 Solve the following sets of equations using Jacobian determinants. solve for the other variable. labour costs $7/𝑢𝑛𝑖𝑡. substitute the functions of 𝑥 in and solve for 𝑥. 𝑧 = −45/13 or . isolate one variable in one equation 2. 𝑧 . plastic costs $3/𝑢𝑛𝑖𝑡. 2. use the bottom equation: 𝑦 − 3𝑧 = 10 5 45 − −3 − = 10 13 13 5 135 − + = 10 13 13 −5 + 135 130 = = = 10 13 13 The solutions are fractions. The solution is also written in coordinate form 𝑥. plastic costs $6/𝑢𝑛𝑖𝑡. City 2: metal costs $5/𝑢𝑛𝑖𝑡.− 13 13 13 2. use this solution to solve for one of the other variables. lastly. 3. For two equation systems. The matrix ready form of an equation is: 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑 To construct a matrix: 1. b) The amount of labour. 2. 𝑝 and 𝑙 for each of the three cities. Substitute the solutions of 𝑥 back into either of the original equations to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the solution. chapter three summary Method 1 to solve two equation systems: 1. substitute one of those equations into the other to solve for one of the unknowns. Solve the following matrices. 1𝑄𝑠 + 97/30 𝑒) 𝑃𝑑 = −2𝑄𝑑 + 24. For the matrix: 𝑎 𝑐 𝑏 𝑑 The determinant is: 𝐹 = 𝑎𝑑 − 𝑏𝑐 The determinant of a 3 × 3 matrix is: 𝐷𝑒𝑡 = + 1𝑠𝑡 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 − 2𝑛𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 + 3𝑟𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × 𝑑𝑒𝑡 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 When finding the determinant using any row or column. 𝐹 = 4.5𝑦 𝑒) 3𝑥 = 3𝑦 − 4 + 𝑧 𝑓) 𝑎 + 𝑐 = 𝑏 − 4 4𝑦 − 15𝑥 = 13 − 𝑏 = 𝑐 − 𝑎 − 1 1.8𝑄𝑠 + 30.75 𝑑) 3𝑥 = −2𝑦 − 4𝑥 + 9 𝑥 + 𝑦 = −3𝑧 + 5 𝑥 + 𝑦 = −4𝑧 + 6 𝑒) 2𝑥 + 2𝑦 + 2𝑧 = 16 3𝑥 + 3𝑧 = 𝑦 + 20 3𝑦 + 𝑧 = 𝑥 + 6 Solve the following sets of equations using matrices: 𝑎) 12𝑥 + 7𝑦 = 19 𝑏) 𝑥 = 𝑦 + 1 8𝑥 + 13𝑦 = 21 3𝑥 + 3𝑦 − 9 = 0 𝑐) 6𝑦 = −8𝑥 + 6 𝑑) 12𝑦 + 22 = 5𝑥 5𝑦 + 6 = −3𝑥 3𝑥 + 8𝑦 + 2 = 0 𝑒) 7𝑥 + 12𝑦 = 319 𝑓) 2𝑥 + 13𝑦 = 140 9𝑦 = 18𝑥 − 63 2𝑥 + 8𝑦 = 80 𝑔) 𝑦 = 3𝑥 + 7 ) 4𝑥 = 𝑦 − 5. and 𝐺 is government spending. and 𝐴𝑏 is the determinant of the “variables” matrix with the 𝑏 column replaced with the solitary constants matrix.5𝑄𝑑 + 59.3𝑄𝑑 + 349/30 𝑃𝑠 = 4. solve the following matrices for a unique solution: 2 1 ⋮ 8 1 4 ⋮ 29 𝑏) −2 2 ⋮ −2 5 3 ⋮ 26 −1 1 ⋮ 3 1 3 ⋮ 10 𝑑) 2 −2 ⋮ −6 3 1 ⋮ 6 −2 −4 ⋮ 14 4 5 ⋮ 100 𝑓) 3 2 ⋮ −9 3 6 ⋮ 84 3 3 ⋮ 24 1 5 ⋮ 42 ) −2 2 ⋮ 18 2 −1 ⋮ −9.5 −3 12 ⋮ 25.6 −9 1 ⋮ −65.75 8𝑥 + 2𝑦 = −10.5 𝑏) 𝑃𝑑 = −1.8 4 1 ⋮ 26.2 −1 5 ⋮ 33. 𝑆 = 𝐼: 𝐼 = 𝑠 1 − 𝑡 𝑌 Government spending (𝐺) is constant regardless of the level of income. Consumption is determined by what is not saved (at a savings rate 𝑠) and not taxed (at the tax rate 𝑡): 𝐶 = 1 − 𝑠 1 − 𝑡 𝑌 Investments comes from savings. 𝐶 is the total consumption expenditure in an economy.07 𝑃𝑠 = 0. 3.41𝑄𝑠 + 3.5 𝑃𝑠 = 2𝑄𝑠 + 28.5 −18 7 ⋮ 23 1 8 ⋮ 36 −1 3 ⋮ 18.7537 Solve the following sets of equations using matrices: 𝑎) 4𝑥 + 𝑦 + 8𝑧 = 21 5𝑥 + 2𝑦 − 𝑧 = −6 6𝑥 + 3𝑦 + 2𝑧 = 3 𝑏) − 2𝑥 + 𝑦 + 2 = 0 2𝑥 + 𝑦 − 3𝑧 + 11 = 0 chapter three questions 1.5 𝑐) 𝑃𝑑 = −1. 71 . the savings rate 𝑠 takes a long time to change. as well as the bottom entry of the second column.5 12 13 ⋮ 104. If possible.5 4𝑥 + 𝑦 = 17.done first).6 𝑃𝑠 = 2.5 𝑗) 1 2 ⋮ 16. 𝐼 is the sum of all investment expenditure in an economy.4𝑦 2𝑦 − 3𝑥 − 1 = 0 11𝑥 − 3 = 1. 6.25 2. the government sets the tax rate 𝑡 and is usually the same for at least a year.5 Use matrices to solve for the following demand and supply equations: 𝑎) 𝑃𝑑 = −2𝑄𝑑 + 62. Using the Jacobian Method to solve a system of equations: 𝐴𝑏 𝑏 = 𝐴 Where 𝐴 is the determinant of the “variables” matrix. 𝑎) 𝑐) 𝑒) 𝑔) 𝑖) 𝑘) 𝑚) 5.5 2𝑥 + 𝑦 = 25.4 𝑙) 1 9 ⋮ 40. Convert the following sets of equations into matrix form: 𝑎) 𝑦 = 3𝑥 − 1 𝑏) − 3 = −𝑦 − 𝑥 𝑦 = −2𝑥 − 17 15 + 𝑦 = −3𝑥 𝑐) 17𝑦 − 13 = 𝑥 + 1 𝑑) 11𝑥 − 3 = 1.4𝑧 − 𝑦 = 𝑥 3𝑎 − 3𝑏 − 𝑐 = 1 Solve the following sets of equations without using matrices: 𝑎) 5𝑥 + 𝑦 = 14 2𝑥 + 8𝑦 + 2 = 0 𝑏) 4𝑦 = 𝑥 + 6 7𝑥 + 3𝑦 = 20 𝑐) 3𝑥 = −7𝑦 + 30.2𝑄𝑠 + 4.4𝑄𝑑 + 72. Matrices can have: A single solution No solutions (if there is no common intersection of all lines) Infinite solutions (if there are not enough equations and/or if some of the equations are the same) The macroeconomic model of an economy is: 𝑌 = 𝐶 + 𝐼 + 𝐺 Where 𝑌 is the total income of an economy. the pattern for the determinant equation is given by the matrix: + − + − + − + − + It is easiest to find the determinant of a matrix using the row/column with the most zeros.2 𝑑) 𝑃𝑑 = −4. so is assumed to be constant.75 𝑃𝑠 = 1. 5𝑥 + 2𝑦 + 𝑧 = 17 3𝑥 − 2𝑦 + 𝑧 = 16 𝑑) 5𝑥 + 2𝑦 + 𝑧 + 4 = 0 1𝑥 + 5𝑦 + 3𝑧 − 5 = −5 3𝑥 + 𝑦 + 4𝑧 + 13 = 0 If possible.12 1 0.1 2 2 2 0 0 𝑗) 𝑘) 𝑙) 2 1.75 −0.1 2.2 1.1 −1.07 2.43 3.5 ⋮ 19 −4. tax rate = 23% and savings rate = 9%.9 ⋮ 28.11 −0.g.5 1 1.21 −1.5 1 ⋮ 9. one or infinite) for the following matrices.8 2. plastic (𝑃).1 4.7.31 ⋮ 10.6 1. 4 7 18 1 −2 −2 𝑎) 𝑏) 𝑐) 1 −1 2 3 −2 3 3 1 3 3 −1 1.5 1 1/3 5/7 8/7 2. 15. Government spending = 600.00 Indonesia: 𝑃 = $9. Indonesia and Taiwan. A mobile phone manufacturer has three identical factories in Malaysia.2 0.00 Indonesia: 𝑇𝐶 = $2873. For the macroeconomic model of an economy with the following facts.5 𝑔) ) 𝑖) 2/3 1/7 2/9 1/3 8 7 1.00.5 9 ⋮ 95 𝑏) 1. To make mobile phones.91 4.1 ⋮ 20.2 2.00 Taiwan: 𝑃 = $4. 13. solve for 𝐶.1 2 2 1 1 𝑐) 10.5 3.75 2. Government spending = 200.51 4.22 4. For the macroeconomic model of an economy with the following facts. These three inputs all have different cost structures in the three countries (e.4 0 ⋮ −17.3 −8. 𝐿 = $25.08 4/3 2 2 ⋮ 26/3 7/9 −11/9 9 ⋮ 254/9 −1/9 1/2 1 ⋮ 43/36 1.5 5.5𝑥 + 3𝑦 + 7𝑧 = 92 1.00. 72 .34 −1. b) The amount of labour.2 12.2 ⋮ −18. 𝐿 = $14.23 1. tax rate = 20%.25 ⋮ −13 3. 𝐼 and 𝑌 using matrices.1 −1. Justify your answer.3 ⋮ 11. 𝑆 = $19. Find the solution to the following matrices using the Jacobian Method: −2 4 1 ⋮ 6 𝑎) −1 1 1 ⋮ 2 −2 1 9 ⋮ 16 7. plastic and silicone chips used in each of the three identical factories. Find the determinant of the following 3 × 3 matrices: −1 5 7 5 2 6 𝑎) 1 3 2 𝑏) 1 1 2 1 1 2 3 −3 2 4 2 4 −1 4 3 𝑐) 1 −2 2 𝑑) 0 8 0 1 0 −3 −2 1 −1 7 6 4 19 23 14 𝑒) 3 −1 6 𝑓) 20 41 18 −1 8 0 0 37 0 11.1 −2.00.00.7 ⋮ 67.5 −14 ⋮ −18 1 5 −3 ⋮ 8 5 −2 −1 ⋮ −3 𝑓) −1 −4 5 ⋮ 7 3 −21 18 ⋮ −9 5 1 3 ⋮ 18 9 −8 3 ⋮ 20 Find the determinant of the following 2 × 2 matrices. For the macroeconomic model of an economy with the following facts.5 𝑑) 𝑒) 𝑓) 1 3 1 1 3.6 4.3 1. due to unions) which are: Malaysia: 𝑃 = $7. 𝑎) 𝑏) 𝑐) 𝑑) 𝑒) 𝑓) 𝑔) ) 8.78 Determine the number of solutions (none.25 ⋮ 60 𝑐) 4. solve for 𝐶. 4𝑦 + 𝑥 = 𝑧 + 12 2. Government spending = 540.3 2.5 ⋮ 21 2.2 1.98 ⋮ 8. solve the following matrices: −3 2 1 ⋮ 4 2 3 3 ⋮ −1 1 −1 −2 ⋮ 4 −2 8 12 ⋮ 49 −1 7 0 ⋮ 23 4 1 2 ⋮ 13.00 Taiwan: 𝑇𝐶 = $3594.23 −1. 𝐼 and 𝑌 using matrices. 𝐼 and 𝑌 in terms of the savings rate 𝑠.24 3.5 4 7 −7 ⋮ −7 1 −12 −11 ⋮ −49 4 5 11 ⋮ 49 1. 𝐿 = $11.1 4.5 7. all running at capacity.31 3.00. silicone chips (𝑆) and labour (𝐿) is required.00 If the total costs of the three factories is: Malaysia: 𝑇𝐶 = $2853. solve for 𝐶.5 1.75 −3.21 ⋮ −0.1 ⋮ −16.3 8 ⋮ 20.5 3. 14.00 Determine: a) Equations relating inputs and costs for the three factories. tax rate = 10% and savings rate = 10%. 𝑆 = $20.5 −2. 𝑆 = $21. 𝑎) 𝑐) 𝑒) 9.00.4 ⋮ −2.25 8.25 1/3 1/2 1/9 ⋮ 1/3 𝑑) 1/6 1/4 2/9 ⋮ 8/30 5/6 3/4 1/3 ⋮ 2/3 12.5 2. 2 1 8 ⋮ 19 2 1 3 ⋮ 15 𝑏) −2 4 10 ⋮ 20 3 1 1 ⋮ 12 5 −2 −3 ⋮ 21 3 −1 −2 ⋮ 5 3 2 7 ⋮ 0 2 1 5 ⋮ 7 8 −3 2 ⋮ 4 𝑑) −8 −3 4 ⋮ 2 5 2 1 ⋮ 1 −10 −1. 8 4.11 Defining Non-Linear Functions Defining a Quadratic Function Quadratic Graphs Sketching Quadratic Functions The Cubic Function The Exponential Function The Logarithmic Function Logarithmic Graphs The Natural Number 𝑒 The Hyperbolic Function Economic Applications 74 75 76 77 80 82 84 88 89 90 91 93 94 Chapter Summary Chapter Four Questions 73 .9 4.4 4.5 4.6 4.Chapter 4 Non-linear functions Moving away from lines towards curves 4.7 4.10 4.1 4.2 4.3 4. 𝑦 has been used when referring to linear functions (lines). 𝐷 𝐸 𝐹 𝐴 𝐺 𝐵 𝑥 𝐶 𝑦 H M 𝑦 J L 𝑥 K Different people denote functions differently. 74 . 𝑟 etc. Graphically. 𝐻) plots two different 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 (𝐽 and 𝐾). there is only one 𝑦 − 𝑣𝑎𝑙𝑢𝑒 plotted by the curve. Theory: to find if a curve is a function. That is. there is only one value for each function (colour). the function may be defined as: 𝑅 𝑄 = 50𝑄 − 0. Common non-linear functions include: Cubic 𝑦 Exponential Logarithmic The curve above is a function as all 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 plot to only one 𝑦 − 𝑣𝑎𝑙𝑢𝑒: 𝐴→𝐵. This does not mean that two different 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 cannot give the same 𝑦 − 𝑣𝑎𝑙𝑢𝑒. for every value on the 𝑥 − 𝑎𝑥𝑖𝑠. This is the only thing that makes it a function. Two different 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 (such as 𝐹 and 𝐺) can plot to a single 𝑦 − 𝑣𝑎𝑙𝑢𝑒. A function is still a function if multiple 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 plot to a sinlge 𝑦 − 𝑣𝑎𝑙𝑢𝑒. and therefore a single 𝑥 − 𝑣𝑎𝑙𝑢𝑒 (e. Hyperbolic 𝑥 Quadratic Look closely at the graphs above and you will see that for each 𝑥 − 𝑣𝑎𝑙𝑢𝑒 (on the horizontal axis). 𝑚. the letters denote the vertical axis. and 𝑥 is the horizontal axis. 𝐶→𝐷. Part of the curve passes the vertical line test. but there are many ways of defining functions: 𝑓(𝑥).4. apply the vertical line test: 𝐴 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑓 𝑎𝑙𝑙 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑙𝑖𝑛𝑒𝑠 𝑐𝑟𝑜𝑠𝑠 𝑡𝑒 𝑐𝑢𝑟𝑣𝑒 𝑜𝑛𝑙𝑦 𝑜𝑛𝑐𝑒 The curve above passes the vertical line test. Sometimes the letters might mean something: if referring to revenue from selling a certain quantity of computers. but unless the whole thing passes.g. (the 𝑦 − 𝑎𝑥𝑖𝑠). then the curve is a function.1 defining non-linear functions Theory: a function is defined as having only one 𝑦 − 𝑣𝑎𝑙𝑢𝑒 for every 𝑥 − 𝑣𝑎𝑙𝑢𝑒. All these mean the same thing: they are all “𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑥”. 𝐷(𝑥). For example. but the following curve does not pass the vertical line test. so it is a function. they are all functions. it is not a function. 𝑔(𝑥). therefore.6𝑄 2 Where 𝑄 is the quantity of computers sold (the 𝑥 − 𝑎𝑥𝑖𝑠) and 𝑅 𝑄 means “𝑟𝑒𝑣𝑒𝑛𝑢𝑒 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑠𝑜𝑙𝑑”. If this occurs for all 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 on a given curve. All these curves pass the vertical line test (try it!). 𝑏 and 𝑐 for 𝑓 𝑥 = −3𝑥 2 − 2𝑥 + 4 Solution: 𝑎 = −3. 𝑏 and 𝑐 for 𝑦 = 2𝑥 2 + 3𝑥 − 5 Solution: 𝑎 = 2. rearrange 𝑥 them into the general form using the crab-claw method from Chapter 1. 𝑏 and 𝑐. and then follow the lines: 𝑦 = 𝑥 + 2 (𝑥 − 3) 𝑦 = 𝑥 2 + 2𝑥 − 3𝑥 − 6 𝑦 = 𝑥 2 − 𝑥 − 6 So 𝑎 = 1. 𝑏 = 3 and 𝑐 = −5. 𝑏 and 𝑐 in: 𝑥 = 2 − 18𝑥 Solution: 𝑥 = −18𝑥 + 2 2 2 Plan: simplify using the crab-claw method. When presented with other forms. Example 4: find 𝑎. Example 2: find 𝑎. Example 3: find 𝑎. 3𝑦 = 2 𝑥 + 1 3𝑥 − 1 Then apply the crab-claw on the right side: 3𝑦 = 2 𝑥 + 1 3𝑥 − 1 3𝑦 = 2 3𝑥 2 + 3𝑥 − 𝑥 − 1 The 2 out front of the square brackets goes into every term in the square brackets. Determine which of the following curves are functions. Example 5: find the values of 𝑎. 𝑏 and 𝑐 for 𝑔 𝑥 = 2 − 7𝑥 2 + 3𝑥 Solution: rearrange: 𝑔(𝑥) = −7𝑥 2 + 3𝑥 + 2 So 𝑎 = −7. 𝑏 = 3 and 𝑐 = 2. The general form of the quadratic is not the only quadratic form. otherwise it would be a linear function. after the inside of the square brackets have been simplified: 3𝑦 = 6𝑥 2 + 4𝑥 − 2 Isolate 𝑦 by dividing both sides by 3: 𝑦 = 6𝑥 2 + 4𝑥 − 2 4 2 = 2𝑥 2 + 𝑥 − 3 3 3 75 . Note that 𝑏 or 𝑐 can equal zero.2 defining a quadratic function The quadratic is by far the most common non– linear function you will come across. Example 6: find the general form of 3𝑦 = 2 𝑥 + 1 (3𝑥 − 1) Plan: simplify and rearrange using the crab-claw Solution: Multiply both sides by (3𝑥 − 1) to get rid of it on the left and have it as a multiplication on the right. It involves the squaring of a variable. 𝑏 = 0 and 𝑐 = 2. 𝑏 = −2 and 𝑐 = 4. 𝑏 and 𝑐 for 𝑦 = 𝑥 + 2 (𝑥 − 3) 𝐵 𝐴 𝑦 𝐷 𝐸 𝐶 4.where 𝑎 = −18. but 𝑎 cannot. Exercises: 1. Example 1: find 𝑎. 𝑏 = −1 and 𝑐 = −6. Solution: draw the ‘crab-claw’. however it is the easiest to work with. It is easier to have the quadratic in the general form to get the constants 𝑎. Theory: the general form of a quadratic is: 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 Where 𝑎. 𝑏 and 𝑐 are constants. The difference between the three graphs is the value of 𝑎. the steeper the curve (black). the turning points (minimum points in the cases so far) have been on the 𝑦 − 𝑎𝑥𝑖𝑠.e. so don’t be one of them! Exercises: 1. Determine the value of 𝑎.5𝑥 2 𝑥 Three different quadratics are drawn above in general form with 𝑏 = 0 and 𝑐 = 0 (i. The shape of the graphs are identical. 𝑏 and 𝑐 in the following quadratics: 𝑎) 𝑦 = 6𝑥 2 − 5𝑥 − 1 𝑏) 𝑦 = 1 − 2𝑥 + 𝑥 2 𝑐) 𝑦 − 6 − 𝑥 2 = 2𝑥 2𝑥 2 1 1 𝑑) 𝑦 = + − 𝑥 3 2 3 𝑒) 𝑦 = 8 − 8𝑥 − 3𝑥 2 𝑦 − 2 𝑓) = 2𝑥 − 1 𝑥 + 1 𝑔) 𝑦 = 𝑥 + 1 2 − 8𝑥 + 23 ) 𝑦 − −𝑥 + 2 2 = 2𝑥 − 1 2 𝑦 = 𝑥 2 − 1 𝑥 𝑦 = 𝑥 2 − 3 4. 𝑦 = 𝑥 2 + 3 𝑦 ≠ 𝑥 2 + 22 You cannot simply put the squared into every term inside the brackets.3 quadratic graphs Quadratics have a distinctive shape which you must recognise. Theory: the value of 𝑐 in the quadratic equation determines the vertical displacement of the graph. The following functions are of the form: 𝑦 = 𝑥 2 + 𝑐 That is. but the 𝑥 + 2 as two separate parts: 𝑦 = 𝑥 + 2 (𝑥 + 2) − 3 𝑦 = 𝑥 2 + 2𝑥 + 2𝑥 + 4 − 3 Simplify: 𝑦 = 𝑥 2 + 4𝑥 + 1 Note: remember from Chapter 1: 𝑥 + 2 2 2 must first be rewritten 𝑦 = 0. similarly. Look at all the graphs so far. 𝑦 = 𝑎𝑥 2 ). However. Theory: The larger the value of 𝑎. 𝑏 is set to zero.Example 7: find the general form of 𝑦 = 𝑥 + 2 2 𝑦 = 2𝑥 2 𝑦 𝑦 = 𝑥 2 − 3 Plan: simplify and rearrange using the crab-claw Solution: this is solved using the crab-claw method. This is where many students stuff up. if the value of 𝑏 is not zero. except for their vertical position (vertical displacement). the smaller the value of 𝑎. 76 . there is a vertical and horizontal shift. the shallower the curve (blue). −6 is the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. this would be: 0 = −2𝑥 2 + 8𝑥 − 6 Rearranging this to isolate 𝑥 is very difficult. and either two. Exercises: 1. To find the 𝑥– 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡(𝑠). Solution: 𝑦 = −2(0)2 + 8 0 − 6 = −6 The values of 𝑎. Finally.𝑦 2. whereas when the value of 𝑎 is negative. 𝑏 and 𝑐 still change the graph in a similar way. In the example above. set all the 𝑦’s equal to zero. but the sign of 𝑎 determines if it is a ‘smiley’ or ‘sad’ face. one or no 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠. The formula is called 𝑦 𝑥 the Quadratic Formula (QF) which you must memorise. Theory: changing the value of 𝑏 in the general quadratic form shifts the graph both horizontally and vertically. 𝑦 𝑦 = −𝑥 2 + 4𝑥 𝑦 = −𝑥 2 − 2𝑥 𝑥 Example 1: Find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 of 𝑦 = −2𝑥 2 + 8𝑥 − 6 Plan: set all the 𝑥’s equal to zero. 𝑦 = −3𝑥 2 𝑦 = 𝑥 2 + 2 2 𝑦 = 𝑥 + 10𝑥 + 25 𝑦 = −𝑥 2 + 8𝑥 Match the graphs to the equations 𝑦 𝑦 = 𝑥 − 2𝑥 𝑥 𝑦 = 𝑥 2 + 4𝑥 𝑥 Notice how the turning point has moved away from the 𝑦 − 𝑎𝑥𝑖𝑠 and there has also been a vertical shift. so a formula is used to solve for 𝑥. set all the 𝑥’s in the equation equal to zero. Theory: To find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 of any function. Match the graphs with the equations. The 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is also simply the value of the constant 𝑐. a note about the sign of 𝑎. the shape becomes inverted. Theory: a positive value of 𝑎 will result in quadratics with a shape of a ‘smiley’ face. To find these three main features of quadratics. 77 .4 sketching quadratic functions All quadratics can be described as having: a 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. So the point 0. theory from Chapter 2 must be applied. 𝑦 = − 3𝑥 2 + 2 𝑦 = 𝑥 − 3 2 + 3 𝑦 = 3𝑥 2 + 10𝑥 + 28 −𝑦 + 5 = 𝑥 2 𝑦 = −𝑥 2 + 2𝑥 2 4. a ‘sad’ face. a turning point. it makes finding the turning point (TP) easier. Plan: plot and label the coordinates. Substitute 𝑥 = 2 into the original equation to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the turning point: 𝑦 2 = −2 2 =2 Giving the turning point (2.Theory: the Quadratic Formula is: 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑠 = −𝑏 ± 𝑏 2 − 4𝑎𝑐 2𝑎 Theory: to find the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the turning point (TP). all over two 𝑎”. Example 3: find the coordinates of the turning point of the quadratic 𝑦 = −2𝑥 2 + 8𝑥 − 6 Plan: use the Quadratic Formula ( but ignore everything under the square root sign) to find the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the turning point.2). substitute this 𝑥 − 𝑣𝑎𝑙𝑢𝑒 into the original function. then substitute this value into the original equation to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒. only look at the circled part): 𝑇𝑃 𝑥 − 𝑣𝑎𝑙𝑢𝑒 = −𝑏 ± 𝑏 2 − 4𝑎𝑐 −𝑏 = 2𝑎 2𝑎 This is a very important formula.0) and (3.0).e. 2 +8 2 −6 = −8 + 16 − 6 78 . so once we find the 𝑥 − 𝑣𝑎𝑙𝑢𝑒. Solution: 𝑇𝑃 𝑥 − 𝑣𝑎𝑙𝑢𝑒 = −𝑏 −8 = =2 2𝑎 2 −2 −8 ± 16 −4 −8 ± 4 = → 𝑥 = 1 𝑂𝑅 𝑥 = 3 −4 The last step is split into two. the corresponding 𝑦 − 𝑣𝑎𝑙𝑢𝑒 needs to be found. it is separated out into a + and a − to get two answers: −8 + 4 −4 −8 − 4 −12 𝑥 = = = 1 𝑂𝑅 = =3 −4 −4 −4 −4 So there are two 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠: (1. If you memorised the quadratic formula. Example 4: Plot the results from Examples 1 − 3. Instead of having the ± sign. and it may be easier to memorise the words “negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐. The only point that remains to be found to determine the nature of a quadratic is the turning point (the point where the function “turns”). A different short-cut is knowing that the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the turning point will always be half way between the two 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠. Remembering that a point has an 𝑥 and 𝑦 value. Example 2: determine the roots of 0 = −2𝑥 2 + 8𝑥 − 6 Plan: use the quadratic formula 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 = Solution: −8 ± 82 − 4 −2 (−6) 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 = 2(−2) = = −8 ± 64 − 48 −4 −𝑏 ± 𝑏 2 2𝑎 − 4𝑎𝑐 Then to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the turning point. This formula is finding where a quadratic function crosses the 𝑥 − 𝑎𝑥𝑖𝑠 and these point(s) are called roots. then join up all the dots with a nice curve. rewrite the Quadratic Function and only look at the things not under the square root sign (i. The value underneath the squareroot sign determines this: 𝑏 2 − 4𝑎𝑐 2 −𝑏 ± 𝑏 2 − 4𝑎𝑐 2𝑎 Use the part of the QF not under the square root sign to find the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the TP. Find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 by setting 𝑦 = 0 and then applying the Quadratic Formula: 2 − 4 2 32 𝑥 𝑦 = 𝑥 − 5 2 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 = Theory: From the Quadratic Formula. then the curve crosses (touches) the 𝑥 − 𝑎𝑥𝑖𝑠 only once (the square root of zero is zero). then the curve will be fine. If it is positive → 2 intercepts.2) (3. If the value of 𝑏 2 − 4𝑎𝑐 is negative. Solution: 𝑏 2 − 4𝑎𝑐 = −16 = 256 − 256 = 0 ∴ only one x − intercept Example 6: Sketch 𝑦 − 8𝑥 2 + 26𝑥 = −7 Plan: rearrange to get the general form. and if negative → 0 intercepts. then the curve will cross the 𝑥 − 𝑎𝑥𝑖𝑠 twice (as the − 26 0 − 7 79 . the red function only just touches the 𝑥 − 𝑎𝑥𝑖𝑠 (cuts it once). and the blue line is a free–hand sketch. then put back into original equation to get the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the TP.Solution: 𝑦 (1.0) 𝑥 square root of a positive number will have two solutions – a positive and negative number). the value of 𝑎 must be negative (𝑎 = −2). then find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting all 𝑥’s to zero.0) (0. it is possible to determine how many times a quadratic crosses the 𝑥 − 𝑎𝑥𝑖𝑠. and goes through the labelled coordinates. Also notice that the turning point is half way between the two 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠. As long as it approximately looks like a quadratic. determine how many times the following equation crosses the 𝑥 − 𝑎𝑥𝑖𝑠 𝑦 = 2𝑥 2 − 16𝑥 + 32 Plan: find the value of the part under the square root sign in the Quadratic Formula: 𝑏 2 − 4𝑎𝑐. on the following graphs notice that the blue function cuts the 𝑥 − 𝑎𝑥𝑖𝑠 twice. Notice that this graph is a ‘sad’ face. On a related topic. then the curve does not cross the 𝑥 − 𝑎𝑥𝑖𝑠 (as the The dotted black line is what the graph should be. which from the last section. 𝑦 𝑦 = 𝑥 2 + 4𝑥 + 8 𝑦 = 𝑥 2 + 2𝑥 + 1 square root of a negative number does not exist). Example 5: using the Quadratic Formula. and the black function does not cross the 𝑥 − 𝑎𝑥𝑖𝑠 at all (cuts it zero times). Solution: rearrange into the general form 𝑦 = 8𝑥 2 − 26𝑥 − 7 𝑦 − 𝑖𝑛𝑡: 𝑦 0 = 8 0 2 If the value of 𝑏 − 4𝑎𝑐 is positive. If the value of 𝑏 2 − 4𝑎𝑐 equals zero. if zero → 1 intercept. −6) (2. 25 − 4.625. 𝑥 3 ).0) 2 𝑦 − 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑇𝑃: 𝑦(1. For example: 𝑦 = 2𝑥 3 − 4𝑥 2 + 3𝑥 − 17 80 . 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡: −𝑏 −(−26) 26 𝑇𝑃 𝑥 − 𝑣𝑎𝑙𝑢𝑒 = = = = 1.5 𝑂𝑅 𝑥 = −0.625) = 8 1.125) Example 7: Sketch 3 𝑥 − 3 = 𝑥 2 − 𝑦 Plan: set all 𝑥’s to zero to find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. then rearrange to get into general form. 𝑎) 𝑦 = −𝑥 2 − 5𝑥 − 6 𝑏) 𝑦 = 𝑥 + 2 2 − 1 𝑐) 𝑦 = 𝑥 − 1 2 + 𝑥 + 2 2 𝑑) 𝑦 = − −𝑥 + 1 1 − 𝑥 + 1 𝑒) 𝑦 = 𝑥 2 + 𝑥 − 1 2 − 3 𝑓) 𝑦 = − 𝑥 − 2 2 − 2 Solution: rearrange into the general form 3𝑥 − 9 = 𝑥 − 𝑦 𝑦 = 𝑥 2 − 3𝑥 + 9 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡: 𝑦 = (0)2 − 3 0 + 9 = 9 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 coordinate: (0. −7) 𝑥 − 𝑖𝑛𝑡: 0 = 8𝑥 2 − 26𝑥 − 7 −(−26) ± (−26)2 − 4 8 (−7) 2(8) 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 = = −(−3) ± (−3)2 − 4 1 (9) 2(1) 3 ± 9 − 36 3 ± −27 = 2 2 𝑥 − 𝑖𝑛𝑡 𝑠 = = = Since there is a negative number under the square root sign.5 the cubic function Theory: The cubic function is a function where the highest power of 𝑥 is 3 (i.5) = 1.e.0) and −0.625.75) 𝑥 𝑥 (0. Find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 (if any) for the following quadratics: 𝑎) 𝑦 = 𝑥 2 + 2𝑥 − 15 𝑏) 𝑦 = 2𝑥 2 − 20𝑥 + 42 𝑐) 𝑦 = 3𝑥 2 + 18𝑥 + 27 𝑑) 𝑦 = 𝑥 2 + 2𝑥 + 9 2.5 + 9 = 6.75) − 3(1.25. then use the QF to find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 and 𝑇𝑃.6. it means there are no 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠. there are no 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 and it is a “smiley” face as 𝑎 is positive: 𝑦 − 26(1.5.25.9) 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠: 2 4.5) + 9 = 𝑦 1. 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 = −𝑏 ± 𝑏 2 − 4𝑎𝑐 2𝑎 Exercises 1.5. and remember.9) 𝑦 (1.125) (3.6.125 (−0. 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡: 𝑥 − 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑇𝑃 = − 𝑏 −3 3 =− = = 1.0 .625 = −28.5 TP coordinate: (1. −7) (1.5. −28.75 Plot all the points. −28.5 = 2.625) − 7 ∴ 𝑇𝑃 𝑎𝑡 (1.5 2𝑎 2 1 2 2 26 ± 676 + 224 16 26 ± 900 16 26 ± 30 = → 𝑥 = 3.= −7 ∴ 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑎𝑡 (0. Sketch the following quadratics.0) (0.25 16 Coordinates of 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 are: (3.5.625 2𝑎 2(8) 16 𝑇𝑃 𝑦 − 𝑣𝑎𝑙𝑢𝑒 : 𝑦(1. which is ‘sharper’ than the red one. but when 𝑎 is negative. count how many times each colour crosses the 𝑥 − 𝑎𝑥𝑖𝑠. 𝑐 and 𝑑. 𝑐 and 𝑑. but the highest power is 3 (i. That is. The number of times a cubic function crosses the 𝑥 − 𝑎𝑥𝑖𝑠 is determined by the values of 𝑎. Just like in quadratics. 𝑏. as 𝑥 becomes more positive (i. 𝑏. 𝑐 and 𝑑 are constants. when the value of 𝑥 becomes more negative (to the left of the origin on the 𝑥 − 𝑎𝑥𝑖𝑠). 𝑏. the ends of the graph extend to the top right and bottom left ↙↗ . the ‘ends’ of the graph go in different directions. the cubed term dominates over all other terms. the value on the 𝑦 − 𝑎𝑥𝑖𝑠 also becomes more negative. with the sign of 𝑎 determining the direction the ends of the graph extends. it is the values of 𝑎. the sign of 𝑎 in a cubic function has an interpretation. Notice the sign of 𝑎 in Here. 𝑐 and 𝑑 that determine the sharpness of this ‘𝑆’ shape. The blue graph is much ‘sharper’ than the black one. a cubic can cross the 𝑥 − 𝑎𝑥𝑖𝑠 once. similarly. and the grey curve cuts the 𝑥 − 𝑎𝑥𝑖𝑠 only once.5𝑥 3 𝑥 Also. Theory: The general form of a cubic function is: 𝑦 = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 Where 𝑎. The reason why 𝑎 is important is that as 𝑥 becomes very large in a positive direction or in a negative direction. to the right of the origin on the 𝑥 − 𝑎𝑥𝑖𝑠). Compare this to a quadratic where both ends of the graph go in the same direction (either both ends go up. These simple cubic functions have an ‘𝑆’ shape: 𝑦 𝑦 = 𝑥 3 𝑦 𝑦 = 𝑥 3 − 𝑥 2 − 21𝑥 + 45 𝑦 = 𝑥 3 + 2 𝑥 𝑦 = 𝑥 3 − 𝑥 2 − 20𝑥 These functions still have the general ‘𝑆’ shape. there is only one 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 (through the origin). or both ends go down). 𝑏. the value on the 𝑦 − 𝑎𝑥𝑖𝑠 also becomes more positive. the 2𝑥 3 ) so this is a cubic function. not a quadratic function. Theory: depending on the values of 𝑎. 𝑐 and 𝑑 are all zero. twice or three times.e. The simplest cubic is: 𝑦 = 𝑎𝑥 3 where 𝑏.Notice that there is an 𝑥 2 . 𝑦 = 0. Theory: when 𝑎 is positive.e. The blue one crosses three times. the ends of the graph extend to the top left and bottom right ↖↘ . More complex cubic functions look like: 81 . the black one crosses twice (touching the 𝑥 − 𝑎𝑥𝑖𝑠 𝑦 = 2𝑥 3 counts as a cross). and unlike the quadratic functions. 𝑏 and 𝑐 are constants. 𝑎 is a number that determines the growth rate (e. After every round (or time-period). the amount of money (𝑚) after 𝑡 turns (assuming you keep winning) can be defined by: 𝑚 = 100 2𝑡 The 2 comes from the fact that earnings are doubled every time-period. You are given $100 to go to the casino and have fun. and the 100 from the initial amount. You win. Every time you play. Theory: an exponential function is one which has a variable in the index. doubling. and end up with $1600. It determines the shape of the graph and where it intersects the 𝑦 − 𝑎𝑥𝑖𝑠. but you don’t win the same amount every time. Now you have $800. you doubled your earnings. This is what exponential growth means.𝑎 = 3 etc. but every round. After the first round.the following graph and the direction the ends extend. You bet it all again. Match the following graphs to the equations: 𝑦 𝑎) 𝑦 = 𝑥 + 4 𝑥 + 2 2 + 𝑥 𝑏) 𝑦 = 𝑥 3 + 4 𝑐) 𝑦 = −3𝑥 3 + 2. the final amount is a certain ratio of the amount of the previous timeperiod. but doesn’t change the shape of the graph. The general form of an exponential function is 𝑥 𝑦 = 𝑏(𝑎 𝑥 ) + 𝑐 Where 𝑎.g. You then have $200. 𝑦 = −𝑥 + 5 𝑥 In the example above. 𝑐 a constant that shifts the graph up or down. an intuitive understanding is needed. You play the coin-toss. which are not the same. tripling. 𝑏 is the initial amount. and the 82 also the value of the asymptote (the value the function will approach but will never reach).6 the exponential function Before defining the mathematical exponential function. you double your earnings. 𝑦 𝑦 = −𝑥 − 𝑥 + 21𝑥 + 45 3 2 last round you won $800. and (somehow) can predict the outcome every single time.1 as any number to the power of 0 equals 1.) 3 Exercises 1. and bet it again. . and now you have $400. and win again. The first round you won $100.5 𝑑) 𝑦 = 𝑥 − 3 𝑥 + 1 −3 − 𝑥 Hint: expand all the brackets before deciding. You bet it once more and win. The simplest exponential is where 𝑏 = 1 and 𝑐 = 0: 𝑦 = 𝑎 𝑥 This exponential function cuts the 𝑦 − 𝑎𝑥𝑖𝑠 at 0. you win $100. It is 4.𝑎 = 2. The shape of an exponential function: 𝑦 𝑦 = 2𝑥 Example 1: sketch the two exponential functions on a single set of axes 1. 𝑦 = 3 2𝑥 + 4 2. 𝑦 = 0.5 2−𝑥 − 2 Plan: to find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠, set the 𝑦= 3(2𝑥 ) 0,3 0,1 0, −2 𝑥− 𝑣𝑎𝑙𝑢𝑒𝑠 to zero. The asymptote is the number added to the end of the exponential. 𝑦 = 2𝑥 − 3 𝑥 Solution: for the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 1. 𝑦 = 3 20 + 4 = 3 + 4 = 7 2. 𝑦 = 0.5 20 − 2 = −1.5 For the asymptotes 1. 𝑦 = 4 2. 𝑦 = −2 Notice that for the first function above, there is a 3 in front which makes the graph much steeper. Conversely, the 0.5 in the second function makes it shallower. There is also a negative sign in front of 𝑥 in the second function above meaning it is flipped along the 𝑦 − 𝑎𝑥𝑖𝑠. 𝑦 The black line is the simplest exponential function, and crosses the 𝑦 − 𝑎𝑥𝑖𝑠 at 𝑦 = 1, as 20 = 1. It is also the value of 𝑏 (= 1). The asymptote is at 𝑥 = 0 as the further left the curve extends, the closer it gets to the 𝑥 − 𝑎𝑥𝑖𝑠 (i.e. 𝑦 = 0). The red line has the 3 out front, meaning that it crosses the 𝑦 − 𝑎𝑥𝑖𝑠 at 𝑦 = 3 but also that the general shape has been changed to be steeper. Similar to the black line, the asymptote is at 𝑥 = 0. The blue line is the same as the black line, except that it has 3 subtracted from it. This shifts the graph down 3 units and also changes the asymptote to 𝑦 = −3. The graph will get closer and closer to 𝑦 = −3 but will never reach it. The other two functions have 𝑐 = 0 which means the asymptote for those two functions is the 𝑥 − 𝑎𝑥𝑖𝑠. Theory: to sketch an exponential function, the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 and the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the asymptote are required (remember, an asymptote is a value that the function approaches but never reaches). The 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is found by setting all 𝑥’s equal to zero. The 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the asymptote is any number added onto the end of the exponential function (i.e. the number 𝑐). 𝑦= 3 2𝑥 + 4 𝑦 = 4 𝑥𝑦 = 0.5 2−𝑥 − 2 𝑦 = −2 A concept similar to growth is that of decay; the reverse of growth. For growth, the more time passes, the faster the thing grows. In decay, the more time passes, the slower things decay. Theory: the general form of the decay function is very similar to that of growth, except there is a negative in front of the index x: 𝑦 = 𝑏(𝑎−𝑥 ) + 𝑐 83 The graph which was shown in the last exercise is one of decay, as are the following: 𝑦 The answer is 𝑥 = 2, as 102 = 100. These are easy, but how about: 10𝑥 = 50 The answer is not 𝑥 = 1.5 because 101.5 ≈ 31.6. It is exactly this sort of situation where logs are used. 𝑦= 2−𝑥 𝑦= 3 2−𝑥 + 1 𝑦 = 1 Theory: the definition of a logarithm is: For 𝑥 𝑎𝑏 = 𝑐 𝑏 = log 𝑐 = log 𝑎 𝑐 log 𝑎 Exercises: 1. Match the following graphs with the equations: Where 𝑏 is called the index, and 𝑎 is called the base. However, with the following log rules, this definition can be easily found. There are some log rules you will have to learn to be able to solve these problems. Without knowing these rules, you will make lots of mistakes. Back to the previous problem: 𝑥 10𝑥 = 50 Theory: RULE 1: taking logs of the whole of both sides separately allows the index to be brought down in front of the log. log 𝑎𝑏 = 𝑏 log 𝑎 Example 1: solve for 𝑥 in 10𝑥 = 50 Solution: log the whole of both sides log 10𝑥 = log 50 Then bring down any exponents log 10𝑥 = log 50 𝑥 log(10) = log(50) This is read as “𝑥 × log(10) is equal to log(50)”. log(10) and log(50) are just numbers (the values can be found using a calculator). Treat it as you would any other number. To isolate 𝑥 in: 5𝑥 = 15 Divide both sides by 5. It is the same for 𝑥 log(10) = log(50) 𝑦 2. 𝑎) 𝑦 = 3 2𝑥 𝑏) 𝑦 = 3 2−𝑥 𝑐) 𝑦 = 5 3𝑥 𝑑) 𝑦 = 4 2−𝑥 + 5 Sketch the following equations on a single set of axes (they must be correct, relative to one another!): 𝑎) 𝑦 = 5 2𝑥 − 4 𝑏) 𝑦 = 2−𝑥 + 3 𝑐) 𝑦 = −2 3𝑥 + 1 𝑑) 𝑦 = 4 2−𝑥 + 5 4.7 the logarithmic function For an exponential function such as: 𝑦 = 𝑎 𝑥 𝑦 is already isolated, but sometimes 𝑥 needs to be isolated. It is impossible to isolate 𝑥 without using logarithms. A logarithm (log for short) is a mathematical method used to solve exponents. Intro Example 1: use common sense to solve: 10𝑥 = 10 Obviously 𝑥 = 1. Now how about: 10𝑥 = 100 84 Divide both sides by log(10): 𝑥 = log(50) log(10) Move the log 73 onto the other side by subtracting it from both sides to get: log 3𝑥 = log 4 − log 73 RULE 1 can now be applied to bring 𝑥 out front on the left side: 𝑥 log 3 = log 4 − log 73 To solve for 𝑥, divide the whole of both sides by log 3: 𝑥 = log 4 − log 73 ≈ −2.6435 (4𝑑. 𝑝. ) log 3 That is the answer. Always give this exact answer, but you can also give a decimal answer; put it into your calculator and you should get 1.69897 (5𝑑. 𝑝. ). The place where most students fail is not realising that something has to be logged. The logarithm changes one thing to another. If you log nothing, then how can it be changed? When travelling in Europe, dollars need to be exchanged for Euros, however if you give the bank nothing, nothing will be exchanged! However, if you give the bank $100, then that $100 will be changed to Euros. The same applies to logs; you have to log something! Example 2: solve for 𝑥 73 3 𝑥 Finally, the last rule is similar to RULE 2, but it is for division instead of multiplication: Theory: RULE 3: the log of a division of two terms is equal to the subtraction of the separate logs. 𝑎 log = log 𝑎 − log 𝑏 𝑏 Example 3: solve for 𝑥 without simplifying first = 4 4𝑥 = 12 56 Solution: log the whole of both sides: log 4𝑥 = log 12 56 Solution: You could divide both sides by 73 and then apply RULE 1. But if both sides were not divided by 73 to begin with, and instead logged: log[ 73 3 𝑥 ] = log[4] The square brackets on the left is 73 × 3𝑥 , so the index of 𝑥 cannot be brought down front. It is not 73 × 3 𝑥 , but rather two separate parts being multiplied together (see Chapter 1). Theory: RULE 2: the log of a multiplication of two terms is equal to the addition of the logs of the separate terms. log(𝑎 ∙ 𝑏) = log 𝑎 + log 𝑏 Example 2 (cont): solving for 𝑥 in log[ 73 3 𝑥 ] = log[4] Solution: by RULE 2, 73 and 3𝑥 can be separated into two logs, with an addition sign between: log 73 + log 3𝑥 = log 4 Use RULE 3 to separate the 4𝑥 and the 56 by subtracting their logs: log 4𝑥 − log 56 = log 12 Use RULE 1 to bring 𝑥 out front and rearrange: 𝑥 log 4 = log 12 + log 56 Use RULE 2 in reverse on the right side: 𝑥 log 4 = log 12 × 56 Rearrange and simplfy: 𝑥 = log 12 × 56 log 672 = ≈ 4.6962 (4𝑑. 𝑝. ) log 4 log 4 Theory: all rules can be used forwards or backwards. It all depends on what is being found. Think about it; if RULE 2 can be used to separate the log of a multiplication into two log additions, then there is no reason why it cannot be used to 85 3 to the other side. Log the whole of both sides: log 7 = log 4.1) Using RULE 3.3 + log(1. and use log rules to solve for 𝑡. then solve using log rules. Solution: 2 = 1. Example 4: inflation is the growth in prices. Similarly. this will be explained later in the chapter) Remember that you can use these rules backwards or forwards.1 The short-cut is to simplify first.3 Then divide both sides by log 1. The estimated inflation rate.02 𝑡−1 + 1 51 The government wants to find out when the price level will double. if two logs are multiplied. Example 3: a country’s Gross Domestic Product (𝐺𝐷𝑃) has the following form (in $billions): 𝐺𝐷𝑃 = 4. then use RULE 1. 𝑝 𝑡 = 2.02 𝑡−1 Use RULE 1 to bring down the two exponents in brackets: 86 . 𝑝.1) 4.3 1.1𝑡 Separate the right side into two logs using RULE 2: + log 1. which is like saying 𝑦 0 = 1.3 𝑡 = ≈ 5. but the following is the harder way. Note: when given: log 672 log 4 many students think they can bring everything inside of a single log: log 672 672 = log log 4 4 This is not correct! Never write this! When two logs are divided. Try find the short-cut.01 51 𝑡 1. It all depends on what is being isolated and where it is. and bring 𝑡 down using RULE 1: log 7 − log 4. relative to the base year 0. so learn them.3 1.02 𝑡−1 ] Using RULE 2.3 = 𝑡 log(1.113 𝑦𝑒𝑎𝑟𝑠 (3𝑑.1𝑡 When will 𝐺𝐷𝑃 reach $7billion? Plan: substitute 7 for the left side in the above equation (as 𝐺𝐷𝑃 is set to $7b).01 51 𝑡 1. Practice with your calculator to make sure. they cannot be simplified further. RULE 4b: log 10 = 1 (ln 𝑒 = 1.1𝑡 ) Move log 4.02 𝑡−1 Log the whole of both sides: log 2 − 1 = log[ 1. bring the left side under a single log: log 7 = 𝑡 log(1. is determined by the following function: 𝑝 𝑡 = 1.1 𝑡 log 7 = log 4. The same applies to both RULE 1 and RULE 3.01 51 𝑡 This could be solved in a number of ways.bring two log additions into a single log multiplication.3 1.02 𝑡−1 + 1 51 Rearrange before applying any log rules: 2− 1 = 1. One last rule is: Theory: RULE 4a: log(1) = 0. Plan: when the price level has doubled.1: 7 log 4. Substitute this into the equation above.01 𝑡 1. separate the square bracket into two separate logs with an addition sign in between: log 2 − 1 = log 1. there are times where you have to use all the rules. they cannot be simplified further. Solution: 7 = 4. However.01 𝑡 1. ) log 1. but instead of log.03 Determine how many cars are produced currently: 𝐶 0 = 48𝑒 0. ln 𝑒 = 1.02 51 Move the log 1. isolating 𝑡 requires RULE 1. and the denominator (bottom) also with RULE 2: 1 51 ≈ 23. Solution: a) Replace 𝐶 𝑡 with 80 and move the 2 to the other side: 80 − 2 = 48𝑒 0.000 cars/year level.63 𝑦𝑒𝑎𝑟𝑠 (2𝑑.02 Simplify the numerator (top) with RULE 2. and then divide both sides by 0.01 + log 1.01 + (𝑡 − 1) log 1. and then decide which rules are needed.03 So 50 doubled is 100: 100 = 48𝑒 0.02 51 Factorise out the 𝑡 on the right side: 1 log 2 − + log 1. 𝑝. bring the 0.03𝑡 Take the natural logarithm (ln) of both sides: ln 78 = ln 48𝑒 0.18 𝑦𝑒𝑎𝑟𝑠 2𝑑. treat it as just another number.02 to the other side: log 2 − 1 + log 1.03𝑡 + 2 The head of Nissan wants you to find out when: a) Output will break the 80.03: 78 ln 48 = t ≈ 16.log 2 − 1 = 𝑡 log 1. There are many bases. and solve using log rules.03𝑡 Use RULE 3 to bring the two logs on the left into one log.03𝑡 Since the square brackets have two things being multiplied. separating out two parts which are logged requires RULE 2. the starting point).01 + 𝑡 log 1. but only use rules when you are trying to do something specific. b) Output will double. ) 𝑡 = log[ 1.03𝑡 ln 𝑒 Use RULE 4b. NOTE: see Section 4.e. 0. That is. it is called ln (the natural logarithm). 𝑝.01 + 𝑡 log 1. and using RULE 1.01 1.02 51 = 𝑡 log 1. RULE 2 will need to be used: ln 78 = ln 48 + ln 𝑒 0. For now.02 2 − This seems very complex.03𝑡 87 0 Half crab-claw the log 1.02 ] log 1. All the same log rules apply.9 below for a definition of 𝑒.01 + log 1.02 51 Divide both sides by the blue part: log 2 − 1 + log 1. but the most commonly used bases are 10 and 𝑒: log10 𝑋 log 𝑒 𝑋 When a logarithm to the base 𝑒 is used. Theory: the logarithmic function log is usually used to the base 10.02 − log 1. but it is easier to use ln when working with exponential functions involving 𝑒. use ln (as the base is 𝑒).02 51 Example 5: An economist has come up with an equation that tells how many cars (in thousands) Nissan will be producing at a given year into the future 𝐶 𝑡 = 48𝑒 0. so: ln 78 − ln 48 = 0. Always think about what you are trying to do as well as the final outcome. + 2 = 50 . b) Find out the number of cars at 𝑡 = 0 (i.02 into the bracket: 1 log 2 − = 𝑡 log 1. then find 𝑡 when this will double.02 = 𝑡 log 1.03𝑡 + 2 98 = 48𝑒 0.02 = 𝑡 log 1. Plan: a) replace 𝐶(𝑡) with 80.03𝑡 down in front: ln 78 − ln 48 = 0.03𝑡 Move ln 48 to the left side. b) Determine when costs will double.03 In a problem with 𝑒. As a general rule.98 = 𝑒 0. ln is used because ln 𝑒 can be cancelled off (as ln 𝑒 = 1).035 𝑡−1 + 207 a) Forecast the price level in 4 years. keep using log in that question. As you can see above. and provide future forecasts. If instead. 4. b) Determine how long (in years) it will take for prices to double. ) 0. If you used log instead. A manufacturer of laptop computers forecasts costs will rise according to the equation: 𝐶 𝑡 = 1.03𝑡 ln 𝑒 48 98 ln 48 = 𝑡 ≈ 23. and 𝑐 shifts of the whole graph up or down. Remember to be consistent. you can still get the same answer. Sketching log graphs is not as important as knowing the four log rules and their applications. the grey line is exactly the same as the black line in shape.2 𝑡 Determine when GDP will reach $1200𝑏𝑖𝑙𝑙𝑖𝑜𝑛. The graph comes from the equation of the general form: 𝑦 = 𝑎 log(𝑥 + 𝑏) + 𝑐 Where 𝑎 determines the sharpness of the ‘r’ shape. and that it is asymptotic (approaches but never touches) to a particular 𝑥 − 𝑣𝑎𝑙𝑢𝑒.02𝑡 − 0. The government has employed you to study the inflation within a particular city. if you use ln for a given problem. if you use a log. 5.03 2 𝑥 2 −2𝑥 3. The red line is also exactly the same as the black one. 𝑝. however a lot more work is involved.79𝑦𝑒𝑎𝑟𝑠 (2𝑑. except it has shifted vertically. Solve for 𝑥 leaving answers in the simplest form: 𝑎) 15𝑥 = 10 𝑏) 15𝑥+1 = 10 𝑐) 8 2𝑥 = 3𝑥−1 𝑑) 25 5𝑥 = 9 1𝑥 5𝑥+1 𝑒) 2𝑥 16𝑥+1 = 31 3𝑥 𝑓) ln(53 ) − ln(25) = 𝑥 ln 5 𝑔) 2. Exercises: 1. 3 𝑥 2 +2𝑥 a) Determine the cost level in 5 years. c) Determine how long (in years) it will take prices to triple. both sides were logged. 𝑏 shifts the whole graph left or right. 𝑦 𝑦 = log 𝑥 + 4 𝑦 = log 𝑥 𝑦 = log 𝑥 − 4 𝑦 = 5log 𝑥 𝑥 = 9 3𝑥 ) 8 = 323−𝑥 Solve for 𝑥 in terms of 𝑦 in the following: 𝑎) 𝑦 = 15𝑥 + 18 𝑏) 𝑦 = 2𝑥 𝑦3𝑥 𝑐) 𝑦 = 13𝑥 2𝑥+1 32𝑥−1 𝑑) 𝑦 = 2𝑥 32 𝑒) 3𝑦 + 42 = 24 3𝑥+1 If a country’s GDP (in $billion) grows according to the following equation: 𝐺𝐷𝑃𝑡 = 700 1. You determine that the historic inflation rate can be accurately estimated using: 7 𝑝 𝑡 = 1. the result would be log 𝑒 and this is NOT equal to 1. 4. use ln as it allows manipulation using all the same log rules. so it will not be covered any further.03𝑒 0. keep using ln throughout that problem. 88 . except it has been shifted right.8 logarithmic graphs A log function has the general shape: 𝑦 𝑥 Theory: the key aspects of this graph are the general shape of a lower case ‘r’.03𝑡 48 98 ln = 0. The blue line shows how 𝑎 determines the “sharpness” of the curve. Theory: any time 𝑒 is used. and at 𝑦 2. it just reaches the doubling point. 2. as say $1000 in an account must be compounded all the time. the 𝐺𝐷𝑃 is £4trillion. but 𝑒 works just as well in decay.).p. a bank offering continuous compounding of any savings will have to use the number 𝑒. things that happen all the time). Theory: the growth of anything continuous has the general form: 𝐴𝑡 = 𝐴0 𝑒 𝑟𝑡 In words. the variable in question must be changing continuously. the amount available at some time 𝑡 in the future 𝐴𝑡 is equal to the initial amount (𝐴0 ) multiplied by the continuous growth base (𝑒) to the power of the rate of growth per time period multiplied by the number of time periods (𝑟𝑡). This equation is very similar to the exponential growth rate equation. The population of the colony is constant throughout the year. which is used for circles.7183 (4d. but 𝑒 is used to describe things that are continuous (i. For example. business examples of continuous decay are not as common as growth and usually very complex. Currently. Substitute the growth rate. which of the following is occurring: 1.5 log(𝑥) + 1 December 31st. It is like the number pi (𝜋). That is one situation of growth.e. 𝑎) 𝑦 = 4 log(𝑥 − 2) 𝑏) 𝑦 = log(𝑥 − 4) 𝑐) 𝑦 = log 𝑥 + 4 𝑑) 𝑦 = 2 log(𝑥) Sketch the following functions on the same set of axes (make sure they are correct relative to each other): 𝑎) 𝑦 = log(𝑥) − 3 𝑏) 𝑦 = log(𝑥 + 2) + 2 𝑐) 𝑦 = 3 log 𝑥 𝑑) 𝑦 = 2. so the 𝑒 growth equation can be used. It is in this sort of situation where the number 𝑒 is used. In very large populations. and then on December 𝑥 31st. A simple example will demonstrate this theory: if a colony of ants doubles every year. The colony of ants obviously grows throughout the year. This number is called 𝑒 and is approximately equal to 2. it suddenly doubles. Plan: 𝐺𝐷𝑃 can be assumed to grow continuously as the citizens of England are constantly earning money throughout the year. 89 . Example 1: the 𝐺𝐷𝑃 of England is estimated to be growing at 2. and it is such a large colony that we can assume that it grows continuously. The population of the colony constantly increases throughout the year. time and initial amount into: 4. Match the graphs to the equations. 𝑒 can be a good approximation of growth and decay. Estimate the 𝐺𝐷𝑃 in 10 years time.9 the natural number 𝒆 This section is all about a special number.3% per year. Exercises: 1. b) How long it takes for the GDP to double.a. Determine the amount you will have to pay back in 3 years.a.1+𝑥 𝑑) 13 = 6.a. The fine-print stated that the amount is compounded every second. 𝑝. If 𝐺𝐷𝑃 of Fiji grows at a rate of 4. 𝑝. 3. and the growth of a population is accurately approximated by continuous growth. the basics are still important. but will be brought back at the end.026𝑡 3. and is currently $50𝑏𝑖𝑙𝑙𝑖𝑜𝑛.1%p. What makes hyperbolas special is that they have two asymptotes. c) How long it takes for the GDP to triple. every year after that. The following is the graph of this simplest hyperbola: 𝑦 𝑦 = 1 𝑥 𝑥 90 . ) Example 2: Woolworths.023 has been used instead of 2.67𝑒 0.67 Take the natural log (ln) of the whole of both sides: 5 = ln 𝑒 0.026𝑡 Divide both sides by 3. An investment portfolio is estimated to grow continuously for 7 years at 6. You have probably seen the function: 𝑦 = 1 𝑥 This is the simplest hyperbola. approximately how many years will it take for the customer base to grow to 5million? Plan: since 400 stores is a large number of stores.5% p. 0. Remember that an asymptote is a line to which the function approaches but never reaches.000. which is the 𝐺𝐷𝑃 in ten years. The trillion has been left out.026 and 𝐴𝑡 = 5. The reason is that for use in most equations.67million customers. Solution: replace the known variables in the exponential growth rate equation: 𝐴10 = 4𝑒 0. 𝑑. and then at 8. has a growth in customers across it’s 400 stores in Australia at a rate of 2. Also.5𝑒 0. c) The value of the portfolio after 8 years.026𝑡 3. Solve for 𝑥 in the following: 𝑎) 15 = 𝑒 𝑥 + 1 𝑏) 15 = 𝑒 𝑥+1 𝑐) 170 = 12𝑒 0. a large company.23 𝐴10 = £5. the growth rate must be in decimal form. determine: a) when the investment will double in value. If the initial investment is $1. the 𝑒 growth formula can be used: 𝐴𝑡 = 𝐴0 𝑒 𝑟 𝑡 Substitute in all known information. 312 days.5𝑒 0. A dishonest bank lends you $5000 to purchase a car. not percentage form. so the equation is: 5 = 3. 4. If they currently have 3.03 𝑡𝑟𝑖𝑙𝑙𝑖𝑜𝑛 (2.67: 5 = 𝑒 0.3%. 𝑟 = 0.026 Exercises: 1. b) how many years until the portfolio reaches a value of $5.𝐴𝑡 = 𝐴0 𝑒 𝑟 𝑡 then solve for 𝐴𝑡 . 0.027 𝑥 − 1 2.67.89𝑦𝑒𝑎𝑟𝑠 2𝑑.67 𝑡 = ≈ 11. at a rate of 4%p.67 5 ln = 0.6% per year.023 10 5 ln 3.03𝑥 𝑒) 21 = 10. Solution: 𝐴0 = 3.10 the hyperbolic function Despite hyperbolic functions not being very common.3% p.a. 𝐴10 = 4𝑒 0. determine: a) The GDP in 5 years.000. then rearrange to solve for 𝑡.67 ln 4.026𝑡 3. with different values of 𝑑. 𝑦 = 2 2𝑥 + 4 𝑦 = −5 − 1 −6 + 𝑥 91 . Lightly draw the vertical and horizontal asymptotes.5. Exercises: 1.5 For the horizontal asymptote. the hyperbola (the two blue lines constitute a single function) has two asymptotes. The value of 𝑑 determines how close the curve gets to the intersection of the two asymptotes. For the 𝑦 − 𝑎𝑥𝑖𝑠 asymptote. 𝑦 𝑦 = 3 𝑥 𝑦 = 1 𝑥 𝑥 The smaller 𝑑 is (blue). set the denominator equal to zero and solve for 𝑥. the closer to the intersection of the asymptotes the graph will lie. To make this last point clear. Match the following graphs to the equations: 𝑦 𝑥 The larger the value of 𝑑 (red). the following graph shows the same hyperbolic function. setting the whole denominator (𝑎𝑥 + 𝑏) equal to zero and solving for 𝑥 is how the vertical asymptote is found.5 𝑥 The dotted vertical line crosses the 𝑥 − 𝑎𝑥𝑖𝑠 at 𝑥 = −1. the further from the intersection of the asymptotes the graph will lie. 𝑦 = 1 +5 2𝑥 + 3 𝑦 𝑦 = 5 𝑥 = −1.In this case. Solution: for the vertical asymptote: 2𝑥 + 3 = 0 2𝑥 = −3 𝑥 = −1. 𝑦 = 5. the 𝑥 − 𝑎𝑥𝑖𝑠 and the 𝑦 − 𝑎𝑥𝑖𝑠. then draw in the general shape of a hyperbola. The value of 𝑐 determines the location of the horizontal asymptote. but never touches it. The horizontal asymptote is the number added on to the fraction. the greater the value of the function on the 𝑦 − 𝑎𝑥𝑖𝑠 becomes (but it never touches the 𝑦 − 𝑎𝑥𝑖𝑠). the closer the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 get to zero (trace it with your fingers by going towards the origin). Similarly. and the horizontal line 𝑐 = 5. Example 1: sketch the following function 𝑦 = 1 +5 2𝑥 + 3 Plan: for the vertical asymptote. the second asymptote is the 𝑥 − 𝑎𝑥𝑖𝑠 and the greater the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 become (both in a positive or negative direction) the closer the function gets to the 𝑥 − 𝑎𝑥𝑖𝑠. That is. Theory: the general form of a hyperbola is: 𝑑 𝑦 = +c 𝑎𝑥 + 𝑏 The denominator (𝑎 and 𝑏) determines the locations of the vertical asymptote. 5 = 0. 𝑟1 and 𝑟2 have been used as 𝐺𝐷𝑃 and population grow at different rates. Set 𝐺𝐷𝑃 per capita to 15. as both grow at different rates. take the decimal (i.000 𝑒 0.e.000 = 60.21𝑡 26 390.000? Answer in years and days. but what is required is when 𝐺𝐷𝑃 per capita will reach 𝑈𝑆$15. one for 𝐺𝐷𝑃 growth and one for population growth.9%.5 = 𝑒 0. to get the solution in years and days.000 𝑒 0.91𝑦𝑒𝑎𝑟 (2𝑑. ) 0.21𝑡 ) Simplify: ln 6.000 6. Example 2: a company that takes care of its employees spends 𝑆 dollars per year on the well- 92 .019𝑡 This is the 𝐺𝐷𝑃 per capita function. and solve: 15. 0.04𝑡−0.000 𝑒 0.04𝑡 26 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑒 0.019𝑡 60.21 Finally.5 = ln(𝑒 0.000. 𝐺𝐷𝑃 per capita is: 𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 = 𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 = 𝐺𝐷𝑃𝑡 𝑃𝑜𝑝𝑡 60 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 𝑒 0. These are then combined to get 𝐺𝐷𝑃 per capita.019𝑡 Simplify the 60billion/26million (billion/million=thousand): 𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 = 60. PNG has a population of 7million. which grows at 1. 333 𝑑𝑎𝑦𝑠 (to the nearest day).91) and multiply it by the number of days in a year (assume 365) to get 𝑡 = 8 𝑦𝑒𝑎𝑟𝑠.000.2. It is defined as: 𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 = 𝐺𝐷𝑃 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 Simplify further using index rules (Chapter 1): 𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 = = 60. 3 3 +3 𝑦 = 3 + 3𝑥 − 9 −3 + 𝑥 Sketch the following functions: 1 𝑎) 𝑦 = −1 2𝑥 2 𝑏) 𝑦 = +2 𝑥 − 1 1 𝑐) 𝑦 = 𝑥 1 𝑑) 𝑦 = − 𝑥 𝑦 = Here.21𝑡 26 Example 1: a developing country like Papua New Guinea (PNG) has a 𝐺𝐷𝑃 of 𝑈𝑆$60billion with a growth of 4% per year. 𝑝.11 applications Economic theory: 𝐺𝐷𝑃 per capita is a measure of the mean income per person in a particular country.000 𝑒 0.21𝑡 60. When will 𝐺𝐷𝑃 per capita reach 𝑈𝑆$15.019𝑡 26 4.5 ≈ 8.21𝑡 Take logs of the whole of both sides: ln 6. Combining the two formulas.04𝑡 The population growth rate is: 𝑃𝑜𝑝𝑡 = 𝑃𝑜𝑝0 𝑒 𝑟2 𝑡 𝑃𝑜𝑝𝑡 = 26 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑒 0. Solution: the growth rate for 𝐺𝐷𝑃 is: 𝐺𝐷𝑃𝑡 = 𝐺𝐷𝑃0 𝑒 𝑟1 𝑡 Substitute what is known: 𝐺𝐷𝑃𝑡 = 60 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 𝑒 0. Plan: 𝐺𝐷𝑃 and populations are both very large so can be assumed to grow at an exponential rate: 𝐴𝑡 = 𝐴0 𝑒 𝑟 𝑡 Two equations are needed.000 = 𝑒 0.04𝑡 26𝑒 0.21𝑡 ln 𝑒 Remember ln 𝑒 = 1: 𝑡 = ln 6. b) Multiply the 𝑆 from a).1𝜋 4 = 1 − 𝑒 −0. The value of 𝑐 in the quadratic equation determines the vertical shift of the graph. Spain’s economy has a 𝐺𝐷𝑃 approximating €800billion and grows at approximately 3.7%p.1 ≈ 15.1 5 Take the natural log of the whole of both sides: ln 𝑒 −0.2 ln 𝑒 −0. c) The profit level when spending on each employee is $2. Then to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the turning point. b) The profit level when spending on each employee will double from that in part a). c) The GDP per capita in 5 years time. and is growing at 0. whereas when the value of 𝑎 is negative.5 −0. d) When the GDP per capita will reach €30. b) The total amount spent on healthcare when profits are $10million and there are 19 employees.5 = 1 − 𝑒−0. a ‘sad’ face. A company invests 𝑆 dollars in the health of each of its employees according to the equation: 𝑆 = 750 1 − 𝑒 −0. Plan: a) Set 𝜋 to 5 and solve for 𝑆. then substitute it in for 𝑆 in the equation. d) The number of employees when profits are $15million and total spending on healthcare is $15.462million c) Set 𝑆 = 2000 and solve: 2000 = 2500 1 − 𝑒 −0.000. c) Set 𝑆 = 2.1 = 2500 1 − 𝑒−0. 𝑏 2 − 4𝑎𝑐 determines the number of solutions: 93 .7%p. b) The current GDP per capita.5 Set this equal to the original equation: 5000 1 − 𝑒−0.being of each employee.5 𝜋 = ln 1 − 2 1 − 𝑒 −0.000.2 𝜋 = ln 0.1𝜋 5 𝑒 −0..1𝜋 = ln 1 − 2 1 − 𝑒 −0. determine: a) When the population will reach 50𝑚𝑖𝑙𝑙𝑖𝑜𝑛. Solution: a) Set 𝜋 to 5: 𝑆 = 2500 1 − 𝑒−0. if found in the Quadratic Formula by evaluating everything not under the square root sign. 𝑝.2 −0. c) The profit level when the amount spent on each employee is double that in part a).5 chapter four summary A function is defined as having only one 𝑦 − 𝑣𝑎𝑙𝑢𝑒 for every 𝑥 − 𝑣𝑎𝑙𝑢𝑒.094million −0. substitute this 𝑥 − 𝑣𝑎𝑙𝑢𝑒 into the original function.a. Determine: a) The amount spent on each individual if the company has a profit of $7million.5 ≈ 983.67 2𝑑.000 and solve for 𝜋. solve for 𝜋.a. The general form of a quadratic is: 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 The value of 𝑎 determines the sharpness of the quadratic.1𝜋 = ln 0.1𝜋 Determine: a) The spending on each employee when profits are $5million.200. A curve is a function if all vertical lines cross the curve only once.5 −0.1𝜋 = ln 0. If the population of Spain is currently 40million. b) Multiply the exact value determined in a) by 2: 2 × 2500 1 − 𝑒−0. The Quadratic Formula is: −𝑏 ± 𝑏 2 − 4𝑎𝑐 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑠 = 2𝑎 The 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the turning point (TP).1𝜋 Divide both sides by 2500: 2 1 − 𝑒−0.1𝜋 Rearrange to isolate 𝜋: 𝑒−0. A positive value of 𝑎 will result in quadratics with a shape of a ‘smiley’ face.1𝜋 = 1 − 2 1 − 𝑒−0.5 = 2500 1 − 𝑒−0. Exercises: 1. 2. by the function: 𝑆 = 2500 1 − 𝑒−0. the shape becomes inverted.1𝜋 = 0.5 = 5000 1 − 𝑒−0. where 𝑆 is related to profits (in millions of dollars).2 ≈ 16.08𝜋 Where 𝜋 is profit in millions of dollars. Changing the value of 𝑏 in the general quadratic form shifts the graph both horizontally and vertically.1𝜋 = ln 1 − 2 1 − 𝑒 −0. the ends of the graph extend to the top right and bottom left ↙↗ . When 𝑎 is positive. but when 𝑎 is negative. Any time the number 𝑒 is used. 𝑏 2 − 4𝑎𝑐 > 0 → two solutions 𝑏 2 − 4𝑎𝑐 = 0 → one solution 𝑏 2 − 4𝑎𝑐 < 0 → no solutions The cubic function is the name of a function where the highest power of 𝑥 is 3. the variable in question must be changing continuously. The logarithmic graph has the general form: 𝑦 = 𝑎 log(𝑥 + 𝑏) + 𝑐 Where 𝑎 determines the sharpness of the “r” shape. 𝑏 the shifting of the whole graph left or right. The general form of the decay function is: 𝑦 = 𝑏(𝑎 −𝑥 ) + 𝑐 The definition of a logarithm is: For 𝑎𝑏 = 𝑐 log 𝑐 𝑏 = = log 𝑎 𝑐 log 𝑎 Where 𝑏 is called the index. determine: The 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is found by setting all 𝑥’s equal to zero. The value of 𝑑 determines how close the curve gets to the intersection of the two asymptotes. The growth of anything continuous has the general form: 𝐴𝑡 = 𝐴0 𝑒 𝑟 𝑡 The general form of a hyperbola is: 𝑑 𝑦 = +c 𝑎𝑥 + 𝑏 Setting the whole denominator (𝑎𝑥 + 𝑏) equal to zero and solving for 𝑥 is how the vertical asymptote is found. LOG RULE 1 log 𝑎𝑏 = 𝑏 log 𝑎 LOG RULE 2: log(𝑎 ∙ 𝑏) = log 𝑎 + log 𝑏 LOG RULE 3: log 𝑎 𝑏 = log 𝑎 − log 𝑏 LOG RULE 4a: log(1) = 0 LOG RULE 4b: log 10 = 1 (ln 𝑒 = 1) All these rules can be used forwards or backwards. The 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the asymptote is 𝑐. 𝑐 a constant that shifts the graph up or down To sketch an exponential function. It is defined as: 𝐺𝐷𝑃 𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 = 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 chapter four questions 1. 94 . 𝑦 = 𝑥 2 − 6𝑥 𝑦 = −3𝑥 2 + 27𝑥 − 54 2 𝑦 = 𝑥 − 25𝑥 + 150 𝑦 = −𝑥 2 + 18𝑥 − 45 Find the roots of the following quadratics: 𝑎) 𝑦 = 𝑥 2 + 3𝑥 − 6 𝑏) 𝑦 = 4𝑥 2 − 9 𝑐) 𝑦 = 𝑥 − 1 𝑥 + 3 𝑑) 𝑦 = 𝑥 + 4 2 𝑒) 𝑦 = 3𝑥 − 6 2 − 9 𝑓) 𝑦 = 𝑥 2 − 2𝑥 + 12 𝑦 + 5 𝑔) = 𝑥 + 4 𝑥 − 3 Sketch the following quadratics: 𝑎) 𝑦 = 𝑥 2 − 3𝑥 − 10 𝑏) 𝑦 = 𝑥 2 − 3𝑥 − 28 𝑐) 𝑦 = 𝑥 2 + 11𝑥 + 30 𝑑) 𝑦 = 𝑥 − 3 2 𝑒) 𝑦 = −4 𝑥 − 5 2 𝑓) 𝑦 = 𝑥 2 − 4𝑥 − 13 𝑦 + 6 𝑔) = 𝑥 + 4 𝑥 − 3 5. Determine if the following curves are functions: 𝐵 𝐴 𝑦 𝐷 𝑦 𝑥 𝑥 𝐸 𝐶 2. An exponential function has the general form: 𝑦 = 𝑏(𝑎 𝑥 ) + 𝑐 𝑎 is a number that determines the growth rate 𝑏 is the initial amount. The value of 𝑐 determines the location of the horizontal asymptote. Determine the values of 𝑎. the ends of the graph extend to the top left and bottom right ↖↘ . 𝑏 and 𝑐 for the following quadratics: 𝑦 = 3 − 4𝑥 − 5𝑥 2 𝑦 = 3 − 𝑥 𝑥 + 2 𝑦 = 𝑥 − 2 𝑥 + 2 𝑦 − 1 = 𝑥 + 1 𝑥 − 3 Match the following quadratics to the functions. 𝑎) 𝑏) 𝑐) 𝑑) 3. twice or three times. 𝐺𝐷𝑃 per capita is a measure of the mean income per person in a particular country. 4. and 𝑐 the shifting of the whole graph up or down. The general form of a cubic is: 𝑦 = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 A cubic can cross the 𝑥 − 𝑎𝑥𝑖𝑠 once. and 𝑎 is called the base. and there are 20 employees.1𝜋 Determine: a) The amount spent on each employee when profits are $5million. 2 𝑥 + 2 −1 Determine the number of solutions for: 𝑎) 𝑦 = 𝑥 2 + 5𝑥 + 6 𝑏) 𝑦 = −𝑥 2 − 6𝑥 + 16 𝑐) 𝑦 = 4𝑥 2 − 3𝑥 + 1 𝑑) 𝑦 = 𝑥 2 − 18 𝑒) 𝑦 = 14𝑥 2 − 2𝑥 − 2 𝑓) 𝑦 = 𝑥 2 + 6𝑥 + 9 Match the following functions to their equations: ) 𝑦 𝑥 − 1 −1 = 11. b) When the price level will reach 1. The price level of an economy is accurately estimated by the function: 1 𝑝 𝑡 = 1.900. b) The total amount spent on relaxation when profits are $10million. d) The number of employees when profits are $12million.1% p. 15.6. Solve for 𝑥 in the following logarithmic functions: 𝑎) 12 = log 4𝑥 𝑏) − 13 = 2 log 3𝑥 𝑥+3 𝑐) 7 = log 2 𝑑) log 2𝑥 = 4𝑥 + 2 𝑥−4 𝑒) log 2 =0 𝑓) 18 = 4 ln 2𝑥 𝑥+1 𝑔) − log 8 = 182𝑥 ) 7 − log 2𝑥 = 𝑥 − 1 𝑖) 25 − ln 3 = − ln 6𝑥−1 13. Determine: a) When the population will reach 140million. c) When the price level will double.05 𝑡−1 + 21 Determine: a) The price level in five years time. d) When 𝐺𝐷𝑃 per capita will reach 𝑈𝑆$10. and is growing at 4.000. 7. c) The profit level when spending on each employee is $1000. 16. The 𝐺𝐷𝑃 of Indonesia is approximately 𝑈𝑆$550billion. Match the following to their equations: 𝑦 𝑥 𝑦 𝑦 = − 1 −4 𝑥 − 2 𝑦 = 1 −6 𝑥 + 4 𝑦 = 2 +5 𝑥 − 5 𝑥 8.5 times that of the current level.03 𝑡 1. Match the following logarithmic functions to their equations: 𝑦 𝑥 𝑦 = 4 log 𝑥 𝑦 = log 𝑥 2 𝑦 = 2 log 𝑥 − 2 + 5 12. 𝑦 = −3𝑥 3 − 3 𝑦 = 2𝑥 3 − 3 3 2 𝑦 = −𝑥 + 5𝑥 + 48𝑥 − 252 𝑦 = 𝑥 3 + 12𝑥 2 + 11𝑥 − 168 Match the following exponential functions to their equations: 𝑦 𝑥 𝑦 = 1. A company invests 𝑟 dollars in the relaxation of each of its employees. c) When 𝐺𝐷𝑃 will reach $1trillion.a. Sketch the following exponential functions on the same set of axes. 95 . 𝑎) 𝑦 = 2𝑥 𝑏) 𝑦 = 3𝑥 + 2 𝑥 𝑐) 𝑦 = 4 2 𝑑) 𝑦 = 3 3−𝑥 − 1 10.5𝑥 𝑦 = 3 2𝑥 𝑦 = 2−𝑥 + 2 9.1% p. and is growing at 1. Solve for 𝑥 in the following exponential functions: 𝑎) 15 = 2𝑥 𝑏) 18 = 4 3𝑥 2 𝑐) 9 − 23 = 4𝑥−1 𝑑) 34𝑥 9 = 3𝑥 2 𝑒) 2𝑥−1 = 32−3𝑥 𝑓) 4 2𝑥−3 = 2𝑥 𝑔) 14 2𝑥−3 = 213𝑥−3 14. b) The 𝐺𝐷𝑃 in 10 years time. and total spending on relaxation is about$10.a. according to the amount of profit (in millions) generated by the firm: 𝑟 = 1200 1 − 𝑒 −0. If the population is currently 120million. Chapter 5 Single Variable Differentiation Finding the slope of curves at any point along the function 5.7 5.10 5.9 5.2 5.1 5.5 5.6 5.3 5.8 5.4 5.11 What is Differentiation? Differentiation by First Principles Differentiation Rules: Power Rule Differentiation Rules: Chain Rule Differentiation Rules: Product Rule Differentiation Rules: Quotient Rule Differentiation Rules: 𝑒 Rule Differentiation Rules: ln Rule The Second Derivative The Gradient Function Graph Simple Applications 97 99 103 104 105 107 109 110 111 112 113 114 115 Chapter Five Summary Chapter Five Questions 96 . 5. On the other hand. Theory: the gradient of a line tangent to a given point is the rate of change of 𝑦 for changes in 𝑥. the gradient would be called 𝑥 marginal profit. Theory: Differentiation is the process of finding the gradient at all points along a function. The following is a profit function: 𝜋 𝜋2 𝜋1 𝐸 𝐹 𝑄1 𝑄2 𝑄 At point 𝐸. because the slope determines the extra profit if output were to increase. so the rate of change of profit (𝑦 − 𝑎𝑥𝑖𝑠) for changes in output (𝑥 − 𝑎𝑥𝑖𝑠) is positive. for point 𝐹. Theory: the rate of change of a function is found by finding the slope of a curve at a given point.1 what is differentiation? Engineers. managers and most professionals use differentiation in everyday tasks. financial analysts. the tangency has a negative gradient meaning the rate of change of profit for changes in output is negative. profit falls below 𝜋2 . Differentiating this will give the rate of change of profit as output changes (or marginal profit). is negative (as the point has a negative gradient). the function is increasing (e. as output increases beyond 𝑄2 . points 𝐴 means “at the margin”.g. If output is increased from 𝑄1 . Four lines have been drawn tangent to the blue function: 𝐴 𝑦 𝐶 and 𝐷). profits will also increase beyond 𝜋1 .g. tangent to that point). Typically. The slope of a curve at a given point is found by determining the slope of a line just touching that point (i. This means the rate of change of 𝑦 as 𝑥 changes is positive (from the slope of the tangent line). 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐𝑎𝑛𝑔𝑒 = 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 For the profit function. Differentiation gives the rate of change of revenue for changes in output (or the marginal revenue. A revenue function is determined by how much output is sold. point 𝐵). it is the rate of change of the 𝑦 − 𝑎𝑥𝑖𝑠 when there is a change in the 𝑥 − 𝑎𝑥𝑖𝑠. economists. it means the function is decreasing (e. If the slope is negative. Theory: differentiation is finding the rate that one thing changes when something else is changed. how much extra revenue will be obtained from selling an extra unit of output). as 𝑥 changes. the tangency has a positive gradient.e. 97 . or “extra”. the rate of change of 𝑦. Marginal 𝐵 𝐷 The gradient at a point along a curve gives information about the function. A profit function is also determined by the quantity of goods sold. If the slope is positive. Interpreting this. Theory: The gradient of a given point along a curve is often called the marginal value. The gradient function is a function that gives the gradient of the original function at all the points along the original function. Exercises: 1. For the functions above. Solution: The profit function is a “sad” quadratic (as 𝑎 < 0). the slope at 𝑄 = 50 must be positive. 𝑥 = 6 Order the following points from lowest (i. but from working with quadratics. For a small change in 𝑥 (the 𝑑𝑥). and the profit function is 𝜋 𝑄 = −𝑄 2 + 110𝑄 − 1000 As a manager. Find the turning point (which is a maximum) using part of the quadratic formula: 𝑥 − 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑇𝑃 = − 𝑏 110 110 =− = = 55 2𝑎 −2 2 2. total profits will increase. Look at points 𝐸 and 𝐹 of the last graph. Determine if the following points have a positive or negative gradient 𝐵 𝐶 𝐷 𝑦 𝐺 𝐻 𝐴 𝑥 𝐸 𝐹 Since 𝑄 = 50 is to the left of the maximum. Example 1: if output is currently 𝑄 = 50. If current output is to the left of the maximum. Find the profit at 𝑄 = 50 and at 𝑄 = 51 to make sure of this result: 3. meaning there is a maximum.e. it will not be so simple. If current output is to the right of the maximum. However. Plan: if the rate of change of profit (marginal profit) is positive. so the marginal profit is positive. This means that if output were to increase to 𝑄 = 51. an extra unit will reduce profits (negative marginal profit). for the function 𝑦 − 7 − 3𝑥 = −𝑥 2 Determine if the following points have a positive or negative gradient: 𝑥 = −1. If the rate of change of profit (marginal profit) is negative. most negative) to highest (i. you would like to know if increasing output will increase profits. most positive): 98 . before this can be done. Theory: mathematicians denote the derivative (gradient) function in a number of ways: 𝑦 ′ is pronounced “𝑦 prime” 𝑑𝑦 𝑑𝑥 is pronounced “dee y dee x”. the gradient function must be denoted by something that distinguishes it from the original function. 𝑦 will change by a certain small amount (𝑑𝑦). 𝜋 50 = − 50 = 2000 2 + 110 50 − 1000 𝜋 51 = −(51)2 + 110(51) − 1000 = 2009 Thus profits will increase if output is increased by one unit. Differentiation provides a way of finding the gradients of any given point along a curve. however. The 𝑑 means very small.e. additional output will reduce profits. The following shows a simple use of gradients. the gradient (or slope) of the tangent changes when moving along the 𝑥 − 𝑎𝑥𝑖𝑠. 𝑥 = 2. The rest of this chapter is devoted to finding the gradient function mathematically. it was known that anything to the left of the maximum must have a positive gradient. The actual value of the gradient at 𝑄 = 50 was not determined. an increase of one unit will increase profits (as the marginal profit is positive). Most of the time. then additional output will generate more profit. Production is to increase from 𝑄 = 50 to 𝑄 = 51. Using your knowledge of quadratics and turning points. The first 𝑥 − 𝑣𝑎𝑙𝑢𝑒 is 𝑥 = 3. Again. Despite this new approximation not being perfect. Two coordinates are still needed: one is still the point (3. the other 𝑥 − 𝑣𝑎𝑙𝑢𝑒 is 3. the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 is found using the Hint: use a ruler and draw in the tangencies.e. substitute 𝑥 = 3 into the original function: 3 3.5 4 𝑥 This red line is a closer approximation to the actual gradient than the black dotted line (which was the first approximation). so 𝑥 = 4. Ignore this for the time being and determine the gradient of this red line. we look back to the start of this problem and instead of having a point 1 unit up from 3. two points are needed (i. say 0. as it is obvious that the gradient of this line (red) and the tangent (blue) line are not the same. two coordinates). Plot these two points on a set of axes (previous graph). say 1 unit up from 3 (i. However.5): 𝑦 3 4 𝑥 This is called a rough approximation of the gradient at 𝑥 = 3.e. but for a coordinate we also need a 𝑦 − 𝑣𝑎𝑙𝑢𝑒.2 differentiation by first principles Differentiation is simply finding the gradient function of an original function. draw a line through this point and another point on the curve. It is just an approximation. a point a little closer to 𝑥 = 3 is used. this red line is not a good approximation to the blue line.5 units up from 3 (i.𝑦 𝐶 𝐵 𝐴 𝑓 𝑥 = 4 𝑥 − 2 𝐷 𝐸 𝐹 𝑥 𝐺 𝑓(3) = 4 3 − 2 𝑓 3 = 4 1 2 2 2 +2 +2 +2=6 The first coordinate is (3. To get a better approximation.18). The other coordinate has the 99 . Intro example 1: beginning with the function 𝑓 𝑥 = 4 𝑥 − 2 2 +2 +2 + 2 = 18 which gives the second coordinate (4. The following is the long method. original function: 𝑓 𝑥 = 4 𝑥 − 2 𝑓(4) = 4 4 − 2 𝑓 4 = 4 2 2 2 2 5. 𝑥 = 4): 𝑦 Thus 12 is the gradient of the red line. then find the gradient: 𝑟𝑖𝑠𝑒 𝑦2 − 𝑦1 18 − 6 12 = = = = 12 𝑟𝑢𝑛 𝑥2 − 𝑥1 4−3 1 +2 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = for which the gradient at 𝑥 = 3 is to be found. the gradient is still found because this approximation is better.e. From Chapter 2. On the graph.6).6). The other point on the red line is 1 unit above 3. To find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒. to find the gradient of a line. but it is necessary that you understand it. like 0. remember that the gradient of the line is not a good approximation of the actual gradient of 3 3. Thus the two points are: 𝑥1 = 𝑥 (this is the base point) +2 2 𝑓(3. the point in question The idea is to get 𝑠 to “go to” zero (but not be zero) because when 𝑠 is very small the approximation becomes better and better.75 4 𝑥 a point if the value of 𝑠 is too large. then finding the gradient of the connecting line: 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝑟𝑖𝑠𝑒 𝑦2 − 𝑦1 11 − 6 = = 𝑟𝑢𝑛 𝑥2 − 𝑥1 3. something smaller could be added.01. a point is chosen a very small distance 𝑠 above this 𝑥 − 𝑣𝑎𝑙𝑢𝑒. Differentiation by First Principles is the formula: 𝑓 ′ = lim 𝑠→0 𝑓 𝑥 + 𝑠 − 𝑓 𝑥 𝑠 Example 1: Find the gradient function of 𝑦 = 5𝑥 + 2 Plan: use the first principles formula.5 − 2 2 +2 𝑥2 = 𝑥 + 𝑠 (an 𝑥 − 𝑣𝑎𝑙𝑢𝑒 a little more than 𝑥1 ) The 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 of these two 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 are found using the original function 𝑓 𝑥 . but it is still not the actual gradient.5. then 0. That is why mathematicians invented something called a limit. The coordinate is then 𝑥 + 𝑠.5 2 plus a very small amount). In the above example. The other point will have an 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of 𝑥 + 𝑠 (i. or 0. 𝑠 was initially 1.5 3. This will not be done mathematically. 𝑓 𝑥 . 𝑓 𝑥 + 𝑠 .5. Since these are not numbers but rather letters. + 2 = 11 Plot the two points on a set of axes (previous graph). How would we get a better approximation? Instead of adding 0.5 − 3 For the first point 𝑥1 .5 This is a closer approximation to the actual gradient at 𝑥 = 3. Having two coordinates allows the gradient of this line to be found using rise/run: 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = = = 𝑟𝑖𝑠𝑒 𝑦2 − 𝑦1 = 𝑟𝑢𝑛 𝑥2 − 𝑥1 𝑓 𝑥 + 𝑠 − 𝑓 𝑥 𝑥 + 𝑠 − 𝑥 𝑓 𝑥 + 𝑠 − 𝑓(𝑥) 𝑠 This generalises what was done graphically. the function will be 𝑓(𝑥1 ). So using very small additions to 𝑥 = 3 gives approximations very close to the tangent at 𝑥 = 3 (blue line).001 etc.𝑥 − 𝑣𝑎𝑙𝑢𝑒 as 𝑥 = 3. the approximations become better and better (red arrow).e.25 3. the same process is done as if they were numbers. the function that is given. The coordinate is then 𝑥. Theory: to mathematically find an approximation to the tangent at a point 𝑥 (this was 3 in the above example). The 𝑦 − 𝑣𝑎𝑙𝑢𝑒 comes from the original function: 𝑓 𝑥 = 4 𝑥 − 2 𝑓 3. which is simply 𝑓(𝑥) .5) = 4 3. but the graph below shows what is meant: 𝑦 For the second point 𝑥2 = 𝑥 + 𝑠.5 = 4 1. Theory: the limit as “𝑠 goes to zero”: lim 𝑠→0 As points closer and closer to 𝑥 = 3 are chosen as the “other coordinate”. However. the function will be 𝑓 𝑥2 = 𝑓(𝑥 + 𝑠). 100 .5 to 3. 5 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = = 10 0. by cancelling it off with an 𝑠 on top. The ultimate goal is to get rid of the 𝑠 on the bottom. this is the gradient function.𝑠 Solution: the trick with this formula is that only the 𝑓(𝑥 + 𝑠) part ever needs to be found. But since there is no 𝑠 left: 𝑓 ′(𝑥) = lim 5 = 5 𝑠 →0 The lim part can now be ignored as there is no 𝑠→0 longer an 𝑠 left. they are related (see section 5. giving the gradient function: 𝑓 𝑥 = 5 ′ 𝑜𝑟 𝑑𝑓 𝑥 =5 𝑑𝑥 To repeat. A line has the same gradient all along it. 𝟑𝒙𝟐 + 6𝑥𝑠 + 𝑠 2 − 𝒙 − 𝑠 − 𝟔 − 𝟑𝒙𝟐 + 𝒙 + 𝟔 𝑠 101 . and it is different from the original function. The original function was a line with equation 𝑦 = 5𝑥 + 2. The easiest way is to look at the numerator (top). and from Chapter 2. and you will see a lot of terms can be cancelled off (colour coded below). This results in something much simpler. however. but the gradient of a quadratic changes along its curve. so replace 𝑓(𝑥) with the original function in the first principles formula: 𝑑𝑦 𝑓 𝑥 + 𝑠 − 𝑓 𝑥 = lim 𝑠 𝑑𝑥 𝑠→0 2 2 3𝑥 + 6𝑥𝑠 + 𝑠 − 𝑥 − 𝑠 − 6 − [3𝑥 2 − 𝑥 − 6] = lim 𝑠→0 𝑠 The tricky part is to simplify this.10). To find this. the gradient of this line is 𝑚 = 5. Example 2: find the gradient function of 𝑦 = 3𝑥 2 − 𝑥 − 6 Plan: use the first principles formula 𝑑𝑦 𝑓 𝑥 + 𝑠 − 𝑓 𝑥 = lim 𝑠 𝑑𝑥 𝑠→0 Solution: The only thing we need to find in this formula is 𝑓(𝑥 + 𝑠) and this is done by replacing every 𝑥 in the original function with (𝑥 + 𝑠) in brackets: 𝑓(𝑥) = 3𝑥 2 − 𝑥 − 6 𝑓 𝑥 + 𝑠 = 3 𝑥 + 𝑠 2 [5𝑥 + 5𝑠 + 2] − 𝑓 𝑥 𝑠 Then replace the 𝑓(𝑥) part. = lim 𝑠→0 = lim 𝑠 →0 5𝑥 + 5𝑠 + 2 − 5𝑥 − 2 Simplify the top by cancelling: 𝑓 ′ = lim 𝑠 →0 𝟓𝒙 + 5𝑠 + 𝟐 − 𝟓𝒙 − 2 𝑠 𝑓 ′ = lim 𝑠 →0 5𝑠 𝑠 And then 𝑠 the 𝑠 part can be crossed off: 𝑓 ′ = lim 𝑠 →0 5𝒔 𝒔 can every other 𝑠 be Only after 𝑠 crossing off 𝑠 replaced with zero. which is the same answer. replace every 𝑥 in the original function with (𝑥 + 𝑠) in brackets: The original function is 𝑓(𝑥) = 5𝑥 + 2 Replace every 𝑥 with 𝑥 + 𝑠 in brackets: 𝑓(𝑥 + 𝑠) = 5(𝑥 + 𝑠) + 2 𝑓(𝑥 + 𝑠) = 5𝑥 + 5𝑠 + 2 Substitute the 𝑓(𝑥 + 𝑠) in the first principles formula with what was found above: 𝑓 ′ = lim 𝑠 →0 𝑓 ′ = lim 𝑠→0 𝑓 𝑥 + 𝑠 − 𝑓 𝑥 Think about the answer 𝑓 ′ = 5. which is simply the original function: 𝑓 ′ = lim 𝑠 →0 5𝑥 + 5𝑠 + 2 − [5𝑥 + 2] 𝑠 𝑠 − (𝑥 + 𝑠) − 6 Write out the “squared” term as two terms: 𝑓 𝑥 + 𝑠 = 3 𝑥 + 𝑠 𝑥 + 𝑠 − 𝑥 + 𝑠 − 6 𝑓 𝑥 + 𝑠 = 3 𝑥 + 𝑠 𝑥 + 𝑠 − 𝑥 + 𝑠 − 6 𝑓 𝑥 + 𝑠 = 3 𝑥 2 + 2𝑥𝑠 + 𝑠 2 − 𝑥 + 𝑠 − 6 𝑓 𝑥 + 𝑠 = 3𝑥 2 + 6𝑥𝑠 + 3𝑠 2 − 𝑥 − 𝑠 − 6 This is the 𝑓(𝑥 + 𝑠) part done. a single fraction can be written: = 3𝑥 − 3𝑥 + 3𝑠 3𝑥 − 3𝑥 − 3𝑠 = 𝑥 + 𝑠 𝑥 𝑥 + 𝑠 𝑥 −3𝑠 𝑥 + 𝑠 (𝑥) Cancel off the 3𝑥 − 3𝑥: = Bring everything back into the first principles formula: 𝑑𝑦 = lim 𝑑𝑥 𝑠→0 −3𝑠 𝑥 + 𝑠 (𝑥) 𝑠 From Chapter 1. The Common Denominator is the multiplication of the two denominators. the gradient of the original function at 𝑥 = 2 would be found by substituting 𝑥 = 2 into the gradient function (i. 𝑑𝑥 2 = 6 2 − 1 = 11). the division of fractions can be written as the multiplication of the inverse: 𝑑𝑦 −3𝑠 1 = lim ∙ 𝑑𝑥 𝑠→0 𝑥 + 𝑠 𝑥 𝑠 Cross out the two red 𝑠: 𝑑𝑦 −3 = lim 𝑑𝑥 𝑠→0 𝑥 + 𝑠 (𝑥) Replace the remaining 𝑠 with zero and remove the limit notation: Plan: use the first principles formula 𝑑𝑦 𝑓 𝑥 + 𝑠 − 𝑓 𝑥 = lim 𝑠 𝑑𝑥 𝑠→0 Solution: find 𝑓(𝑥 + 𝑠) 𝑓(𝑥) = 3 𝑥 3 𝑥 + 𝑠 This cannot be simplified right now. all other 𝑠’s can be replaced with zero.Note that the negative sign must be put into all the terms of 𝑓(𝑥) when it is taken away from 𝑓(𝑥 + 𝑠). Example 3: using first principles. an 𝑠 in the numerator should be factorised out first.e. 𝑑𝑦 6𝑥𝑠 + 𝑠2 − 𝑠 = lim 𝑑𝑥 𝑠→0 𝑠 To cancel off the 𝑠 in the denominator. find the derivative function of 𝑦 = 3 𝑥 𝑑𝑦 Isolate the numerator (top) of this fraction. For example. 𝑑𝑦 𝑠(6𝑥 + 𝑠 − 1) = lim 𝑑𝑥 𝑠→0 𝑠 𝑑𝑦 𝒔(6𝑥 + 𝑠 − 1) = lim 𝑑𝑥 𝑠→0 𝒔 Having cancelled off the 𝑠 on the bottom. and the limit notation removed: 𝑑𝑦 = 6𝑥 + 0 − 1 = 6𝑥 − 1 𝑑𝑥 This is the derivative function. It is the function that gives the derivative (gradient) at every point on the original function. so if you’re not comfortable spend some time revising Chapter 1. so substitute 𝑓(𝑥 + 𝑠) = everything into the first principles formula: 3 3 𝑑𝑦 𝑥 + 𝑠 − 𝑥 = lim 𝑑𝑥 𝑠→0 𝑠 𝑑𝑦 −3 −3 3 = = =− 2 𝑑𝑥 𝑥 + 0 (𝑥) 𝑥 (𝑥) 𝑥 Differentiation by first principles requires knowing how to simplify and factorise. Get rid of the brackets: = = 3 𝑥 3 − 𝑥 + 𝑠 𝑥 𝑥 𝑥 + 𝑠 𝑥 + 𝑠 3𝑥 3𝑥 + 3𝑠 − 𝑥 + 𝑠 (𝑥) 𝑥 + 𝑠 (𝑥) Having the Common Denominator present. 102 . so simplification is easier (it will be brought back in later): 3 3 3 𝑥 3 (𝑥 + 𝑠) − = − 𝑥 + 𝑠 𝑥 𝑥 + 𝑠 𝑥 𝑥 (𝑥 + 𝑠) This is finding a Common Denominator (Chapter 1). Only then can it be cancelled off with the bottom one. The original function must first be simplified (see Chapter 1): 𝑦 = −3𝑥 −4 + 5𝑥 2 − 𝑥 + 15 Then differentiate each section separately: 𝑑𝑦 = 2𝑥 𝑑𝑥 𝑑𝑦 = −4 −3 𝑥 −4−1 + 2 5𝑥 2−1 − 1 𝑥 1−1 + 0 𝑑𝑥 = 12𝑥 −5 + 10𝑥 − 1 Watch out for the negative signs! Example 2: find the derivative of 𝑦 = 2𝑥 3 Plan: use the power rule to bring the power out front then subtract one from the power. so the 𝑥 is no longer present. Theory: the power rule states that to find the derivative function. Solution: Exercises: 1. It means the gradient of a horizontal line is zero. Example 4: differentiate 𝑦 = − 3 + 5𝑥 2 − 𝑥 + 15 𝑥 4 5. This is why.Exercises: 1. Using First Principles. you bring the power to the front and then subtract 1 from the power. but at least now you’ll know what you are actually doing. Theory: the derivative of a constant is zero. Differentiate the following using the power rule: 𝑎) 𝑦 = 3𝑥 2 𝑏) 𝑦 = −2𝑥 3 + 𝑥 2 3𝑥 3 𝑐) 𝑦 = − 7 103 . Using Differentiation by First Principles. it is really 𝑥 1 and when this is differentiated. one section at a time. Example 1: use the power rule to differentiate: 𝑦 = 𝑥 2 Solution: bring the 2 out front of the 𝑥.3 differentiation rules: power rule This is a rule which makes the last 4 pages redundant. determine the gradient function of the following: 𝑎) 𝑦 = 3𝑥 2 − 𝑥 3 5𝑥 2 12 𝑏) 𝑦 = − + 𝑥 − 1 7 5 5 𝑐) 𝑦 = 𝑥 3 𝑑) 𝑦 = 7𝑥 1 𝑒) 𝑦 = 2 𝑥 𝑦 ′ = (3)2𝑥 2 Multiply the 2 and 3 together to give the solution: 𝑦 ′ = 6𝑥 2 Example 3: find the gradient function of 𝑦 = 4𝑥 3 − 2𝑥 2 + 7𝑥 − 1 Plan: use the power rule to differentiate each section. Sorry. the −1 differentiated to give zero. Solution: the four sections are: 𝑦 = 4𝑥 3 − 2𝑥 2 + 7𝑥 − 1 Apply the power rule to each of the sections: 𝑑𝑦 = 12𝑥 2 − 4𝑥 + 7 𝑑𝑥 Remember: when you see 𝑥 by itself. one at a time. then subtract 1 from the power: Step 1: Step 2: 𝑦 ′ = = 𝑥 2 = 2𝑥 2−1 Solution: you cannot simply differentiate using the power rule if the term is not up top. For the original function: The derivative function is: 𝑦 = 𝑎𝑥 𝑛 𝑑𝑦 𝑑𝑥 = 𝑛𝑎𝑥 𝑛−1 Plan: use the power rule. it gives 𝑥 1−1 = 𝑥 0 = 1. determine the gradient function given the following original functions: 𝑎) 𝑦 = 7𝑥 − 3 𝑏) 𝑦 = −9𝑥 + 1 𝑐) 𝑦 = 𝑥 2 + 2𝑥 𝑑) 𝑦 = −3𝑥 2 + 14𝑥 − 11 2. in the last example. Solution: Differentiate as if using the power rule: 3(4𝑥 + 2)3−1 = 3 4𝑥 + 2 2 Then multiply this by the derivative of the inside of the brackets: 𝑑𝑦 = 3 4𝑥 + 2 2 (4) 𝑑𝑥 𝑑𝑦 = 12 4𝑥 + 2 2 𝑑𝑥 The 12 came from multiplying the 3 and 4 together. and the second underlined part is the derivative of inside the brackets. the chain rule. it would is only multiplied by 2𝑥. then multiply by the derivative of the inside of the brackets” Solution: the first underlined part of the following is the as if part of the chain rule. and then differentiate the sections: 𝑦 = 4𝑥 + 2 4𝑥 + 2 4𝑥 + 2 𝑦 = 16𝑥 2 + 16𝑥 + 4 4𝑥 + 2 𝑦 = 64𝑥 3 + 80𝑥 2 + 48𝑥 + 8 Differentiate this: 𝑑𝑦 = 192𝑥 2 + 160𝑥 + 48 𝑑𝑥 This is a very long method (especially when you show all working) but there is an easier method. Example 2: Find the gradient function of 𝑦 = (𝑥 2 + 3𝑥)5 Plan: use the chain rule to differentiate: “differentiate the function as if using the power 5. which is not the case! This last example can be simplified to: 𝑑𝑦 = (10𝑥 + 15) 𝑥 2 + 3𝑥 𝑑𝑥 which is to the power of 1. Theory: the informal method is: differentiate the function as if using the power rule.4 differentiation rules: chain rule Intro example 1: differentiate 𝑦 = (4𝑥 + 2)3 Solution: to solve this.𝑑) 𝑒) 𝑓) 𝑔) ) 𝑖) 𝑗) 𝑘) 5 𝑥 2 − 𝑥 4 5 𝑥 2 𝑥 3 1 𝑦 = −𝑥 − − + 2 3 𝑥 1 1 1 𝑦 = + 2 + 3 𝑥 𝑥 𝑥 3 −3 𝑦 = 3 𝑥 + 𝑥 2 4 15 𝑦 = 6 + −𝑥 −4 1 𝑦 = 𝑥 14 + 14 𝑥 𝑦 = −13𝑥 −13 + 15 − 𝑥 𝑦 = 12 𝑦 = rule. There is the formal and informal way of remembering the chain rule. We cannot put the 12 into the brackets because the bracket is squared. then multiply by the derivative of the inside of the brackets.5 4 As the 5 can only go into every term in the bracket Plan: use the informal chain rule method: differentiate the function as if using the power 104 . 𝑑𝑦 = 5 𝑥 2 + 3𝑥 4 (2𝑥 + 3) 𝑑𝑥 The 2𝑥 + 3 must be in brackets as 5 𝑥 2 + 3𝑥 imply that the 5 𝑥 2 + 3𝑥 4 4 is multiplied by all of 2𝑥 + 3 . Otherwise. you might expand it using the crab-claw. Example 1: differentiate the above example 𝑦 = (4𝑥 + 2) 3 rule. and then 3 is added onto the whole thing. then multiply by the derivative of the inside of the brackets. Example 3: differentiate 𝑦 = 2 3𝑥 3 − 4𝑥 2 + 1 −5. 105 .5−1 (9𝑥 2 − 8𝑥) Simplify the constants (2 ∙ 5.5 𝑑𝑦 𝑑𝑢 𝑑𝑦 𝑑𝑢 𝑑𝑦 × = × = 𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑥 With the equation above: 𝑑𝑦 = 7𝑢6 × 2𝑥 𝑑𝑥 = 14𝑥𝑢6 = 2𝑥 7 𝑥 2 + 1 = 14𝑥 𝑥 2 + 1 6 6 Replace 𝑢 with what it was originally defined as (i. the chain rule is: 𝑑𝑦 𝑑𝑦 𝑑𝑢 = × 𝑑𝑥 𝑑𝑢 𝑑𝑥 To demonstrate this.5) into the only brackets which are to the power of 1 (9𝑥 2 − 8𝑥): 𝑦 = 99𝑥 2 − 88𝑥 3𝑥 3 − 4𝑥 2 + 1 4. Then replace the brackets in the original equation with the function 𝑢: 𝑦 = (𝑢)7 Differentiate both of these new functions: Exercises: 1. Theory: formally.5 The formal method of the chain rule is temporarily inventing a new function. Chain rule: differentiate the function as if using the power rule.Plan: move everything up top so the chain rule can then be easily used. the following function will be used: 𝑦 = (𝑥 2 + 1)7 Invent the function 𝑢 = 𝑥 2 + 1. so a new rule is required: the product rule. The following shows examples of two separate functions: 𝑦 = 𝑥 + 2 2 −4 𝑥 4 + 2 6 𝑑𝑦 = 7𝑢6 𝑑𝑢 And: 𝑡 = 𝑒 𝑥 (4𝑥 − 1) 3𝑥 𝑟 = +2 4 2 1 +1 𝑥 −2 𝑑𝑢 = 2𝑥 𝑑𝑥 If these two derivatives are multiplied together. then multiply by the derivative of the inside of the brackets. it was defined as (𝑥 2 + 1) in this question).e. the wanted derivative is obtained: These cannot be differentiated using the rules learnt so far. 𝑑𝑦 = 2 5. however always write out the theory used. which is whatever is inside the brackets.5 3𝑥 3 − 4𝑥 2 + 1 𝑑𝑥 5. Differentiate the following using the chain rule: 𝑎) 𝑦 = 𝑥 + 1 3 𝑏) 𝑦 = 2𝑥 − 3 4 𝑐) 𝑦 = −3𝑥 + 1 4 𝑑) 𝑦 = 𝑥 2 + 2 3 𝑒) 𝑦 = 2𝑥 2 − 3 4 𝑓) 𝑦 = −4𝑥 2 − 𝑥 3 𝑔) 𝑦 = 12𝑥 − 𝑥 3 4 ) 𝑦 = 4 𝑥 2 + 𝑥 + 1 5 𝑖) 𝑗) 𝑘) 𝑦 = 5 − 𝑦 = 3𝑥 3 1 + 8 𝑥 3 4 1 1 + 𝑥 𝑥 2 𝑦 = 12 𝑥 2 + 𝑥 + 5 + 1 5 + 𝑥 𝑥 2 2 5. Solution: use index rules to bring the denominator up top and change the sign of the index: 𝑦 = 2 3𝑥 3 − 4𝑥 2 + 1 5. You get the same answer using the formal or informal method. Apply the chain rule: the first underlined part below is the as if part.5 differentiation rules: product rule This rule applies whenever two separate functions are multiplied together. and the second underlined part is “the derivative of the inside of the brackets”. it is possible to factorise out some of those parts (underlined below) to make it a bit simpler: 𝑑𝑦 = 4𝑥 𝑑𝑥 2 −3𝑥2 − 1 5 12 −3𝑥 2 − 1 + −36 4𝑥 Simplify the square brackets: 6 = −28𝑥 𝑑𝑦 = 4𝑥 𝑑𝑥 2 −3𝑥2 − 1 5 [−36𝑥2 − 12 − 144𝑥] Put all the parts in brackets and substitute into the power rule formula: 𝑓 ′ 𝑥 = (30𝑥 + 9𝑥 2 )(−4𝑥 7 + 6) + (−28𝑥 6 )(15𝑥 2 + 3𝑥 3 ) 𝑑𝑒𝑟1𝑠𝑡 2𝑛𝑑 𝑑𝑒𝑟2𝑛𝑑 1𝑠𝑡 That is the extent of simplifying. But in harder examples. (2) the chain rule within our base rule. You should get: 𝑓 ′ 𝑥 = −120𝑥 9 − 540𝑥 8 + 54𝑥 2 + 180𝑥 You could have crab-clawed the original function and then applied the power rule four times. Example 2: find the gradient function of 𝑦 = (4𝑥)3 −3𝑥 2 − 1 1𝑠𝑡 2𝑛𝑑 6 Exercises: 1. The product rule is not complicated. but it is essentially in two parts. Solution: Use the product rule to find all the parts needed (with the chain rule used within these parts): 𝑑𝑒𝑟 1𝑠𝑡 = 3 4𝑥 = 12(4𝑥)2 𝑑𝑒𝑟 2𝑛𝑑 = 6 −3𝑥 2 − 1 = −36𝑥 −3𝑥 2 − 5𝑥 5 5 2 where 𝑔(𝑥) and (𝑥) are separate functions. 𝑑𝑦 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 𝑑𝑥 Example 1: find the gradient function of 𝑓(𝑥) = (15𝑥 2 + 3𝑥 3 )(−4𝑥 7 + 6) Plan: use the product rule: 𝑓 ′ 𝑥 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 Solution: defining the two functions: 𝑓(𝑥) = (15𝑥 2 + 3𝑥 3 )(−4𝑥 7 + 6) 2𝑛𝑑 1𝑠𝑡 Following the product rule. at times. but determining the two functions is sometimes difficult. it would take too much time. and the simplifications can.25 𝑥 + 0. plus the derivative of the second part multiplied by the first part”. be complex. If you look carefully in both sections.5𝑥 7𝑥 3 + 7𝑥 3 106 . Try simplifying this last line using the crab claw (twice). 𝑥 1𝑠𝑡 2𝑛𝑑 Plan: two rules will be required: (1) the product rule as our “base rule”.Theory: given the function 𝑓 𝑥 = 𝑔 𝑥 . Differentiate the following using the product rule: 𝑎) 𝑦 = 5𝑥 4 3𝑥 2 + 1 𝑏) 𝑦 = 𝑥 3 + 1 𝑥 2 + 𝑥 5 𝑐) 𝑦 = −3𝑥 2 − 7 13𝑥 4 − 𝑥 𝑑) 𝑦 = 12𝑥 4 + 12𝑥 5 𝑥 7 2 𝑒) 𝑦 = 15𝑥 3 + 7𝑥 13𝑥 7 + 4 1 2 𝑓) 𝑦 = 15𝑥 3 + +5 𝑥 𝑥 𝑥 1 1 𝑔) 𝑦 = 17 𝑥 5 − 4 + 𝑥 𝑥 𝑥 2 ) 𝑦 = 0. do one section at a time: 𝑑𝑒𝑟 1𝑠𝑡 = 30𝑥 + 9𝑥 2 𝑑𝑒𝑟 2 𝑛𝑑 4 −6𝑥 Substitute into the product rule formula: 𝑑𝑦 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 𝑑𝑥 = 12 4𝑥 2 −3𝑥 2 − 1 6 + −36𝑥 −3𝑥 2 − 1 5 4𝑥 3 This looks very big. separated by an addition sign. The derivative function is then: 𝑓 ’(𝑥) = 𝑔’(𝑥) ∙ (𝑥) + 𝑔(𝑥) ∙ ’(𝑥) Informally: “the derivative function is the derivative of the first part multiplied by the second part. Do it yourself. except now. and that is what is done. differentiate the following: 𝑎) 𝑦 = 2𝑥 + 1 3 7𝑥 2 − 1 𝑏) 𝑦 = 1 − 𝑥 5 2𝑥 2 − 1 2 3 1 1 2 𝑐) 𝑦 = + 𝑥 2 𝑥 𝑥 −4 5 5 𝑑) 𝑦 = 4 𝑥 + 𝑥 7 − 𝑥 3 −1 −𝑥 −3 𝑑𝑦 𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇𝑇𝑂𝑀 − 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 × 𝑇𝑂𝑃 = 𝑑𝑥 [𝐵𝑂𝑇𝑇𝑂𝑀]2 Solution: using the quotient rule. 𝑑𝑦 𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇𝑇𝑂𝑀 − 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 × 𝑇𝑂𝑃 = 𝑑𝑥 [𝐵𝑂𝑇𝑇𝑂𝑀]2 = 225𝑥 4 + 60𝑥 3 + 15𝑥 2 + 6𝑥 [5𝑥 + 1]2 𝑑𝑦 3𝑥 75𝑥 3 + 20𝑥 2 + 5𝑥 + 2 = 𝑑𝑥 [5𝑥 + 1]2 3𝑥 could also be factorised: If simplification is obvious. especially if you need to use the crab-claw method. differentiate the two parts: 𝑑𝑒𝑟 𝑇𝑂𝑃 = 60𝑥 3 + 6𝑥 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 5 Substitute each section in brackets into the quotient rule: 5.6 differentiation rules: quotient rule This rule is similar to the product rule. in this last case by using the crab-claw method. Example 1: differentiate 𝑦 = 15𝑥 + 3𝑥 5𝑥 + 1 4 2 Plan: use the quotient rule: 107 . Theory: the quotient rule applies for functions of the form: 𝑦(𝑥) = 𝑡(𝑥) 𝑏(𝑥) 2 𝑑𝑦 60𝑥 3 + 6𝑥 5𝑥 + 1 − 5 (15𝑥 4 + 3𝑥 2 ) = 𝑑𝑥 [5𝑥 + 1]2 You don’t usually need to expand the denominator.2. always simplify the numerator. these can look like: 𝑦 = 3𝑥 1 + 𝑥 −3 𝑒 2𝑥 + 𝑥 𝑦 = 4 3 𝑥 + 3𝑥 Theory: formally. the quotient rule is 𝑑𝑦 𝑡 ′ 𝑥 × 𝑏 𝑥 − 𝑏 ′ (𝑥) × 𝑡(𝑥) = 𝑑𝑥 𝑏(𝑥) 2 Informally. This is because sometimes you might be able to cancel it off. all divided by the bottom squared”. then it should be done. as the following is only the answer: 𝑑𝑦 300𝑥 4 + 60𝑥 3 + 30𝑥 2 + 6𝑥) − 75𝑥 4 + 15𝑥 2 = 𝑑𝑥 5𝑥 + 1 2 For example. Using all the differentiation rules learnt so far. Example 2: find the gradient function of 𝑓 𝑥 = 3𝑥 2 − 𝑥 −4 2𝑥 + 1 Plan: use the quotient rule 𝑑𝑦 𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇𝑇𝑂𝑀 − 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 × 𝑇𝑂𝑃 = 𝑑𝑥 [𝐵𝑂𝑇𝑇𝑂𝑀]2 Solution: find the derivative of each of the two parts: 𝑑𝑒𝑟 𝑇𝑂𝑃 = 6𝑥 + 4𝑥 −5 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 2 Substitute back into the quotient rule formula (remember the brackets): 𝑑𝑦 6𝑥 + 4𝑥 −5 2𝑥 + 1 − 2 3𝑥 2 − 𝑥 −4 = 𝑑𝑥 2𝑥 + 1 2 Ask yourself. would you naturally differentiate the top or bottom first? Most people say they would differentiate the top first. For the numerator (top part of the fraction) the order of what you differentiate first is important as there is a minus sign. we have one function divided by another function. However. it is: “the derivative function is the derivative of the top times the bottom minus the derivative of the bottom times the top. Exercises: 1. but the more you practice. Differentiate the following using any of the rules learnt so far: 3𝑥 4 − 𝑥 3 𝑎) 𝑦 = 5𝑥 − 1 2𝑥 − 1 7 𝑏) 𝑦 = 3𝑥 + 1 3 15𝑥 3 + 3𝑥 2 𝑐) 𝑦 = 1 − 𝑥 2 12𝑥 2 𝑥 8 𝑥 −9 𝑑) 𝑦 = 1 + 𝑥 5 8 13𝑥 2 + 14 𝑒) 𝑦 = 18𝑥 − 3 2 𝑓) 𝑔) ) 𝑦 = − 5𝑥 4 +1 3 12𝑥 2 − 𝑥 2 𝑑𝑦 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 𝑑𝑥 Solution: find the derivative of each section of the base rule: 𝑑𝑒𝑟 𝑇𝑂𝑃 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 = 8𝑥 −3𝑥 −5 + 15𝑥 −6 4𝑥 2 + 1 = 36𝑥 −4 + 15𝑥 −6 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 10𝑥 4 Substitute back into the Quotient Rule: 𝑑𝑦 𝑑𝑥 = 36𝑥 −4 + 15𝑥 −6 2𝑥 5 − 10𝑥 4 4𝑥 2 + 1 (−3𝑥 −5 ) [2𝑥 5 ]2 Factorise out 2𝑥 4 from the top. Differentiate the following using the quotient rule: 𝑥 + 1 𝑎) 𝑦 = 3𝑥 − 1 𝑥 2 𝑏) 𝑦 = 𝑥 + 1 2𝑥 3 + 1 𝑐) 𝑦 = 5𝑥 + 𝑥 2 3𝑥 2 + 4𝑥 3 𝑑) 𝑦 = 𝑥 2 14𝑥 4 + 5 𝑒) 𝑦 = 13𝑥 3 − 𝑥 2 2. and also simplify the denominator (which we wouldn’t usually do. the easier it will become. then don’t bother. Keeping track of all the separate sections as well as simplification are the hardest parts of using the quotient rule. except here. Example 3: find the derivative of 4𝑥 2 + 1 (−3𝑥 −5 ) 𝑦 = 2𝑥 5 Plan: use the quotient rule as the base rule and the product rule within the quotient rule: 𝑑𝑦 𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇𝑇𝑂𝑀 − 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 × 𝑇𝑂𝑃 = 𝑑𝑥 [𝐵𝑂𝑇𝑇𝑂𝑀]2 𝑑𝑦 [𝑥 36𝑥 −4 + 15𝑥 −6 − 5 4𝑥 2 + 1 −3𝑥 −5 ] = 𝑑𝑥 2𝑥 6 Crab-claw the numerator (do it yourself!) and you should get: 𝑑𝑦 96𝑥 −3 + 30𝑥 −5 = 𝑑𝑥 2𝑥 6 A rule of thumb for simplification using the quotient rule is: if you can’t factorise anything out or do a simple crab-claw.Then use the crab claw method (you should do it!) to simplify the top: 𝑑𝑦 12𝑥 2 + 8𝑥 −4 + 6𝑥 + 4𝑥 −5 − 6𝑥 2 + 2𝑥 −4 = 𝑑𝑥 2𝑥 + 1 2 Bring together “like terms”: 𝑑𝑦 6𝑥 2 + 10𝑥 −4 + 6𝑥 + 4𝑥 −5 = 𝑑𝑥 2𝑥 + 1 2 Factorise out the number 2 from the top: 𝑑𝑦 2 3𝑥 2 + 5𝑥 −4 + 3𝑥 + 2𝑥 −5 = 𝑑𝑥 2𝑥 + 1 2 This next example combines the quotient rule with other differentiation rules. we can see that part of it will cancel off with the denominator): 𝑑𝑦 2𝑥 4 [𝑥 36𝑥 −4 + 15𝑥 −6 − 5 4𝑥 2 + 1 −3𝑥 −5 ] = 𝑑𝑥 4𝑥 10 3 1 + 𝑥 2 𝑦 = 𝑥 𝑥 − 1 2 1 1 + 𝑥 𝑥 2 𝑦 = 1 1 3 − 𝑥 𝑥 2 Cancel off the 2𝑥 4 with part of the bottom: 108 . by putting the 3 into each of the terms in the bracket: 𝑑𝑦 2 = (6𝑥 + 6)𝑒 𝑥 +2𝑥 𝑑𝑥 Solution: the two functions are underlined 𝑦 = 4𝑥 3 + 𝑥 𝑒 −3𝑥 𝑑𝑒𝑟 1𝑠𝑡 = 12𝑥 2 + 1 𝑑𝑒𝑟 2𝑛𝑑 = (15𝑥 −6 )𝑒 −3𝑥 −5 −5 Substitute into the base (product) rule: 𝑑𝑦 −5 −5 = 12𝑥 2 + 1 𝑒 −3𝑥 + (15𝑥 −6 𝑒 −3𝑥 )(4𝑥 3 + 𝑥) 𝑑𝑥 𝑑𝑦 −5 = 𝑒 −3𝑥 (12𝑥 2 + 1 + 60𝑥 −3 + 15𝑥 −5 ) 𝑑𝑥 A step was deliberately missed. Exercises: 1. bring it out front in brackets. but it could just as easily have been 𝑦 = 𝑒 𝑥 . not a variable. Example 1: differentiate the function 𝑦 = 𝑒 2𝑥+1 Simplify the 14 into the first bracket to get: 𝑔′ 𝑥 = 42𝑥 2 + 154 𝑒 𝑥 could the 𝑒 𝑥 3 +11𝑥−1 3 +11𝑥−1 The 14 was simplified into the front bracket as . then multiply it by the original function. This is also an exponential function because 𝑒 is a number. In Chapter 4. then multiply by the original function 𝑔′ 𝑥 = 3𝑥 2 + 11 14𝑒 𝑥 3 +11𝑥−1 𝑥 𝑑𝑦 = 𝑓 ′ 𝑥 𝑒 𝑓 𝑑𝑥 𝑥 The derivative is 𝑦 = 𝑓 𝑥 𝑒 The informal 𝑒 rule is: differentiate the exponent. Example 4: find the gradient function of 𝑦 = (4𝑥 3 + 𝑥)𝑒 −3𝑥 the 𝑒 rule within it: −5 Plan: use the product rule as the base rule with 𝑑𝑦 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 𝑑𝑥 𝑦 ′ = 𝑓 ′ 𝑥 𝑒 𝑓 𝑥 Plan: use the exponential rule “differentiate the exponent. bring it out front in brackets. there is no reason why it could not have an index.5. Either way is fine. there is no point. Bring this out front and then multiply it by the original function: 𝑑𝑦 = (2)𝑒 2𝑥+1 𝑑𝑥 Example 2: differentiate 𝑔 𝑥 = 3𝑒 𝑥 2 +2𝑥 we do not. Theory: the formal 𝑒 rule is: For the function 𝑦 = 𝑒 𝑓 ′ ′ 𝑥 𝑓 𝑥 The 𝑒 𝑥 2 +2𝑥 could be crab-clawed into the brackets but in this case. Differentiate the following using the 𝑒 rule: 𝑎) 𝑦 = 𝑒 2𝑥−3 𝑏) 𝑦 = 𝑒 −4𝑥−2 109 . then multiply by the original equation” Solution: differentiate the exponent: 𝑑𝑒𝑟 𝐸𝑋𝑃𝑂𝑁𝐸𝑁𝑇 = 2𝑥 + 2 Then bring it out front in brackets and multiply by the original equation: 𝑑𝑦 2 = 2𝑥 + 2 (3𝑒 𝑥 +2𝑥 ) 𝑑𝑥 This can be simplified a bit. Example 3: find the derivative of 𝑥 = 14𝑒 𝑥 3 +11𝑥−1 Plan: use the 𝑒 rule of differentiation 𝑦 = 𝑒 𝑓 Solution: 𝑑𝑒𝑟 𝐸𝑋𝑃𝑂𝑁𝐸𝑁𝑇 = 3𝑥 2 + 11 Put this out front in brackets. an exponential function of the form 𝑦 = 2𝑥 was discussed.7 differentiation rules: 𝒆 Remembering that 𝑒 is just a number. however it looks “cleaner” if Solution: differentiatie the exponent to get 2. so you should do it and see if you can get the same answer. If you don’t understand this. just the inside of the brackets). ′ 𝑓 ′ 𝑥 𝑓 𝑥 = 4𝑥 3 𝑒 𝑥 𝑒 𝑥 4 = 4𝑥 3 In 𝑑𝑒𝑟 2𝑛𝑑 both the ln rule and the 𝑒 rule were used. the derivative function would be: 𝑑𝑦 2𝑥 = 2 𝑑𝑥 𝑥 + 1 Ignore the ln when differentiating the function. but it’s not entirely vital for this section. The 𝑒 𝑥 cancelled off so only 4𝑥 3 was left. Example 2: find the derivative of 𝑦 = 3𝑥 5 + 1 ∙ ln(𝑒 𝑥 ) Plan: use the product rule (as there are two functions being multiplied together) as the base rule 𝑑𝑦 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 𝑑𝑥 then the ln rule and the 𝑒 rule within the base rule. Substitute into the base (product) rule: 𝑑𝑦 4 = 15𝑥 4 ∙ ln 𝑒 𝑥 + 4𝑥 3 (3𝑥 5 + 1) 𝑑𝑥 Factorise out 𝑥 3 and simplify inside the brackets: 𝑑𝑦 4 = 𝑥 3 15𝑥 ∙ ln 𝑒 𝑥 + 4 3𝑥 5 + 1 𝑑𝑥 𝑑𝑦 4 = 𝑥 3 15𝑥 ∙ ln 𝑒 𝑥 + 12𝑥 5 + 4 𝑑𝑥 4 110 . then divide by the inside of the brackets” Solution: simplify the inside of the brackets: 4 𝑦 = ln 3𝑥 −4 + 16𝑥 − 5 Differentiate using the ln rule: 𝑑𝑦 −12𝑥 −5 + 16 = −4 𝑑𝑥 3𝑥 + 16𝑥 − 5 That is the extent of the ln rule. go back to Chapter 4.𝑐) 𝑑) 2. It only gets more complex when other rules also need to be used. then use the ln rule: “differentiate the inside of the brackets. then divide by the inside of the brackets (NOT the whole function. 𝑦 = 𝑒 𝑥 𝑦 = 𝑒 2 +1 Example 1: find the gradient function of 𝑦 = ln 3 + 16𝑥 − 5 𝑥 4 3𝑥 2 +𝑥 1 +𝑥 𝑥 𝑒) 𝑦 = 𝑒 Differentiate the following using any of the rules learnt so far: 2 𝑎) 𝑦 = 𝑒 3𝑥 +1 1 + 𝑒 3𝑥 𝑏) 𝑐) 𝑑) 𝑒) 𝑓) 𝑔) ) 𝑦 = 𝑒 10 − 𝑥 15𝑥 2 + 𝑥 𝑦 = 𝑒 9𝑥−2 𝑦 = 𝑒 2𝑥−1 + 𝑥 − 2 𝑦 = 12 𝑦 = 1 𝑒 𝑥 2 3−2𝑥 2 3 Plan: simplify the brackets. Solution: the two functions for the product rule are underlined 𝑦 = 3𝑥 5 + 1 ∙ ln(𝑒 𝑥 ) 𝑑𝑒𝑟 1𝑠𝑡 = 15𝑥 4 𝑑𝑒𝑟 2 𝑛𝑑 4 4 4 + 5𝑥 4 1 + 𝑥 𝑒 3−2𝑥 1 1 5 𝑦 = + 4𝑥 2 𝑥 𝑒 4 𝑒 𝑥 −1 𝑦 = 𝑥 + 𝑒 3−3𝑥 5. Theory: formally. So in this case.8 differentiation rules: 𝐥𝐧 rule The natural log is the log of a function with 𝑒 as the base rather than 10. If you are to find the derivative function of 𝑦 = ln(𝑥 2 + 1) Differentiate the inside of the brackets. then divide by the inside of the brackets”. the ln rule is: For the function 𝑦 = ln 𝑓(𝑥) The gradient function is 𝑦 = Informally: To differentiate the log of something in brackets: “differentiate the inside of the brackets. Exercises: 1. Differentiate the following using the ln rule: 𝑎) ln 𝑥 𝑏) ln 15𝑥 𝑐) ln 𝑥 2 𝑑) ln 𝑥 2 + 3 𝑒) ln 3𝑥 4 + 17𝑥 2 𝑓) ln 𝑒 𝑥+1 − 1 2. Differentiate the following using any of the rules learnt so far: 𝑎) 𝑦 = ln 𝑥 2 + 1 3 − 𝑥 4 𝑏) 𝑐) 𝑑) 𝑒) 𝑓) 𝑔) ) 𝑦 = 𝑒 𝑥 2 +3𝑥 𝑑𝑒𝑟2𝑛𝑑 = 2 2𝑥 2 + 1 4𝑥 = 8𝑥 2𝑥 2 + 1 Substitute it into the product rule: 𝑦 ′′ = 12 2𝑥 2 + 1 Factorise out 12 2𝑥 2 + 1 : 𝑦 ′′ = 12 2𝑥 2 + 1 2𝑥 2 + 1 + 8𝑥 2 𝑦 ′′ = 12 2𝑥 2 + 1 10𝑥 2 + 1 Example 2: find the second derivative of 𝑦 = 𝑒 3𝑥−6 1 − 𝑥 3 −2 2 + 8𝑥 2𝑥 2 + 1 12𝑥 ln 𝑥 2 + 3𝑥 8 + 𝑥 3 𝑥 𝑦= ln −4𝑥 3 + 7𝑥 2 2 𝑦 = ln 5 − 𝑥 𝑒 6−3𝑥 𝑥 1 2 𝑦 = + ln 3𝑥 2 − 1 𝑥 𝑥 2 2 𝑦 = ln 𝑒 𝑥 + 𝑥 4𝑥 3 ln 5𝑥 − 3 𝑦 = 𝑥 2 + 1 𝑦 = ln ln 3𝑥 Plan: differentiate, then differentiate the derivative. Solution: simplify first: 𝑦 = 𝑒 3𝑥−6 1 + 0.5𝑥 3 Differentiate using the product rule: 𝑑𝑒𝑟1𝑠𝑡 = 3𝑒 3𝑥−6 𝑑𝑒𝑟2𝑛𝑑 = 1.5𝑥 2 Substitute into the base (product) rule: 𝑦 ′ = 3𝑒 3𝑥−6 1 + 0.5𝑥 3 + 1.5𝑥 2 𝑒 3𝑥−6 Now to find the second derivative, you must use the product rule twice (for the two sections shown below). You should do this one section at a time to avoid confusion: 𝑦 ′ = 3𝑒 3𝑥−6 1 + 0.5𝑥 3 + 1.5𝑥 2 𝑒 3𝑥−6 Differentiate the red section first: 𝑑𝑒𝑟1𝑠𝑡 = 9𝑒 3𝑥−6 𝑑𝑒𝑟2𝑛𝑑 = 1.5𝑥 2 Substitute into the product rule formula: 𝑦 ′′ 𝑟𝑒𝑑 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = 9𝑒 3𝑥−6 1 + 0.5𝑥 3 + 1.5𝑥 2 3𝑒 3𝑥−6 𝑦 ′′ 𝑟𝑒𝑑 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = 9𝑒 3𝑥−6 1 + 0.5𝑥 3 + 4.5𝑥 2 𝑒 3𝑥−6 5.9 the second derivative Theory: the second derivative of a function is obtained by differentiating the function once, and then differentiating the derivative. Similar to the first derivative, there are several different notations for a second derivative, the common ones being: 𝑦 ′′ 𝑎𝑛𝑑 𝑑2 𝑦 𝑑𝑥 2 Example 1: find the second derivative of 𝑦 = 2𝑥 2 + 1 3 Plan: differentiate the function, then differentiate that derivative. Solution: find the first derivative: 𝑦 = 3 2𝑥 + 1 ′ 2 2 4𝑥 2 ′ Differentiate the blue section: 𝑑𝑒𝑟1𝑠𝑡 = 3𝑥 𝑑𝑒𝑟2𝑛𝑑 = 3𝑒 3𝑥−6 Substitute into the product rule formula: 𝑦 ′′ 𝑏𝑙𝑢𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = 3𝑥𝑒 3𝑥−6 + 3𝑒 3𝑥−6 1.5𝑥 2 𝑦 ′′ 𝑏𝑙𝑢𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = 3𝑥𝑒 3𝑥−6 + 4.5𝑒 3𝑥−6 𝑥 2 𝑦′ = 12𝑥 2𝑥 2 + 1 To find the second derivative, 𝑦 must be differentiated, and the product rule is required. 𝑑𝑦 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 𝑑𝑥 𝑑𝑒𝑟1𝑠𝑡 = 12 111 The second derivative is the sum of the derivative of the red section and the blue section: 𝑦 ′′ = 9𝑒3𝑥−6 1 + 0.5𝑥3 + 4.5𝑥2 𝑒3𝑥−6 + 3𝑥𝑒 3𝑥−6 + 4.5𝑒 3𝑥−6 𝑥 2 Factorise out 𝑒 3𝑥−6 2. Find the second derivative of: 2 𝑎) 𝑦 = 𝑒 2𝑥 5𝑥 2 𝑥 2 𝑏) 𝑦 = ln +1 3 𝑐) 𝑦 = ln 4𝑥 3 + 2𝑥 2 𝑑) 𝑒) 𝑦 = 𝑒 4𝑥 7 − 𝑒 𝑥 1 3 2 because each term has 𝑒 3𝑥−6 : 𝑦′′ = 𝑒 3𝑥−6 9 1 + 0.5𝑥3 + 4.5𝑥2 + 3𝑥 + 4.5𝑥 2 Then simplify everything inside the square brackets: 𝑦 ′′ = 𝑒 3𝑥−6 4.5𝑥 3 + 9 + 9𝑥 2 + 3𝑥 After the first derivative was found, the 𝑒 3𝑥−6 could have been factorised out, and then the derivative found that way. Both methods will give the same solution. Try it. Example 3: find the second derivative of 𝑦 = ln(3𝑥 4 + 𝑥) Plan: differentiate the function, then apply differentiation rules to differentiating the first derivative. Solution: the first derivative is 12𝑥 3 + 1 𝑦 ′ = 3𝑥 4 + 𝑥 To differentiate this, the quotient rule is needed. 𝑑𝑒𝑟 𝑇𝑂𝑃 = 36𝑥 2 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 12𝑥 3 + 1 Replacing the base (quotient) rule: 𝑦 ′′ = 36𝑥 2 3𝑥 4 + 𝑥 − 12𝑥 3 + 1 12𝑥 3 + 1 3𝑥 4 + 𝑥 2 2 𝑦= 𝑒 2𝑥−1 ln 𝑒 2𝑥 + 1 5.10 the gradient function graph Just like any function, a gradient function can be graphed. 𝑦 7 𝑑𝑦𝑑𝑥 1 3 𝑥 Theory: The gradient function is related to the original function: any value of the gradient function at a given 𝑥 − 𝑣𝑎𝑙𝑢𝑒 is the gradient at that same 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the original function. That is, to find the gradient of any point on the original function (which has not been shown above), you go up from that point on the 𝑥 − 𝑎𝑥𝑖𝑠 to hit the gradient function, then across to the 𝑦 − 𝑎𝑥𝑖𝑠 which gives a value; this value is the gradient of the original function. On the graph above, the gradient of the original function at 𝑥 = 1, is 2. At 𝑥 = 3, the gradient of the original function is 7 (Note: the original function is not shown). Shown below is shown a function as well as its gradient function, to reinforce the relationship. Try simplifying this last equation to see if you get the following: 𝑦 ′′ = −36𝑥 6 + 12𝑥 3 − 1 3𝑥 4 + 𝑥 2 Exercises: 1. Find the second derivative of the following functions: 𝑎) 𝑦 = 𝑥 3 + 𝑥 2 − 1 𝑥 4 𝑏) 𝑦 = + 𝑥 2 −4 1 𝑐) 𝑦 = 2 𝑥 + 𝑥 𝑑) 𝑦 = 3 5 − 3𝑥 5 𝑒) 𝑦 = 2𝑥 2 1 + 3𝑥 4 112 𝑦 𝑑𝑦𝑑𝑥 is from the 𝑥 − 𝑎𝑥𝑖𝑠. Similarly, the smaller the slope of the original function, the closer the gradient function is to the 𝑥 − 𝑎𝑥𝑖𝑠 at that point. −2 1 𝑥𝑦 = 𝑓(𝑥) Re-read this section before trying the following exercises, as it takes a little time to fully understand. Whenever the original function (blue) is sloping upwards (i.e. a positive gradient), the gradient function (red) is above the 𝑥 − 𝑎𝑥𝑖𝑠 (i.e. is positive). For any 𝑥 − 𝑣𝑎𝑙𝑢𝑒 to the left of 𝑥 = −2, the gradient of the original function is positive, and so the gradient function to the left of 𝑥 = −2 is above the 𝑥 − 𝑎𝑥𝑖𝑠. Similarly for any 𝑥 − 𝑣𝑎𝑙𝑢𝑒 to the right of 𝑥 = 1. Whenever the original function is sloping downwards (i.e. negative gradient), the gradient function is below the 𝑥 − 𝑎𝑥𝑖𝑠. That is, any 𝑥 − 𝑣𝑎𝑙𝑢𝑒 between 𝑥 = −2 and 𝑥 = 1, the slope of the original function is negative, and so the gradient function is below the 𝑥 − 𝑎𝑥𝑖𝑠. Where there is a maximum or minimum on the original function, the gradient is zero, and this is where the gradient function crosses the 𝑥 − 𝑎𝑥𝑖𝑠. That is, at 𝑥 = −2, 𝑥 = 1. Theory: a positive gradient on the original function gives a value above the 𝑥 − 𝑎𝑥𝑖𝑠 on the gradient function. A negative gradient on the original function gives a value below the 𝑥 − 𝑎𝑥𝑖𝑠 on the gradient function. When the original function is at a stationary point, the gradient function crosses the 𝑥 − 𝑎𝑥𝑖𝑠. An inflection point on the original function maps to a turning point on the gradient function (i.e. either a maximum or a minimum). The steeper the slope at a particular point on the original function, the further the gradient function Exercises: 1. Sketch the original function (i.e. the quadratic): 𝑦 = 3 𝑥 + 3 2 − 5 (using theory from Chapter 4) then sketch the gradient function on the same set of axes. 2. Given the function graphed below, sketch in the gradient function. 𝑦𝒚 = 𝒇(𝒙) 𝑥 3. Given the gradient function below, sketch in the original function. 𝑦 𝒅𝒚𝒅𝒙 𝑥 5.11 simple applications Remember that the gradient function gives us a method of finding the gradient of a point on the original function. Theory: to find the gradient of a point 𝑥 = 𝑎 where 𝑎 is a number, differentiate the original function to get the gradient function, then put the number 𝑎 into the gradient function. The value 113 ) 𝑒 −12 − 1 2 Thus the gradient of the original function at 𝑥 = −3 is approximately 8.02𝑄 − 30𝑄 − 400 Will increasing output lead to increased profits? Justify your answer.e. profits will also increase. 𝑑. Determine the gradient of the function 2 𝑦 = 𝑒 𝑥 𝑥 3 + 2 2 At the points 𝑄 = 2. more output will increase 4𝑥 Substitute into the quotient rule: − 1 − 4 𝑒 𝑒 4𝑥 − 1 2 𝑥 − 3𝑥 2 profits. Substitute the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 (one at a time) into the gradient function. To get this number. Exercises: 1. Plan: use the quotient rule as the base rule 𝑑𝑦 𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇 − 𝑑𝑒𝑟 𝐵𝑂𝑇 × 𝑇𝑂𝑃 = 𝑑𝑥 [𝐵𝑂𝑇𝑇𝑂𝑀]2 then use the 𝑒 rule to find the gradient function. Determine the gradient of the function 1 𝜋 = 5𝑄4 + − 7 𝑄 At the points 𝑄 = 4. Simplification is not required as only the final number is needed (i. A firm manufacturing cranes has current output of 10 and a profit function: 𝜋 = −207𝑄 log 0.03𝑄2 + 0. Solution: 𝑑𝜋 = −2𝑄 + 150 𝑑𝑄 Substitute 𝑄 = 70 into this gradient function: 𝑑𝜋 70 = −140 + 150 = 10 > 0 𝑑𝑄 Since the gradient is positive. 3. 𝑄 = 7. substitute 𝑥 = −1 into the gradient function: 𝑑𝑦 −2 − 3 𝑒 −4 − 1 − 4 𝑒 −4 1 + 3 −1 = 𝑑𝑥 𝑒 −4 − 1 2 Simplify: 114 . 𝑝. If the gradient is negative. substitute 𝑥 = −3 into the gradient function: 𝑑𝑦 −3 𝑑𝑥 = −6 − 3 𝑒 −12 − 1 − 4 𝑒 −12 9 + 9 𝑒 −12 − 1 2 𝑑𝑦 −9𝑒 + 9 − 72𝑒 −3 = −12 − 1 2 𝑑𝑥 𝑒 = −12 −12 Simplifying gives: −81𝑒 −12 + 9 ≈ 8. 2. For the gradient at 𝑥 = −1. then increased output will reduce profits. ) 𝑒 −4 − 1 2 which is the gradient of the original function at 𝑥 = −1. if output is increased. If the gradient is positive.9996. then substitute 𝑄 = 70 into the gradient function.7892 (4𝑑. the gradient at a point). Example 1: find the gradient of 𝑥 2 − 3𝑥 𝑦 = 4𝑥 𝑒 − 1 when 𝑥 = −3 and also when 𝑥 = −1.9996 (4𝑝. Plan: differentiate the profit function. Use the properties of the gradient function to determine if it is more profitable for the firm to increase production. Example 2: a car manufacturer has a monthly profit function 𝜋 = −𝑄 2 + 150𝑄 − 5000 and is currently producing 70 cars per month. Give exact answers and approximations.from the gradient function is the gradient of the original function at 𝑥 = 𝑎. Give exact answers. 𝑄 = 5. Solution: 𝑑𝑒𝑟 𝑇𝑂𝑃 = 2𝑥 − 3 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 4 𝑒 𝑑𝑦 2𝑥 − 3 𝑒 = 𝑑𝑥 4𝑥 4𝑥 𝑑𝑦 −5𝑒 −4 + 5 − 16𝑒 −4 𝑥 = −1 = 𝑑𝑥 𝑒 −4 − 1 2 = −21𝑒 −4 + 5 ≈ 4. 𝑦 = 8𝑥 −4 + 𝑥 3 − 6 𝑦 = 12𝑥 −2 + 𝑥 −3 + 𝑥 3 + 2 1 𝑔) 𝑦 = 2 𝑥 15 ) 𝑦 = 2 + 𝑥 2 3𝑥 1 1 1 𝑖) 𝑦 = + 2 − 3 𝑥 𝑥 𝑥 Using the chain rule. Use differentiation by first principles to determine the gradient of the following functions: 𝑎) 𝑦 = 18𝑥 𝑏) 𝑦 = 𝑥 2 𝑐) 𝑦 = (𝑥 + 3)(𝑥 + 4) 𝑑) 𝑦 = 𝑥 3 + 4𝑥 2 5 1 𝑒) 𝑦 = 𝑓) 𝑦 = 2 𝑥 𝑥 Using the power rule. then put the number 𝑎 into the gradient function. differentiate the functions: 𝑎) 𝑦 = 𝑥 + 1 2 𝑏) 𝑦 = 3𝑥 − 4 5 𝑐) 𝑦 = 2 𝑥 + 5 6 𝑑) 𝑦 = 12 1 − 4𝑥 3 𝑒) 𝑦 = 75 𝑥 2 + 𝑥 4 3 4𝑥 − 1 4 𝑓) 𝑦 = − 5 Using the product rule. For the function: 𝑡(𝑥) 𝑦(𝑥) = 𝑏(𝑥) The gradient function is: The second derivative of a function is obtained by differentiating the function once. 𝐷 𝐸 3. When the original function is at a stationary point. differentiate the following functions (Hint: underline the two functions first): 𝑎) 𝑦 = 𝑥 3 15𝑥 + 2 𝑏) 𝑦 = 𝑥 3 − 1 𝑥 9 + 1 𝑐) 𝑦 = 15𝑥 4 1 − 𝑥 3 𝑑) 𝑦 = 4𝑥 −7 𝑥 2 + 𝑥 4 1 𝑒) 𝑦 = − 𝑥 3 + 𝑥 4 𝑥 5 3 3 𝑒) 𝑓) chapter five questions 1. Differentiation by First Principles is the formula: 𝑓 𝑥 + 𝑠 − 𝑓 𝑥 𝑓 ′ = lim 𝑠→0 𝑠 The power rule: For the original function 𝑦 = 𝑎𝑥 𝑛 The derivative function is 𝑦 ′ = 𝑛𝑎𝑥 𝑛−1 The derivative of a constant is zero The chain rule: differentiate the function as if using the power rule. then multiply by the derivative of the inside of the brackets. and then differentiating the derivative. 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐𝑎𝑛𝑔𝑒 = 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 The gradient function is a function that gives the gradient of the original function at all the points along the original function. The gradient function is related to the original function: any value of the gradient function at a given 𝑥 − 𝑣𝑎𝑙𝑢𝑒 is the gradient at that same 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the original function. (𝑥) + 𝑔(𝑥). A positive gradient on the original function gives a value above the 𝑥 − 𝑎𝑥𝑖𝑠 on the gradient function. Order the points on the following graph from lowest gradient to highest gradient: 𝐶 𝐵 𝑦 𝐹 𝐺 𝑥 𝐻 4. A negative gradient on the original function gives a value below the 𝑥 − 𝑎𝑥𝑖𝑠 on the gradient function. either a maximum or a minimum).e. 115 . The gradient of a line tangent to a given point is the rate of change of 𝑦 for changes in 𝑥. ’(𝑥) 𝑑𝑦 𝑜𝑟 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 𝑑𝑥 The quotient rule is. 𝐴 2. (𝑥) The derivative function is: 𝑓 ’(𝑥) = 𝑔’(𝑥). differentiate the following functions: 𝑎) 𝑦 = 3𝑥 2 𝑏) 𝑦 = 15𝑥 2 + 𝑥 𝑐) 𝑦 = 2 + 𝑥 2 + 𝑥 3 17𝑥 3 𝑑) 𝑦 = + 5𝑥 2 2 5. 𝑑𝑦 𝑑𝑦 𝑑𝑢 = × 𝑑𝑥 𝑑𝑢 𝑑𝑥 The product rule: For the function: 𝑑𝑦 𝑡 ′ 𝑥 × 𝑏 𝑥 − 𝑏 ′ (𝑥) × 𝑡(𝑥) = 𝑑𝑥 𝑏(𝑥) 2 𝑑𝑦 𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇𝑇𝑂𝑀 − 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 × 𝑇𝑂𝑃 = 𝑑𝑥 [𝐵𝑂𝑇𝑇𝑂𝑀]2 The formal 𝑒 rule is: For the function The derivative is Formally. An inflection point on the original function maps to a turning point on the gradient function (i. The value from the gradient function is the gradient of the original function at 𝑥 = 𝑎. the ln rule is: For the function The gradient function is 𝑦 = 𝑒 𝑓 𝑥 𝑦 ′ = 𝑓 ′ 𝑥 𝑒 𝑓 𝑥 𝑦 = ln 𝑓(𝑥) 𝑦 ′ = 𝑓 ′ 𝑥 𝑓 𝑥 𝑓(𝑥) = 𝑔(𝑥).chapter five summary Differentiation is finding the rate the 𝑦 − 𝑎𝑥𝑖𝑠 changes when there is a change in the 𝑥 − 𝑎𝑥𝑖𝑠. the gradient function crosses the 𝑥 − 𝑎𝑥𝑖𝑠. differentiate the original function to get the gradient function. To find the gradient of a point 𝑥 = 𝑎 where 𝑎 is a number. e) Sketch the total cost function from the marginal cost function. determine the derivative of the functions: 𝑎) 𝑦 = 2𝑥 3 − 1 2 3𝑥 + 5 3 𝑏) 𝑦 = 1 − 𝑒 −5𝑥−1 𝑥 − 1 4 3 𝑐) 𝑦 = ln 𝑒 0.e. 𝑦 𝑓 ′ 𝑥 𝑥 9. sketch in the gradient function: 𝑦 𝑦 = 𝑓(𝑥) 𝑥 13. sketch in the original function: 8. For the total cost function: 𝑇𝐶 = 𝑄3 − 15𝑄2 + 75𝑄 − 5 Determine: a) The gradient function (i.1𝑥 2 + 2 3 + 15𝑥 − 18 Determine: a) The marginal cost function. A company producing fishing rods faces a cost function of the form: 𝑇𝐶 = 50 ln 0. find the derivative of the following functions: 𝑎) 𝑦 = ln 5𝑥 𝑏) 𝑦 = ln 𝑥 2 𝑐) 𝑦 = 2 ln 5𝑥 − 1 𝑑) 𝑦 = −3 ln 𝑥 2 − 𝑥 4 𝑒) 𝑦 = − ln 𝑥 −3 + 𝑥 −3 1 𝑓) 𝑦 = 2 ln 𝑥 2 − 𝑥 −3 + 𝑥 𝑔) 𝑦 = 0.25 + 4𝑥 12. For the following gradient function. b) The gradient of the marginal cost function when 𝑄 = 30 and when 𝑄 = 60. b) When the marginal cost function is at a minimum.2𝑥 +𝑥 2 1 − 𝑒 𝑥 𝑑) 𝑦 = 2 1 + 𝑒 𝑥 𝑒) 𝑦 = 15 ln 12 − 𝑥 −3 4 1 𝑓) 𝑦 = 5 1 + ln 𝑥 𝑥 − 1 2 3𝑥−1 𝑔) 𝑦 = 5𝑥 𝑒 1 𝑒 𝑥 + 1 ) 𝑦 = ∙ −3 2𝑥 −3 − 𝑥 4 2 𝑖) 𝑦 = 2𝑥 3 − 1 7 1 + ln 𝑒 𝑥 + 1 𝑒 7𝑥−1 + 𝑥 4 𝑗) 𝑦 = + 18𝑥 4 − ln 𝑥 7 − 1 ln 𝑥 5 + 𝑥 −7 −3 𝑥 10. Find the first and second derivatives of the following functions: 𝑎) 𝑦 = 𝑥 4 + 15𝑥 3 − 1 17 𝑏) 𝑦 = 4 + ln 8𝑥 𝑥 𝑐) 𝑦 = 0. c) The gradient of the marginal cost function at 𝑄 = 15. 14. marginal cost function).03𝑥 9 + 𝑥 2 − 1 4 2 𝑑) 𝑦 = 𝑒 2 1 − 𝑥 4 + 𝑒 𝑥 −1 1 − 𝑥 12 − 𝑥 𝑒) 𝑦 = 3𝑥 2 + 1 𝑓) 𝑦 = ln 5𝑥 2 + 𝑥 11. Determine the gradient at 𝑄 = 10 and at 𝑄 = 35 for the production function: 𝑄 = ln 𝑥 0. 15. 7.04 ln 8𝑥 − 2𝑥 −2 Using any of the rules. For the following original function. 116 .6. 𝑥 −6 + 2𝑥 −1 − 𝑥 −3 −4 𝑔) 6 𝑥 −6 − 6 𝑥 6 − 6𝑥 Use the quotient rule to find the derivative of the following functions: 𝑥 + 1 2𝑥 + 1 𝑎) 𝑦 = 𝑏) 𝑦 = 𝑥 − 1 1 − 2𝑥 3𝑥 − 2 𝑥 2 + 1 𝑐) 𝑦 = 𝑑) 𝑦 = 2 1 − 𝑥 2 𝑥 − 1 13 +1 𝑥 2 − 𝑥 3 2 𝑒) 𝑦 = 𝑥 𝑓) 𝑦 = 2 5 − 𝑥 𝑥 − 2 13𝑥 −3 − 𝑥 3 𝑔) 𝑦 = +9 1 + 𝑥 −4 2 3 𝑥 + 1 + 𝑥 ) 𝑦 = 1 − 3𝑥 2 Use the 𝑒 rule to find the derivative of the following functions: 𝑎) 𝑦 = 𝑒 2𝑥 𝑏) 𝑦 = 𝑒 5𝑥−1 2 1−4𝑥 𝑐) 𝑦 = 𝑒 𝑑) 𝑦 = 𝑒 𝑥 −1 2 4 −3 𝑒) 𝑦 = 𝑒 13𝑥−𝑥 𝑓) 𝑦 = 2𝑒 𝑥 −1+𝑥 𝑓) 𝑦 = 2𝑒 𝑥 −2 4/𝑥 𝑔) 𝑦 = 12𝑒 ) 𝑦 = −7 Using the ln rule. d) Sketch the marginal cost function using quadratic theory. 6 6.3 6.Chapter 6 Applications of differentiation The usefulness of differentiation 6.2 6.4 6.1 6.8 6.10 Graphical Optimisation Mathematical Optimisation The Nature of an Optimal Point Inflection Points Combining All Theory Applications – Profits Applications – Break-Even Applications: Marginal and Average Values Differentiation and Elasticity Elasticity and Total Revenue 118 118 120 123 124 126 129 130 133 135 137 138 Chapter Six Summary Chapter Six Questions 117 .5 6.9 6.7 6. such as 25 or 90 units. on the 𝑥 − 𝑎𝑥𝑖𝑠. Point 𝐴 is a local maximum (a maximum over a small range around the point). This is where differentiation becomes very useful. does not give maximum profit. the consumer is finding a minimum. These are two optimisation problems. The following is a typical profit curve. 𝐵 𝐷 𝐸 𝐹 𝑥 118 .e. and in the other case.6.2 mathematical optimisation Finding optimal points graphically is quite simple. there are two maximum points and one minimum point. whereas point 𝐶 is a local and global maximum (a maximum over the entire number range: 𝑥 − 𝑎𝑥𝑖𝑠). 25 50 90 𝑄 𝑦 𝐴 𝐶 𝐺 Any rational person would aim to sell 50 units. but you have to make sure). the firm is finding a maximum. A local optimum is an optimal point only over a small range around that point. Any other quantity. Below is a graph where the gradients of optimal point have been drawn in. As a consumer. The maximum above is both a local and global maximum. A shop selling mobile phones aims to maximise profit. Exercises: 1. which may be difficult to draw.1 graphical optimisation Theory: optimisation is the process of finding an optimal point. with 𝜋 on the 𝑦 − 𝑎𝑥𝑖𝑠 and quantity. in one case. 𝑥 − 𝑎𝑥𝑖𝑠). determine which points are local maxima/minima and global maxima/minima. you would like to minimise how much you pay for a given phone. Point 𝐵 is a local minimum. The profit of a firm (usually denoted by 𝜋) is a function of the quantity of goods sold to consumers. Be aware that an optimal point may not always be the globally optimal point (most of the time it will be. 𝑄. For the graph below. but is not a global minimum as the ends of the curve are lower. however most of the time only the original function is given. with that point usually being a maximum or minimum. Theory: a global optimum is a point that is the optimal point over the entire number range (i. so these are not optimal points. How many units would you produce? 𝜋 𝜋 𝑄 𝐵 𝑦 𝐴 𝐶 𝑥 Above. 6. e. Theory: the First Order Condition (FOC) is found by setting the first derivative equal to zero: 𝑑𝑦 =0 𝑑𝑥 That is. This feature allows the original function to be differentiated to obtain the gradient function. FOC): 𝑑𝑦 = 3𝑥 2 + 6𝑥 − 24 = 0 𝑑𝑥 Solving for 𝑥 requires solving this quadratic. A point also requires a 𝑦 − 𝑣𝑎𝑙𝑢𝑒 which comes from the original function: 119 𝑦 = 𝑥 3 + 3𝑥 2 − 24𝑥 + 5 Plan: use the FOC to find the optimal point(s). Example 2: find the optimal point(s) for the function 𝑥 Theory: the gradient at any maximum or any minimum is always zero. Solve for the unknown(s). . the Quadratic Formula must be used: 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = −𝑏 ± 𝑏 2 − 4𝑎𝑐 2𝑎 −6 ± 62 − 4 3 (−24) 2(3) −6 ± 324 6 OR 𝑥 − 𝑣𝑎𝑙𝑢𝑒 = 2 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = −4 This gives two optimal points.85 and 2. Example 1: Find the optimal point(s) of the quadratic 𝑦 = 2𝑥 2 − 8𝑥 − 1 Plan: use the FOC to find the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 𝑠 of the optimal point(s) 𝑑𝑦 =0 𝑑𝑥 Solution: differentiate the original function: 𝑑𝑦 = 4𝑥 − 8 𝑑𝑥 Set it equal to zero: 𝑑𝑦 = 4𝑥 − 8 = 0 𝑑𝑥 Rearrange to solve for 𝑥: 4𝑥 = 8 𝑥 = 2 There is an optimal point at 𝑥 = 2. there should be one local maximum and one local minimum. but they also require 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 which come from the original function: 𝑦 −4 = −4 𝑦 −4 = 85 𝑦 2 = 2 3 3 + 3 −4 2 2 − 24 −4 + 5 +3 2 − 24 2 + 5 𝑦 2 = −23 Both −4.𝑦 𝑓 𝑥 𝑦 2 = 2 2 𝑦 2 = −9 2 −8 2 −1 Thus 2. 𝑑𝑦 =0 𝑑𝑥 Solution: differentiate the original function and set it to zero (i. −9 is an optimal point. From the cubics section in Chapter 4. To solve the quadratic. Determining which value corresponds to the maximum and which to the minimum is left to the next section. the gradient function is set equal to zero to solve for the optimal point(s). differentiate the function and set it equal to zero. −23 are optimal points. 𝑒 + 7 are optimal points. The nature of an optimal point is found (i. 2𝑥 = 0. but this will never occur as this never crosses the 𝑥 − 𝑎𝑥𝑖𝑠. or 3. 𝑒 2𝑥+3 = 0. Maximum if 𝑑2 𝑦 <0 𝑑𝑥 2 1 + 𝑥 = 0 For the solution. Profit is something which is usually maximised and if an optimal point was chosen using the FOC without checking it is a maximum. 120 .Example 3: find the optimal points of the function 𝑓 𝑥 = 𝑥 2 𝑒 2𝑥+3 + 7 Plan: differentiate using the product rule as the base rule: 𝑎) 𝑏) 𝑐) 𝑑) 𝑒) 𝑓) 𝑔) ) 𝑑𝑦 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 𝑑𝑥 Use the 𝑒 rule within the product rule: 𝑦 ′ = 𝑓 ′ 𝑥 𝑒 𝑓 Then apply the FOC: 𝑑𝑦 =0 𝑑𝑥 Solution: define the two functions: 𝑓 𝑥 = 𝑥 2 𝑒 2𝑥+3 + 7 Find the derivative of each of the two parts: 𝑑𝑒𝑟 1𝑠𝑡 = 2𝑥 𝑑𝑒𝑟 2𝑛𝑑 = 2𝑒 2𝑥+3 Substitute into the product rule: 𝑑𝑓 𝑥 = 2𝑥𝑒 2𝑥+3 + 2𝑒 2𝑥+3 𝑥 2 𝑑𝑥 Simplify and factorise: 𝑑𝑓 𝑥 = 2𝑥𝑒 2𝑥+3 1 + 𝑥 𝑑𝑥 Apply the FOC: 2𝑥𝑒 2𝑥+3 𝑥 𝑦 = 𝑥 2 − 13𝑥 + 15 𝑦 = 3𝑥 3 + 2𝑥 2 − 5𝑥 + 15 7𝑥 3 𝑦 = − + 12𝑥 − 1 8 𝑦 = 3𝑥 2 𝑒 4𝑥−1 𝑥 2 + 1 𝑦 = 𝑥 − 3 𝑦 = 𝑒 3𝑥+1 + 2𝑒 −2𝑥+1 − 6 𝑦 = −𝑥 2 + 3𝑥 + 5 𝑒 𝑥+5 𝑦 = 𝑒 𝑥 𝑥 2 − 3 2 6. giving 𝑥 = −1. There are three options: 1. Find any stationary point(s) for the functions: The process is quite simple and is easily demonstrated in an example. Exercises: 1.e. if an optimal point is a maximum or minimum) using the second derivative. one of the products on the left side must equal zero. 𝑑2 𝑦 Minimum if >0 𝑑𝑥 2 A maximum if the second derivative evaluated at the optimal point is less than zero. The point in question is: A minimum if the second derivative evaluated at the optimal point is greater than zero. 𝑑2 𝑦 Stationary inflection point if =0 𝑑𝑥 2 +7 +7 𝑓 −1 = 𝑒 1 + 7 = 𝑒 + 7 Thus 0. 1 + 𝑥 = 0. For the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠: 𝑓 0 = 0 2 𝑒 2 𝑓 0 = 7 𝑓 −1 = −1 2 𝑒 2 −1 +3 0 +3 A stationary inflection point if the second derivative evaluated at the optimal point is equal to zero. so 𝑥 = 0. 2. Theory: the Second Order Condition (SOC) is used to find the nature of an optimal point.3 the nature of an optimal point After finding an optimal point. There are two optimal points with 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 𝑥 = 0 and 𝑥 = −1. it is important to determine if that point is a minimum or maximum. profit might in fact be minimised.0 and −1. 56 is a maximum. Then apply the SOC to determine if it is a maximum or minimum.56) and (6.Example 1: find the optimal point and the nature of this optimal point for the function 𝑦 = −𝑥 2 + 2𝑥 + 15 Plan: use the FOC to find any stationary points 𝑑𝑦 =0 𝑑𝑥 Find the second derivative and evaluate at the optimal point. Example 3: find and define any optimal points for 𝑦 = 𝑒 𝑥 2 −3 3 − 12 2 − 12 6 2 + 36 2 + 24 + 36 6 + 24 3 2 + 2 1 + 15 + ln 4 − 𝑥 2 Plan: find any optimal points using the FOC 𝑑𝑦 =0 𝑑𝑥 121 . 2. To summarise the process: 1. and evaluate it at any optimal points. Solution: find the optimal points 𝑑𝑦 = 3𝑥 2 − 24𝑥 + 36 = 0 𝑑𝑥 Solve using the Quadratic Formula: 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = −𝑏 ± 𝑏 2 − 4𝑎𝑐 2𝑎 2 2 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = −(−24) ± (−24)2 − 4 3 (36) 2(3) 24 ± 576 − 432 6 24 ± 12 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = 6 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = 2 𝑂𝑅 𝑥 − 𝑣𝑎𝑙𝑢𝑒 = 6 The two optimal points have 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 of 𝑥 = 2 and 𝑥 = 6.16 . a maximum is expected. From Chapter 4. Applying the SOC.24 is a minimum. Example 2: find any optimal points and their nature for the function 𝑦 = 𝑥 3 − 12𝑥 2 + 36𝑥 + 24 Plan: use the FOC to find any optimal points. and 6.24). Then find the second derivative. Solution: to find any stationary points: 𝑑𝑦 = −2𝑥 + 2 = 0 𝑑𝑥 −2𝑥 = −2 𝑥 = 1 The 𝑦 − 𝑣𝑎𝑙𝑢𝑒 is obtained from the original function: 𝑦 1 = − 1 𝑦 1 = 16 The second derivative is: 𝑑2 𝑦 = −2 𝑑𝑥 2 Apply the SOC: 𝑑 𝑦 = −2 < 0 ∴ maximum 𝑑𝑥 2 Thus there is a maximum at 1. The corresponding 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 are: 𝑦 2 = 2 𝑦 2 = 56 𝑦 6 = 6 𝑦 6 = 24 Thus the two points are (2. apply the SOC to the results. 3. Find the second derivative: 𝑑2 𝑦 = 6𝑥 − 24 𝑑𝑥 2 Evaluate the second derivative at 𝑥 = 2: 𝑑2 𝑦 2 = 6 2 − 24 = −12 < 0 𝑑𝑥 2 ∴ maximum at 𝑥 = 2 Evaluate the second derivative at 𝑥 = 6: 𝑑2 𝑦 6 = 6 6 − 24 = 12 > 0 𝑑𝑥 2 ∴ minimum at 𝑥 = 6 Thus 2. Find the optimal 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 from the FOC and their corresponding 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠. since the value of 𝑎 in the quadratic is negative. Determined if the point(s) are maxima or minima by substituting those 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 from the FOC into the second derivative. 9 can be obtained from a calculator. 𝑒 −3 + ln 4 is a maxima. The two minimums are flanking a maximum (the shape of a W). If the second derivative at the point in question is greater than + ln 4 − 0 2 zero. For 𝑥 = 3: + 3 −3 2 =1 The only time when an exponential is equal to 1 is when the exponent is equal to zero. For 𝑥 = − 3: 3)2 −3 − 2𝑥 = 0 − 1) = 0 2 −3 2𝑥(𝑒 𝑥 2 −3 So either 2𝑥 = 0 or 𝑒 𝑥 If 2𝑥 = 0. 2. it’s not what is intuitively expected). Then find and evaluate the second derivative at the optimal points. it is a maximum (again. and apply the SOC. For 𝑥 = 0: 𝑑2 𝑦 (0) = 2𝑒 0−3 + 4(0)2 𝑒 0−3 − 2 𝑑𝑥 2 = 2𝑒 −3 − 2 ≈ −1. Use the second derivative to find the nature of the three points: 𝑑2 𝑦 2 2 = 2𝑒 𝑥 −3 + 2𝑥 𝑒 𝑥 −3 2𝑥 − 2 𝑑𝑥 2 = 2𝑒 𝑥 2 −3 3. 0. and 0. not what is intuitively = ln 4 − 2 𝑦 3 = 𝑒 3 −3 2 + ln 4 − 3 expected). ln 4 − 2 and 3. then determine the nature of these stationary points.9 < 0 ∴ maximum The −1. If 𝑒 𝑥 2 −3 −1=0 𝑑 2 𝑦 (− 3) = 2𝑒 (− 𝑑𝑥 2 + 4(− 3)2 𝑒 (− 3)2 −3 −2 − 1 = 0 then 𝑒 𝑥 2 −3 = 2𝑒 0 + 12𝑒 0 − 2 = 12 > 0 ∴ minimum 3. Exercises: 1. then 𝑥 = 0.The 𝑒 rule and power rule need to be used. if the second = 𝑒 −3 + ln 4 𝑦 − 3 = 𝑒 − 3 −3 + ln 4 − − 3 2 2 derivative at the point in question is less than zero. it is a minimum (see. Note: a good way of remembering the SOC is that it is not what is intuitively expected. ln 4 − 2 and + 4𝑥 2 𝑒 𝑥 2 −3 −2 The product rule was used to evaluate the second derivative however simplification is not needed as 122 . So 𝑥 2 − 3 = 0 is required: 𝑥 2 = 3 𝑥 2 = 3 𝑥 = − 3 and 𝑥 = + 3 There are three solutions for optimal points: 𝑥 = 0. Find any stationary points for the functions below. Solution: find the optimal points using the FOC: 𝑑𝑦 2 = 2𝑥𝑒 𝑥 −3 − 2𝑥 = 0 𝑑𝑥 2𝑥𝑒 𝑥 2 −3 all that is required is to evaluate the second derivative at each of the three optimal points: 1. 𝑥 = − 3 and 𝑥 = + 3 For the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠: 𝑦 0 = 𝑒 0 2 −3 2 𝑑 2 𝑦 + 3 = 2𝑒 𝑑𝑥 2 +4 + 3 2 𝑒 + 3 −3 2 −2 = 2𝑒 0 + 12𝑒 0 − 2 = 12 > 0 ∴ minimum Thus − 3. ln 4 − 2 are minima. 𝑎) 𝑦 = 3𝑥 2 − 11𝑥 + 10 𝑏) 𝑦 = 𝑥 3 + 4𝑥 2 − 5𝑥 + 1 2𝑥 3 𝑐) 𝑦 = − + 2𝑥 − 19 9 𝑑) 𝑦 = 5𝑥 2 𝑒 −7𝑥+4 4𝑥 2 + 3 𝑒) 𝑦 = 5 − 𝑥 5 𝑓) 𝑦 = 𝑒 −3𝑥−6 + − 𝑒 7𝑥+1 − 5 6 𝑔) −3𝑥 2 + 5𝑥 + 12 𝑒 𝑥+4 ) 𝑦 = 3𝑒 𝑥 2𝑥 2 − 8 2 = ln 4 − 2 Giving points − 3. ln 4 − 2 . 𝑒 −3 + ln 4 . Similarly. but are not maxima nor minima. A generic way of finding any inflection point (not just a stationary inflection point) is a continuation of the SOC: 𝑑2 𝑦 =0 𝑑𝑥 2 inflection point SOC for any inflection point: That is. Concave means that the curve looks like part of a cave when drawn on a set of axes. 𝑦 stationary inflection point non-stationary inflection point 𝑥 The blue function has a non-stationary inflection point. Solution: differentiate the function twice: 123 . The point at which this change occurs is the inflection point. and convex is when a “cave” cannot be drawn (for a technical definition. so they are convex. 2. The mathematical way of finding inflection points is a continuation of the SOC.4 inflection points Theory: Inflection points are points of a function where a function changes from concave to convex (or vice versa). 𝑥 Notice that the function is initially concave (red) and then changes to convex (blue). the red lines are concave. then solve for 𝑥. that point is a stationary inflection point. Stationary inflection points have a gradient of zero at the inflection point. This finds all inflection points. as they could be drawn into a smooth “cave” shape. consult a second year mathematics for economists text book). Below. set the second derivative equal to zero. Example 1: find any inflection point(s) (if they exist) for the function 𝑦 = 𝑥3 − 3𝑥2 − 28𝑥 + 60 Plan: find the second derivative. whereas the blue lines cannot. Non-stationary inflection points do not have a gradient of zero at the inflection point. whereas the red function has a stationary inflection point. 𝑥 Theory: an inflection point is where a function changes from convex to concave (or concave to convex). 𝑦 Theory: if the SOC is evaluated for a stationary point and is found to be zero. But a much easier way of understanding this is to draw an inflection point. 𝑦 Theory: There are two main types of inflection points: 1.6. set it to zero then solve for 𝑥. Finally. Theory: if the FOC and SOC (inflection point) have the same 𝑥 − 𝑣𝑎𝑙𝑢𝑒. as then 3 0 2 2 3 +2=2 There is a stationary inflection point at (1. Solution: find any stationary points using the FOC: 𝑑𝑓 𝑥 = 3 𝑥 − 1 𝑑𝑥 zero. 𝑥 = 1. To find the corresponding 𝑦 − 𝑣𝑎𝑙𝑢𝑒 use the original function. So if 𝑥 − 1 = 0.25𝑥 4 − 𝑥 3 − 10𝑥 2 + 12𝑥 − 71 𝑥 2 + 1 𝑓) 𝑦 = 𝑥 − 3 𝑔) 𝑦 = −2𝑥 2 + 4𝑥 + 10 𝑒 𝑥+3 ) 𝑦 = ln 𝑥 2 − 9 +2 Plan: find any stationary points using the FOC 6. 𝐹𝑂𝐶: 𝑑𝑦 =0 𝑑𝑥 Evaluating the second derivative at any stationary points.2).5 combining all theory Theory summary: The FOC finds the stationary point(s) of a function. There is only one stationary point at 𝑥 = 1. then it must be a stationary inflection point. Exercises: 1. 𝑦(1) = 1 3 2 To find any stationary points: 𝑑2 𝑓 𝑥 = 6 𝑥 − 1 = 0 𝑑𝑥 2 Again.𝑑𝑦 = 3𝑥 2 − 6𝑥 − 28 𝑑𝑥 𝑑 𝑦 = 6𝑥 − 6 𝑑𝑥 2 Set the second derivative equal to zero. 𝑎) 𝑦 = 𝑥 3 + 5𝑥 2 − 2𝑥 + 1 𝑏) 𝑦 = −3𝑥 3 + 7𝑥 − 13 −13𝑥 3 𝑐) 𝑦 = + 12𝑥 2 − 54 + 𝑥 4 𝑑) 𝑦 = 2𝑥 4 + 13𝑥 3 − 12𝑥 𝑒) 𝑦 = 0. (𝑥 − 1) must equal = 0. 124 . then applying the SOC finds the nature: 𝑑2 𝑦 > 0 is a minimum 𝑑𝑥 2 𝑑2 𝑦 < 0 is a maximum 𝑑𝑥 2 𝑑2 𝑦 = 0 is a stationary inflection point 𝑑𝑥 2 To find any inflection points. for the left side to be equal to zero (𝑥 − 1) must equal zero. then it must be a stationary inflection point. set the second derivative to zero and solve for 𝑥: =0 For the left side to equal zero. 6𝑥 − 6 = 0 𝑥 = 1 There is an inflection point at 𝑥 = 1. Since 𝑥 = 1 is a stationary point (from the FOC) and at that same point it is also an inflection point (from the SOC).30). Determine any inflection points for the functions below (Note: some functions do not have inflection points). Expanding the brackets and applying the Quadratic Formula will give the same solution. This means 𝑥 = 1. and that it is a stationary inflection point 𝑓 𝑥 = 𝑥 − 1 𝑑𝑓 𝑥 =0 𝑑𝑥 Then find any inflection points by setting the second derivative equal to zero 𝑑2 𝑓 𝑥 =0 𝑑𝑥 2 If a single 𝑥 − 𝑣𝑎𝑙𝑢𝑒 is both an inflection and stationary point. substitute 𝑥 = 1 in to the original function: 𝑓 1 = 1 − 1 3 −3 1 2 − 28(1) + 60 = 1 − 3 − 28 + 60 = 30 There is an inflection point at (1. Example 2: prove the following function has one inflection point. then that inflection point is a stationary inflection point. to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 at 𝑥 = 1. 6 =3 6 𝑑𝑥 2 ∴ minimum 2 2 2 inflection point at − 12.25𝑥4 − 18𝑥2 + 31 Plan: use the FOC to find any stationary points: 𝑑𝑦 =0 𝑑𝑥 Then evaluate the second derivative at these points. The 125 . −293) − 36 = 72 > 0 Then simply extend the lines to form a curve. Find the nature of these points: 𝑑2 𝑦 = 3𝑥 2 − 36 𝑑𝑥 2 1.𝑑2 𝑦 =0 𝑑𝑥 2 To find 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠. substitute the respective 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 into the original function. the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 can now be found using the original function (see if you get the same 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠): minimum at (−6. one at 𝑥 = − 12 and another at 𝑥 = + 12. set the second derivative equal to zero to find any inflection points. −149) minimum at (6. Solution: use the FOC to find any stationary points: 𝑑𝑓 𝑥 = 𝑥 3 − 36𝑥 = 0 𝑑𝑥 𝑥 3 − 36𝑥 = 0 𝑥(𝑥 2 − 36) = 0 So either 𝑥 = 0 𝑂𝑅 𝑥 2 − 36 = 0 Separate the second part into two answers: 𝑥 2 − 36 = 0 𝑥 2 = 36 𝑥 = ±6 The stationary points have 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 𝑥 = 0. Finally. but as long as the general shape is correct and the important − 36 = 72 > 0 points labelled. 𝑥 = −6 and 𝑥 = 6. A maximum at 𝑥 = 0 is flanked by minima at 𝑥 = −6 and 𝑥 = 6. and apply the SOC. Obviously. 𝑑2 𝑦 −6 = 3 −6 𝑑𝑥 2 ∴ minimum 𝑑2 𝑦 3. and at each minimum draw a small happy face. −293) (6.31 inflection point at ( 12. Having figured out the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 of all the important points. that is all that matters. this sketch is not to scale. −149) − 36 = −36 < 0 (−6. −149) ( 12. and then at each maximum draw a small sad face: 𝑦 (0. plot these 5 points. −293) Sketching this function is quite simple. −293) Example 1: find the nature of all stationary and inflection points for the function 𝑓 𝑥 = 0. To find the inflection points: 𝑑2 𝑦 = 3𝑥 2 − 36 = 0 𝑑𝑥 2 3𝑥 2 = 36 𝑥 2 = 12 𝑥 = ± 12 There are two inflection points. 𝑑2 𝑦 0 =3 0 𝑑𝑥 2 ∴ maximum 2.31) 𝑥 (− 12. −149 maximum at 0. so you open a small shop on the sand and you sell surfboards. 𝑎) 𝑦 = 3𝑥 4 + 5𝑥 3 − 2𝑥 2 − 14 𝑏) 𝑦 = 4𝑥 − 3 4 𝑐) 𝑦 = ln 5𝑥 + 3 + 𝑥 − 2 𝑥 − 3 𝑑) 𝑦 = ln 8𝑥 + 7 + 𝑥 − 1. The inflection point is not a stationary inflection point as there is no corresponding 𝑥 − 𝑣𝑎𝑙𝑢𝑒 in the FOC. set the second derivative equal to zero: 𝑔′′ 𝑥 = 0 −3𝑒 −𝑥−4 2 − 𝑥 = 0 For this to be true. But you need to make money.0202 𝑎𝑠 − 0. as this exponential never crosses the 𝑥 − 𝑎𝑥𝑖𝑠. then determine if there are any inflection points. 3𝑒 −𝑥−4 = 0. which is not possible. 𝑒 6 ). either −3𝑒 −𝑥−4 = 0 which we know from our work with exponentials is never 6. Solution: the FOC using the product and 𝑒 rules: 𝑑𝑒𝑟 1𝑠𝑡 = 3 𝑑𝑒𝑟 2 𝑛𝑑 the case. Example 2: determine the nature of any stationary points.5 0. find the second derivative (using the product rule on 𝑔′ 𝑥 above): 𝑔′′ 𝑥 = 3 −1 𝑒 −𝑥−4 1 − 𝑥 + −1 3𝑒 −𝑥−4 Simplify this to the extreme (easier to work with): 𝑔′′ 𝑥 = −3𝑒 −𝑥−4 1 − 𝑥 − 3𝑒 −𝑥−4 𝑔′′ 𝑥 = −3𝑒 −𝑥−4 1 − 𝑥 + 1 𝑔′′ 𝑥 = −3𝑒 −𝑥−4 2 − 𝑥 Evaluate the second derivative at the 𝑥 = 1: 𝑔′′ 1 = −3𝑒 −1−4 2 − 1 ≈ −0. that is why the 𝑒’s become denominators. then set the second derivative equal to zero to find any inflection points. You obviously want to minimise your time in the shop and also maximise your profits.5𝑥 − 2. Thus there is an inflection point at 𝑥 = 2. Determine the nature of any stationary points for the following functions. but they are beyond the scope of this book. 2.75 𝑒) 𝑦 = −4𝑥 2 + 4𝑥 + 12 𝑒 5+𝑥 2.0202 < 0 ∴ maximum To find any inflection points. Finally. 𝑎) 𝑦 = 𝑥 𝑥 − 3 𝑥 + 7 𝑏) 𝑦 = 𝑥 − 5 𝑥 − 3 2 𝑥 + 5 𝑐) 𝑦 = 5𝑥 4 − 3𝑥 3 + 12𝑥 2 − 8𝑥 + 18 6 3 𝑒 = −1 𝑒 −𝑥−4 Substituting into the product rule: 𝑔′ 𝑥 = 3𝑒 −𝑥−4 + 3𝑥 −1 𝑒 −𝑥−4 𝑔′ 𝑥 = 3𝑒 −𝑥−4 1 − 𝑥 = 0 There are two possibilities: 1. Sketch the following functions after the nature of all stationary points and inflection points have been found. There is only one stationary point.𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 could also be found.6 applications – profit Intro example: you have just finished your business degree and all you want to do is lie on the beach all day and not work. You’ve done a business degree so you sit down one day and determine the approximate demand function for surfboards to be 𝑃𝑑 = 350 − 5𝑄𝑑 126 . as well as any inflection points for 𝑔 𝑥 = 3𝑥𝑒 −𝑥−4 Plan: use the FOC and SOC to find the nature and location of any stationary point(s). 1 − 𝑥 = 0 therefore 𝑥 = 1. There is a maximum at (1. Exercises: 1. or 2 − 𝑥 = 0 which is true when 𝑥 = 2. To determine the nature of this point. 5 ) and an inflection point at (2. the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 of the two points are obtained from the original function: 𝑔 1 = 3 1 𝑒 −1−4 = 𝑔 2 = 3 2 𝑒 −2−4 3 𝑒 5 6 = 6 𝑒 Remember index rules. the $400/month to rent the area on the sand is a fixed cost. Theory: the general form of profit is: 𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 − 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡𝑠 𝜋 = 𝑇𝑅 − 𝑇𝐶 Total revenue can also be simplified: 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑃𝑟𝑖𝑐𝑒 × 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑠𝑜𝑙𝑑 𝑇𝑅 = 𝑃 ∙ 𝑄 If you sold 10 surfboards at $200 each. and it costs you $400/month to rent the sand from the local government. you as the owner would have bought that from a supplier. Theory: in general. and the 50𝑄 is the cost of buying 𝑄 surfboards from the supplier (i. 𝜋 = 𝑇𝑅 − 𝑇𝐶 𝜋 = 350𝑄 − 5𝑄 2 − 400 + 50𝑄 𝜋 = 350𝑄 − 5𝑄 2 − 400 − 50𝑄 𝜋 = −5𝑄2 + 300𝑄 − 400 Note: the 𝑇𝐶 function was written in brackets. This is where many students screw up. costs need to be defined. The quantity of surfboards to be sold to maximise profit is found by determining the maximum of this profit function using the FOC and SOC: 𝑑𝜋 = −10𝑄 + 300 = 0 𝑑𝑄 −10𝑄 = −300 𝑄 = 30 To make sure this is a maximum: 𝑑2 𝜋 = −10 𝑑𝑄 2 127 . To solve the above example: 𝑇𝑅 = 𝑃 ∙ 𝑄 Insert the demand function into the 𝑇𝑅 function to substitute away the 𝑃 and have 𝑇𝑅 in terms of 𝑄: 𝑇𝑅 = 350 − 5𝑄 𝑄 Simplify: 𝑇𝑅 = 350𝑄 − 5𝑄 2 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡𝑠 are: 𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶 𝑇𝐶 = 400 + 50𝑄 The 400 is the rent. Above.e. 𝑄). the profit function can be determined. that $400 still needs to be paid to the local government. $50 for each surfboard multiplied by the quantity of surfboards sold. What price should you set to maximise your profits? This is a very common application of optimisation. Theory summary: 𝜋 = 𝑇𝑅 − 𝑇𝐶 𝑇𝑅 = 𝑃 ∙ 𝑄 𝑑𝑒𝑚𝑎𝑛𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑡𝑒 𝑓𝑜𝑟𝑚 𝑃 𝑄 𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶 All functions are in terms of 𝑄 as profits are directly determined by the quantity of goods sold. But there is another layer: the quantity of goods sold depends on the demand for the goods. To find profits. total costs are represented by: 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡𝑠 = 𝐹𝑖𝑥𝑒𝑑 𝐶𝑜𝑠𝑡𝑠 + 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝐶𝑜𝑠𝑡𝑠 𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶 Variable costs are usually represented by: 𝑉𝐶 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑄 Where the constant is how much a single item costs from the supplier.You purchase the surfboards from a supplier for $50 each. total revenue would be 10 × $200 = $2000. Thus 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 depends on the demand function. The variable costs are those costs which are affected when a good is sold. Now that both 𝑇𝑅 and 𝑇𝐶 have been found. because it does not matter how many surfboards are sold. When a surfboard is sold at the beach. so don’t be one of them. to emphasise the negative sign goes into all terms inside that bracket. 𝑄 𝑇𝑅 = (−5𝑄 + 400). and the profit he will make at this price. Find the number of cakes he needs to sell to maximise profit.𝑑2 𝜋 30 = −10 < 0 ∴ maximum 𝑑𝑄 2 Thus 30 surfboards need to be sold to maximise profit. and one cakes costs him approximately $100 to make. the SOC must be used: 𝑑2 𝜋 = −10 𝑑𝑄 2 𝑑2 𝜋 30 = −10 < 0 ∴ maximum 𝑑𝑄 2 What price should Mr Watkins charge for each of these cakes? Using the (rearranged) demand function: 𝑃 = −5𝑄 + 400 𝑃(30) = −5(30) + 400 + 300(30) − 400 + 300(30) − 400 128 . The demand function gives the price required to sell 30 surfboards per month. how much profit will be made per month if 30 boards are sold? Using the profit function. Lastly. you can never control how many surfboards you sell. and he specialises in wedding cakes. it determines profit: 𝜋 = −5𝑄 + 300𝑄 − 400 𝜋(30) = −5 30 𝜋(30) = −5 30 𝜋 30 = 4100 Example 1: Mr Watkins opens a cake store called Sugar Overload. the price per cake. then apply the SOC to determine if it is a maximum. Substitute the optimal quantity back into the profit function to determine the maximum profit. as a supplier. To make sure this quantity will give maximum profit. 𝑃 = 350 − 5𝑄 𝑃 30 = 350 − 5 30 𝑃 30 = 200 The surfboards should be priced at $200 each. He figures out his monthly demand is 𝑃 𝑄 = 80 − 5 The rent for the bakery he works in is $500/month. However. which maximises profit. 𝑄 𝑇𝑅 = −5𝑄 2 + 400𝑄 Find the 𝑇𝐶 function: 𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶 𝑇𝐶 = 500 + 100𝑄 Set up the profit function: 𝜋 = 𝑇𝑅 − 𝑇𝐶 𝜋 = −5𝑄2 + 400𝑄 − 500 + 100𝑄 𝜋 = −5𝑄2 + 400𝑄 − 500 − 100𝑄 𝜋 = −5𝑄2 + 300𝑄 − 500 Apply the FOC to find any stationary points: 𝑑𝜋 = −10𝑄 + 300 = 0 𝑑𝑄 −10𝑄 = −300 𝑄 = 30 Mr Watkins needs to make 30 cakes per month. because as its name implies. but you can control the price. then 2 2 2 into the rearranged demand function to get the price. Solution: the demand function needs to be rearranged to isolate 𝑃: 𝑄 = 80 − 𝑃 5 𝑃 5 𝑄 − 80 = − (−5)(𝑄 − 80) = 𝑃 𝑃 = −5𝑄 + 400 Substitute this into the 𝑇𝑅 equation: 𝑇𝑅 = 𝑃. Plan: use the profit equation 𝜋 = 𝑇𝑅 − 𝑇𝐶 Replace 𝑇𝑅 with 𝑃 ∙ 𝑄 where 𝑃 is the rearranged demand function (𝑃 in terms of 𝑄). Apply the FOC to find any optimal points. This implies that at break-even. 𝜋𝐵𝐸 = 0 Example 2: A firm produces wide-screen TV’s and their estimated demand is 𝑄𝑑 = 440 − 4𝑃𝑑 2 If they faces the increasing cost function 𝑇𝐶 = 0.𝑃 30 = −150 + 400 = 250 Looking back to what the question is asking. the profit he will make is: 𝜋 = −5𝑄2 + 300𝑄 − 500 𝜋(30) = −5 30 2 This is very similar to the previous section.000 If the costs labour and capital which average out to $15. and when on-selling the motorbikes.05 𝑄 + 200 𝑄 − 200 + 20. the profit function is simply set equal to zero. they face a domestic demand function of: 𝑃𝑑 = − 0. 129 . which implies profit is zero.500/𝑤𝑒𝑒𝑘. determine: a) The yearly revenue function b) The annual cost function c) The annual profit function d) The maximum yearly profit attainable 2.5𝑄 2 + 1800 find the break-even output levels for this firm. A manufacturer of surfboards faces a monthly demand function of form: 𝑃𝑑 = −0. Example 1: find the break-even quantities for a firm with a profit function 𝜋 = −𝑄 2 + 200𝑄 − 7500 Plan: set the profit function equal to zero and solve. it means the revenues just meet the costs.0) 6. Exercises (remember to prove they are optimal points): 1.0) (150.2𝑄𝑑 2 + 110 If costs of making a board are $60 and the factory has fixed costs of $745/𝑚𝑜𝑛𝑡.0) (100.000. determine: a) The revenue function b) The cost function c) The profit function d) The maximum profit attainable 3.1𝑄 + 15 2 0.1𝑄 + 13 (𝑄 − 22) If there are no other costs.7 applications – break-even When a business “breaks even”.500/𝑡𝑟𝑎𝑦 and fixed costs of $7. except that instead of maximising profit. Theory: break even is when 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 equals 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒. Solution: set 𝜋 = 0: 0 = −𝑄 2 + 200𝑄 − 7500 Use the Quadratic Formula: 𝑄𝐵𝐸 = 𝑄𝐵𝐸 = 𝑄𝐵𝐸 = 𝑄𝐵𝐸 = −𝑏 ± 𝑏 2 − 4𝑎𝑐 2𝑎 −200 ± (200)2 − 4 −1 (−7500) 2(−1) −200 ± (200)2 − 4 −1 (−7500) 2(−1) −200 ± 10000 −200 ± 100 = −2 −2 𝑄𝐵𝐸 = 50 + 300(30) − 500 𝜋 30 = −5 900 + 9000 − 500 = 4000 Thus Mr Watkins should sell 30 cakes per month at $250 each and he will make a maximum profit of $4000 per month. A company imports motorbikes from China and sells them onto consumers at an inflated price. if Mr Watkins sells 30 cakes at $250 each. A company manufacturing mining truck trays has a yearly demand of: 𝑃 = −0. If they purchase each motorbike for $5. determine: a) The profit function b) The profit maximising quantity. profit is zero. 𝑄𝐵𝐸 = 150 or Thus the two break-even quantities are 150 and 50. these are simply the roots of the profit function: 𝜋 𝜋𝑚𝑎𝑥 𝐵𝐸1 𝐵𝐸2 𝑄 (50. Mathematically. If they purchase the golf-clubs for $5 each.g.Plan: find 𝑇𝑅 in terms of 𝑄. Solution: rearrange the function to isolate 𝑃: 𝑄 = 440 − 4𝑃 2 2. financial planning) has a monthly demand for the services (on a per minute basis) according to the function: 𝑃 = ln 7𝑄 + 130 If the costs are commissions ($100/𝑢𝑛𝑖𝑡) and fixed costs of $500/𝑚𝑜𝑛𝑡. the number of units sold). When the firm hires the first worker. they face a domestic demand function of: 𝑃 = 𝑄 + 800 𝑄 − 700 If there are no other costs.5𝑄 + 110 Replace the 𝑇𝑅 equation: 𝑇𝑅 = 𝑃. Set this equal to zero and solve for break-even output levels.5𝑄 2 + 1800 𝜋 = −0. A company selling financial services (e. The “marginal” value of a variable is the addition to the total from an extra (marginal) unit.5𝑄 + 110𝑄 − 0. marginal productivity of labour is the additional output obtained from an extra (marginal) person being employed (which is found using the first derivative). that worker produces three cabinets per day. 6.g. determine: a) The profit function b) The break-even quantities. cost) is the sum of all the individual variables (e. then replace the profit equation 𝜋 = 𝑇𝑅 − 𝑇𝐶 to find the profit function. the average output of labour is the total output divided by the number of workers.5𝑄2 + 110𝑄 Having 𝑇𝑅 and 𝑇𝐶. the 𝜋 function can be formed: 𝜋 = 𝑇𝑅 − 𝑇𝐶 𝜋 = −0. determine: a) The revenue function b) The cost function c) The profit function d) The break-even profit e) The maximum profit attainable 130 . c) The profit maximising quantity. The firm then hires another worker. 𝑄 𝑇𝑅 = −0.g. and −110 ± 4900 −110 ± 70 = −2 −2 Thus the firm breaks even at 𝑄𝐵𝐸 = 90 or 𝑄𝐵𝐸 = 20. sum all the costs) divided by the number of units (e.5𝑄 2 + 110𝑄 − 0. it means the derivative. and when on-selling the clubs. For example. Exercises (remember to prove they are optimal points): 1. Intro example: a firm manufactures wooden cabinets in a small factory. 2𝑄 = 440 − 4𝑃 2𝑄 − 440 = −4𝑃 𝑃 = −0. determine: a) The revenue function b) The cost function c) The profit function d) The break-even profit e) The maximum profit attainable A company imports cheap golf-clubs from China and sells them onto consumers at an inflated price. set 𝜋 = 0: 𝜋𝐵𝐸 = 0 −𝑄 2 + 110𝑄 − 1800 = 0 Use the quadratic formula to solve this quadratic: 𝑄𝐵𝐸 = 𝑄𝐵𝐸 = 𝑄𝐵𝐸 = −𝑏 ± 𝑏 2 2𝑎 −110 ± (110)2 − 4 −1 (−1800) 2(−1) − 4𝑎𝑐 2 2 2 3.8 applications – marginal and average values Theory: the average of a group of variables (e.5𝑄 − 1800 𝜋 = −𝑄 + 110𝑄 − 1800 To find the break even quantities. Similarly. In mathematical terms: 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 = 𝑠𝑢𝑚 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑑𝑦 𝑑𝑥 𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 = It is easier to remember that whenever you read the word “marginal”.g.5𝑄 + 110 𝑄 𝑇𝑅 = −0.5𝑄+6 If costs of making a single mobile phone are $10 and the factory has fixed costs of $1412/𝑚𝑜𝑛𝑡. A manufacturer of mobile phones faces a monthly demand function of form: 𝑃 = 5𝑒 −0. When output is increasing at a decreasing rate (i. The point at which it changes from convex to concave is the inflection point. convex).e. and differentiate 𝑇𝐶 with respect to 𝑄 to get marginal cost. no additional output can be obtained from additional labour (and in theory. When output begins to fall.). Example 1: find the average cost and marginal cost functions if the total cost function is 1 5 𝑇𝐶 = 𝑄 3 + 𝑄 2 + 50 3 2 Plan: divide the 𝑇𝐶 function by 𝑄 to get average cost. and now the three employees produce 12 cabinets. 3. Eventually. as some of them are always waiting to use the equipment.5 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = 14 − 12 = 2 𝑂𝑢𝑡𝑝𝑢𝑡 3 1 2 3 4 𝐿𝑎𝑏𝑜𝑢𝑟 1 3 5 2 𝑄 + 2 𝑄 + 50 𝐴𝐶 = 3 𝑄 1 5 50 𝐴𝐶 = 𝑄 2 + 𝑄 + 3 2 𝑄 For marginal costs: 𝑀𝐶 = 𝑑𝑇𝐶 𝑑𝑄 Economic theory: as more labour (or other input) is added to a production process. When one worker wants to use the saw. 𝐼𝑛𝑝𝑢𝑡 There are three distinct regions: 1. additional labour adds less and less to total output. produce eight cabinets (as they are specialising: one cuts the timber.e. another worker is using it. together. The firm hires another person. In summary: Worker 1: o o 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = = 3 1 3 𝑂𝑢𝑡𝑝𝑢𝑡 1. The theory of marginal and average values can be 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = 8 − 3 = 5 Worker 3: o o 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = 12 3 =4 applied to mathematical functions. However. After hiring yet another worker. as the Law of Diminishing Marginal Returns. Hiring any more workers will not at all increase production.these two workers. 2. This theory is 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = 3 Worker 2: o o 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = = 4 2 8 known. output initially increases greatly. 𝑀𝐶 = 𝑄 2 + 5𝑄 131 . Solution: for average costs: 𝐴𝐶 = 𝑇𝐶 𝑄 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = 12 − 8 = 4 Worker 4: o o 14 12 8 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = 14 4 = 3. 2. When output is increasing at an increasing rate (i. output may start to fall). in general terms. the other sands it etc. past some point. concave) 3. all four people make 14 cabinets. 64 132 . Solution: 𝑀𝐶 = 𝐴𝐶 = 𝐴𝐶 = 𝑑𝑇𝐶 = 0.02𝑄 2 − 0.02𝑄 3 − 0. for the cost function: 𝑇𝐶 = 0.64 𝑑𝑄 𝑇𝐶 𝑄 0.72𝑄 + 8. We know from Section 6. Example 4: find the point where average costs are equal to marginal costs. find 1. an expression for average cost. 2.44𝑄 + 8. Then find the minimum of the marginal cost function using FOC. the FOC must be applied to the marginal cost function: 𝑑𝑀𝐶 6 6 = 𝑄 − =0 𝑑𝑄 100 10 Solve this for 𝑄: 6 6 𝑄 = 100 10 𝑄 = 10 To prove this is a minimum: 𝑑2 𝑀𝐶 6 = > 0 ∴ mimimum 2 𝑑𝑄 100 The 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 at this point is: 10 3 −3 𝑇𝐶 10 = + 10 100 10 An expression of average costs is: 𝑄 3 −3 2 𝑇𝐶 100 + 10 𝑄 + 3𝑄 + 7 𝐴𝐶 = = 𝑄 𝑄 𝐴𝐶 = 𝑄 3 7 − 𝑄 + 3 + 100 10 𝑄 2 2 Finding the inflection point of an original function is the same as finding the maximum/minimum for the first derivative (Chapter 5). So using the FOC on the 𝑀𝑃𝐿 function: 𝑑𝑀𝑃𝐿 𝑑𝑄 2 = 2 = −6𝐿 + 20 = 0 𝑑𝐿 𝑑 𝐿 Solving this gives: 6𝐿 = 20 𝐿 = 20 ≈ 3.72𝑄 2 + 8. for the production function: 𝐴𝐶 = 0. then set them equal to each other. The inflection point needs to be found (as it is at this point where the function changes from increasing at an increasing rate to increasing at a decreasing rate). 𝑇𝐶 = Plan: 1.06𝑄 2 − 1. 2.02 𝑄 3 − 36𝑄 2 + 432𝑄 Plan: find 𝑀𝐶 and 𝐴𝐶.33 6 Thus between the third and forth unit of labour.Example 2: for the given function.4 that this is done by setting the second derivative equal to zero. the productivity of labour changes from increasing at an increasing rate to increasing at a decreasing rate. make sure it is a minimum using the SOC. Solution: differentiate the function 𝑀𝑃𝐿 = 𝑑𝑄 = −3𝐿2 + 4 + 20𝐿 𝑑𝐿 𝑄 3 −3 2 + 𝑄 + 3𝑄 + 7 100 10 Solution: differentiating 𝑇𝐶: 𝑑𝑇𝐶 3 2 6 = 𝑀𝐶 = 𝑄 − 𝑄 + 3 𝑑𝑄 100 10 To find where marginal cost is at a minimum.64𝑄 𝑄 + 3 10 + 7 𝑇𝐶 10 = 10 − 30 + 30 + 7 = 17 Example 3: find the level of output when productivity of labour changes from increasing at an increasing rate to increasing at a decreasing rate. differentiate 𝑇𝐶 to get 𝑀𝐶. the point where marginal costs are at a minimum. Average cost 𝑇𝐶 is 𝑄 𝑄 = −𝐿3 + 4𝐿 + 10𝐿2 + 20 Plan: remember 𝐿 is the independent variable and 𝑄 the dependent variable. 64 = 0. then substitute into the elasticity equation.04𝑄 − 0. or 2. 𝑄 = 0.Set 𝑀𝐶 = 𝐴𝐶: 0.72 = 0 %∆𝑄 %∆𝑃 𝑑𝑄 𝑃 × 𝑑𝑃 𝑄 𝑂𝑅 𝜀 = ∆𝑄 𝑃 × ∆𝑃 𝑄 Bring everything onto one side: 0.02𝑄 2 − 0.9 differentiation and elasticity Before reading this section.01𝐾𝑒 4𝐾−1 𝑐) 𝑄 = 𝐾 ln 𝐾 + 1 4. Justify your interpretation.44𝑄 + 8. Remember that the formula for elasticity is either: 𝜀 = 𝑄 0. and find the quantity where profit is maximised.5 𝑏) 𝑄 = 0.05𝑄𝑒 0. revise the concept of elasticity in Chapter 2. 2.e. Sketch the function of marginal productivity of an extra lawn mower (i. Plan: use the elasticity formula: 𝜀𝑠 = 𝑑𝑄 𝑃 × 𝑑𝑃 𝑄 Differentiate the supply equation.04𝑄 2 − 0. Initially the firm has no lawnmowers.64 6. Solution: differentiate the supply equation: 𝑑𝑄𝑠 = 2𝑃 + 1 𝑑𝑃 Substitute this into the elasticity formula: 𝜀𝑠 = 2𝑃 + 1 𝑃 [2𝑃 + 1]𝑃 = 𝑄 𝑄 Find 𝑃 when 𝑄 = 50 from the supply equation: 𝑄𝑠 = 𝑃2 + 𝑃 − 6 50 = 𝑃2 + 𝑃 − 6 0 = 𝑃2 + 𝑃 − 56 Solve for 𝑃 using the quadratic formula: 𝑃 = 𝑃 = = −𝑏 ± 𝑏 2 − 4𝑎𝑐 2𝑎 −1 ± 12 − 4 1 −56 2 1 −1 ± 225 2 𝑃 = −8 𝑜𝑟 𝑃 = 7 133 . 𝑎) 𝐶 = 15𝑄2 𝑏) 𝐶 = 16𝑄3 + 12𝑄2 + 5𝑄 𝑐) 𝐶 = 0. b) Find the profit function.06𝑄2 − 1. Find the marginal productivity of capital in the following production functions: 𝑎) 𝑄 = 14𝐾 1. 𝑎) 𝑄 = − 𝐿 + 5 2 𝑏) 𝑄 = 𝐿3 − 20𝐿2 + 125𝐿 𝑐) 𝑄 = 𝐿𝑒 𝐿−5 − 𝐿2 This makes finding elasticities a great deal easier and allows us to find elasticities of non-linear functions.72 = 0 0.1𝑄+4 3. find where the average product of labour is equal to the marginal product or labour.72𝑄 = 0 Either factorise out 𝑄 or use the quadratic formula to solve for 𝑄: The formula on the right can be rewritten as: 𝜀 = So either: 1. Find 𝑃 by substituting 𝑄 = 50 into the supply equation.72 𝑄 = 18 Marginal cost is equal to average cost when 𝑄 = 0 and when 𝑄 = 18. For the revenue and cost functions: 𝑅 𝑄 = 𝑄2 − 257𝑄 + 12.04𝑄 − 0. capital) for a lawn mowing firm with four employees. Example 1: find and interpret the elasticity of supply for the supply function 𝑄𝑠 = 𝑃2 + 𝑃 − 6 when output is 50 units. 5.04𝑄 = 0.72𝑄 + 8. Exercises: 1. c) Compare the answers from a) and b) above.222 𝐶 𝑄 = 5. For the following functions.000 + 180𝑄 a) Determine when marginal revenue and marginal costs are equal. 0. Find expressions for the marginal and average costs of the following functions. Having been given 𝑄𝑑 = 480. if price is increased by a certain percentage. as the derivative of 𝑄𝑑 with respect to 𝑃 is needed for the elasticity formula): 𝑑𝑃 = −0.875 Substitute 𝑃 = 823. Interpret the result.000 −200 2 𝑃 = −0. Find 𝑃 when 𝑄 = 480 from the demand function. Exercises: 1.000 = 𝑄𝑑 + 5 2 + 2000 2 𝑃 = −0. isolate the other variable 𝑃 (which is a lot easier): −200𝑃 = 𝑄𝑑 + 5 𝑃 = 2 − 400.005 480 + 5 + 2000 He wants to test your skills so asks you to find the elasticity of demand at the current sales of 480 computers/month. and your boss gives you the following demand function of one of his client’s products (a type of computer): −200𝑃 + 400.7 2 𝑏) 𝑃𝑠 = 𝑄 + 5 𝑎𝑡 𝑄 = 13 𝑐) 𝑃𝑠 = 7 ln 𝑄 + 5 + 3 𝑎𝑡 𝑄 = 6 Differentiate the demand function.1) and apply the elasticity rules.05 480 2 Example 2: you have just got a new job at a big finance firm.01𝑄𝑑 − 0.Price can never be negative so disregard 𝑃 = −8.000 𝑄𝑑 + 5 2 − 400. and what it means.01𝑄𝑑 − 0. quantity demanded would be reduced by only 0.354 This means demand is inelastic.05 𝑑𝑄𝑑 Inverting the whole of both sides of this last equation gives the derivative that is required: 𝑑𝑄𝑑 1 = 𝑑𝑃 −0. Solution: rearranging this demand function to have 𝑄𝑑 by itself is very difficult (try for yourself using the reverse of BIMDAS).1 Take the absolute value of 2.1 (it is still 2. say 10%. That is. 𝑃 is obtained from the rearranged demand equation: 𝑃 = −0.005 𝑄𝑑 + 5 𝑃 = 823.05 𝑄 𝜀𝑠 = 2.05 Substitute everything into the elasticity formula: 𝜀𝑑 = 1 𝑃 × −0.25𝑄 + 4𝑄 + 121 𝑎𝑡 𝑄 = 28 2. then substitute it into the above equation.01 480 − 0. 𝑃𝑠 − 3 𝑎) = 𝑄𝑠 𝑎𝑡 𝑄 = 2 0. Determine the elasticity of the following demand functions at the shown output. Interpret the result.005 𝑄𝑑 + 5 + 2000 134 .01𝑄𝑑 − 0.354 of that percentage. Substitute 𝑃 = 7 and 𝑄 = 50. supply would increase by 21%. 1 𝑎) 𝑄𝑑 = − 𝑃𝑑 + 48 𝑎𝑡 𝑃 = 20 4 −2𝑃𝑑 𝑏) 𝑄𝑑 = + 71 𝑎𝑡 𝑃 = 23 7 𝑐) 𝑃𝑑 = −0.27𝑄𝑑 + 77 𝑎𝑡 𝑄 = 13 𝑑) 𝑃𝑑 = −𝑄2 + 4𝑄 + 54 𝑎𝑡 𝑄 = 5 2 𝑒) 𝑃𝑑 = −0. the interpretation is that this good is elastic in supply. Determine the elasticity of the following supply functions at the price indicated. That is. if price was to increase by. Plan: use the elasticity formula: 𝜀𝑑 = 𝑑𝑄 𝑃 × 𝑑𝑃 𝑄 𝜀𝑑 ≈ −0.005 2 𝑄𝑑 + 5 𝑑𝑄𝑑 𝑑𝑃 = −0. Price is therefore 𝑃 = 7.875 and 𝑄𝑑 = 480 into the elasticity formula above: 𝜀𝑑 = 1 823.875 × −0. To make things simpler. into the elasticity formula: 𝜀𝑠 = 𝜀𝑠 = [2𝑃 + 1]𝑃 𝑄 2 7 + 1 (7) (50) Differentiate 𝑃 with respect to 𝑄𝑑 (the opposite of what is required. 7 20 This gives an elasticity of: 𝜀𝑑 = −0.2% For a 20% price increase. Theory: given the definition of total revenue: 𝑇𝑅 = 𝑃 ∙ 𝑄 Take the log of the whole of both sides: log 𝑇𝑅 = log 𝑃 ∙ 𝑄 log 𝑇𝑅 = log 𝑃 + log 𝑄 Taking the logs of an equation is a good approximation to the percentage change of the two sides. But this is not always the case.2%.7 ∙ 5 = −8. by what percentage will total revenue change? Plan: find %∆𝑄 from the elasticity formula. apply the %∆𝑇𝑅 formula: 135 .𝑑) 𝑒) 𝑃𝑠 = 𝑒 3𝑄+4 + 19 𝑃𝑠 = 𝑄 + 5 𝑄2 + 4𝑄 + 4 𝑎𝑡 𝑄 = 14 𝑎𝑡 𝑄 = 15 %∆𝑇𝑅 = 20 + −13. Find the %∆𝑄: −1.5 × 𝑃 20 To find 𝑃.10 elasticity and total revenue Most retail businesses have a relatively constant supply curve (especially when they import goods). then applying the %∆𝑇𝑅 formula. Plan: find the elasticity of demand at 𝑄 = 20 using 𝜀𝑑 = 𝑑𝑄 𝑃 × 𝑑𝑃 𝑄 6. use the demand equation: 20 = −0. which is determined by the quantity of goods sold.69 = 20 𝜀𝑑 = %∆𝑄 = 20 −0. demand is inelastic so this means that for a given price increase. Example 2: determine the percentage change in total revenue if the demand function for a retail good is 𝑄𝑑 = −0.69 = −13.5 68 = −1.5𝑃 + 54 𝑃 = 20 − 54 = 68 −0. The elasticity concept can be used to determine if changes in price will lead to an increase (or decrease) in total revenue.8% Then applying the percentage change in total revenue formula: %∆𝑇𝑅 = %∆𝑃 + %∆𝑄 Then use the original definition of elasticity to determine %∆𝑄 𝜀𝑑 = %∆𝑄 %∆𝑃 which can then be used in the %∆𝑇𝑅 formula %∆𝑇𝑅 = %∆𝑃 + %∆𝑄 Solution: find the elasticity: 𝜀𝑑 = −0. and the owner wants to raise prices by 5%. Such firms focus mainly on maximising total revenue.8 = 6. Even though the price has increased. so: %∆𝑇𝑅 = %∆𝑃 + %∆𝑄 Since from elasticity. Example 1: given the elasticity of demand is −0.5 × This means that demand is elastic at this point (why?). total revenue will increase by 6. Solution: %∆𝑄 %∆𝑃 %∆𝑄 −0. %∆𝑃 and %∆𝑄 can be found easily.5𝑃 + 54 when initial quantity sold is 20.5% Finally. then the %∆𝑇𝑅 can also be determined. and prices rise by 20%. which in turn comes from the demand function. quantity demanded will be reduced less than the increase in price so overall revenue will increase.69.7 = %∆𝑄 5 %∆𝑄 = −1. total revenue can either increase or decrease. Solution: find the derivative of the demand function: 𝑑𝑄 = −0. 𝑇𝑅 must decrease). the 𝑇𝑅 rectangle changes from the green and red. which can then be used in the %∆𝑇𝑅 formula: %∆𝑇𝑅 = %∆𝑃 + %∆𝑄. Initially. Theory: for an inelastic demand function: A price increase will increase total revenue (the diagram below shows that as price rises from 𝑃1 to 𝑃2 .5 = −3. which is also 𝑇𝑅. the percentage change could have been found using the original demand equation. but backwards). however this is difficult. It all depends on the elasticity of demand.%∆𝑇𝑅 = 5 + −8. 136 𝜀𝑑 = −0. a change in price will leave total revenue unchanged.162 37 . Since the red rectangle is larger than the blue one. as price increases from 𝑃1 to 𝑃2 . the area 𝑃 ∙ 𝑄. it is the green and blue area. Example 3: given the demand function 𝑄 = −0. A price decrease will decrease total revenue (work through the same logic as above.5% So for a 5% price increase.8𝑄 𝑑𝑃 Find the value of 𝑃 when 𝑄 = 37 by rearranging 𝑃1 𝑄2 𝑄1 𝑄 the demand equation: 37 − 77 = −0. These last two examples show that for a price increase.5% in total revenue.8 ∙ 10 × 𝜀𝑑 = − 80 ≈ −2. In Example 2.4𝑃2 + 77 and current sales of 𝑄 = 37. increases. there would be a fall of 3.4 𝑃 = ±10 Price must be positive. Substitute this in to find the elasticity: 𝜀𝑑 = −0.4𝑃2 −40 = 𝑃2 = 100 −0. to the green and blue. A price decrease will increase total revenue (using the same diagram. it is the red and green area. 𝑇𝑅 must increase). 𝑃 𝑃2 𝐷𝐷 𝑃 𝐷𝐷 𝑃2 𝑃1 𝑄2 𝑄1 𝑄 When elasticity is equal to 1 (in absolute terms). work the other way to make clear that a decrease in price will increase 𝑇𝑅). Since the blue area is larger than the red area.8𝑄 × 𝑃 𝑄 10 37 For an elastic demand function: A price increase will decrease total revenue (in the diagram below. so 𝑃 = 10. then after the price rise. how will a 15% decrease in prices affect total revenue? Plan: find the elasticity of demand at 𝑄 = 37 using 𝜀𝑑 = 𝑑𝑄 𝑃 × 𝑑𝑃 𝑄 %∆𝑄 %∆𝑃 then use the original definition of elasticity 𝜀𝑑 = to determine %∆𝑄. 6 and sales of 55 units per week. %∆𝑇𝑅 = %∆𝑃 + %∆𝑄 For an inelastic demand function: A price increase will increase total revenue A price decrease will reduce total revenue. A local optimum is an optimal point only over a small range around that point. 2. total costs are represented by: 𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶 Variable costs are usually represented by: 𝑉𝐶 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑄 All functions are in terms of 𝑄 as profits are directly determined by the quantity of goods sold. 𝑥 − 𝑎𝑥𝑖𝑠). a change in price will leave total revenue unchanged. 𝑎) 𝜀𝑑 = −0. Non-stationary inflection points do not have a gradient of zero at the inflection point.9 𝑑) 𝜀𝑑 = −1. with that point usually being a maximum or minimum. as the Law of Diminishing Marginal Returns. 137 . Exercises: Given the following elasticities. The nature of an optimal point is found using the Second Order Condition (SOC): 𝑑 2 𝑦 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑖𝑓 >0 𝑑𝑥 2 2 𝑑 𝑦 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑖𝑓 <0 𝑑𝑥 2 𝑑 2 𝑦 𝑆𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡 𝑖𝑓 =0 𝑑𝑥 2 Any inflection point is using the SOC: 𝑑 2 𝑦 𝑆𝑂𝐶 𝑓𝑜𝑟 any 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡: =0 𝑑𝑥 2 Inflection points are points of a function where a function changes from concave to convex (or vice versa). output may start to fall). The two types of inflection points: 1. 2. 3. %∆𝑄 = −15 1200 %∆𝑄 = ≈ 32. determine how a 5% price decrease will affect total revenue. which implies profit is zero. 𝜋𝐵𝐸 = 0 𝑠𝑢𝑚 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑑𝑦 𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 = 𝑑𝑥 As more labour is added to a production process. This theory is known. 4. The general form of profit is: 𝜋 = 𝑇𝑅 − 𝑇𝐶 Total revenue can also be simplified: 𝑇𝑅 = 𝑃 ∙ 𝑄 In general. past some point. determine how a 15% price increase will affect total revenue. apply the %∆𝑇𝑅 formula: %∆𝑇𝑅 ≈ −15% + 32.01 𝑒) 𝜀𝑑 = −1 For the weekly demand function 1000 2𝑃𝑑 𝑄𝑑 = − 3 0. When elasticity is equal to 1 (in absolute terms). output initially increases greatly. For an elastic demand function: A price increase will reduce total .43% Thus. determine how a 24% price increase will affect total revenue. determine how 𝑇𝑅 will change for a 17% price decrease. use the original elasticity formula: − 80 %∆𝑄 = 37 −15 − 80 37 1. The gradient at any maximum or any minimum is always zero.To find %∆𝑄 when price decreases by 15%.43% for a 15% price decrease. For the demand function 𝑃𝑑 = 𝑄2 + 10𝑄 − 3000 and sales of 33 units per week. Break even is when 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 equals 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒. Stationary inflection points have a gradient of zero at the inflection point. additional labour adds less and less to total output. total revenue will increase by approximately 17.43% 37 Finally. However.9 𝑐) 𝜀𝑑 = −0. The First Order Condition (FOC): 𝑑𝑦 =0 𝑑𝑥 Then solving for the unknowns.4 𝑏) 𝜀𝑑 = −1. A price decrease will increase total . The point at which it changes from convex to concave is the inflection point. no additional output can be obtained from additional labour (and in theory. A global optimum is a point that is the optimal point over the entire number range (i.43% = 17. Eventually. in general terms. chapter six summary Optimisation is the process of finding an optimal point.e. For the monthly demand function 𝑃𝑑 = −5 ln 𝑄2 + 5 + 20 And sales of 5 units per week. b) The quantity that maximises profit. at output of 20 units: 𝑃𝑠 = 100 ln 0. c) The quantity corresponding to maximum profit.5 𝑄 + 1 𝑑) 𝑇𝐶 = 0. d) The price corresponding to maximum profit. For the following demand function 𝑃𝑑 = − 𝑄 + 50 2 𝑄 − 40 𝑒 −𝑥 If output is at 4.2𝑄2 + 40 Determine: a) The total revenue function.5𝑥 1 − 𝑥 𝑓) 𝑓 𝑥 = 2 𝑥 + 1 Find all stationary and inflection points for the following functions. e) The output level where average profit and marginal profit are equal. 10. 11. Determine and interpret the elasticity for the following supply function. Sketch the profit function. b) The profit function. 15. c) The output for maximum profit.1𝑄 + 4 + 0. 4. A company faces a demand function of the form 1 𝑃𝑑 = + 0.9𝑄 The owner of a firm producing car rims wants to increase price by 5% from a level of 𝑄 = 20.1 𝑞 + 20 𝑞 + 5 𝑐) 𝑇𝑅 = 𝑄0. The factory faces a cost function of the form: 𝑇𝐶 = 0. e) The maximum profit.5 + 𝐿2 + 4𝐿 A firm manufacturing microwave dishes determines its profit functions to be 𝜋 = −1.5𝑄3 + 50𝑄2 Determine: a) An expression for average and marginal profit. c) The break-even quantities. 7.000𝑄 − 47.500 Find: a) The break-even quantities.01𝑥 2 ln 𝑥 + 1 𝑒) 𝜋 = −0. 3. How will total revenue be affected? Give a numerical answer. 16. e) The maximum profit attainable.1 𝑒 𝑄−5 + 𝑒 5−𝑄 + 10 𝑓) 𝜋 = 𝑞𝑒 −𝑞+1 For the profit function 𝜋 = −3𝑄3 + 75𝑄2 Determine the values of 𝑄 where average profit is equal to marginal profit. 138 . 𝐹 𝐸 𝑥 8. c) The maximum profit. 14. A production function has the form: 1 𝑄 = 15𝐿3 + 3𝐿2 + 4𝐿 Determine the levels of labour where the marginal productivity of labour is double the average productivity of labour. 13. Determine which of the following points are local/global maxima/minima: And a total cost function of the form: 𝑇𝐶 = 0. Determine the FOC for the following functions: 𝑎) 𝑦 = 𝑥 2 + 3𝑥 − 1 𝑏) 𝑦 = 15𝑥 3 + 1 − 𝑥 4 2 𝑐) 𝑦 = 12𝑥 2 𝑒 1−𝑥 2 𝑑) 𝑦 = 2 𝑥 − 1 𝑥 3 − 1 2 𝑥 − 1 𝑒) 𝑦 = 𝑥 + 1 Determine the nature of any stationary points for the following functions: 𝑎) 𝑦 = 𝑥 − 1 𝑥 + 3 𝑏) 𝑦 = 𝑥 3 + 3𝑥 2 − 45𝑥 + 18 𝑐) 𝑦 = −2𝑥𝑒 𝑥+3 𝑑) 𝑦 = 𝑒 4−𝑥 𝑥 − 3 2 1 2 𝑒) 𝑦 = 𝑒 𝑥 +3 + 19 14. and the owner of the factory wants to increase price by 25%.chapter six questions 1.2𝑒 0.01𝑄+4 Determine: a) The total revenue function. for the function: 𝑄 = 5𝐿0.01𝑄 −0. 6. Given the demand function is 100 𝑃𝑑 = 𝑄 + 1 Determine if total revenue will increase or decrease with the 5% increase in price. and relate it 𝑦 𝐵 𝐷 𝐶 𝐴 2. d) The maximum profit attainable. b) The break-even quantities. Determine the average and marginal values for the following functions: 𝑎) 𝜋 = −𝑄2 + 20𝑄 𝑏) 𝑇𝐶 = 0. b) The profit function.25 0.05𝑄+3 + 180 Determine and interpret the elasticity at 𝑄 = 8. d) The quantity for profit maximisation. and determine their nature where applicable: 𝑎) 𝑦 = 𝑥 3 − 𝑥 2 + 15𝑥 − 79 𝑏) 𝑦 = 𝑥 − 1 𝑥 − 5 𝑥 + 4 𝑐) 𝑦 = 𝑥 + 3 2 1 − 𝑥 𝑑) 𝑓 𝑥 = −𝑥𝑒 4−𝑥 𝑒) 𝑓 𝑥 = 𝑥 2 𝑒 1−𝑥 2 𝑓) 𝑦 = ln 𝑒 4𝑥 −3𝑥+2 A company has determined their profit function to be: 𝜋 = −100𝑄2 + 10. and plot all the important points.1𝑄 9. Determine when the average product of labour is equal to the marginal product of labour. 5. 12. Given the demand function 2 𝑃𝑑 = 𝑒 −0. A factory manufactures music players and sells them at $350. c) The profit function. Determine: a) The total cost function b) The total revenue function. j) Where the marginal and average profit functions are equal. f) The maximum profit. and the owner is thinking of raising prices by 10%. 17. i) The average profit function. f) The maximum profit. 18. j) Where the marginal and average profit functions are equal. A firm produces top quality surfboards and has a demand function of the form: 𝑃𝑑 = −0. h) The marginal profit function. and the owner is thinking of raising prices by 25%. If output is initially 4000. i) The average profit function. and the factory in which they are manufactured has a rent of $200. If output is initially 1000.5𝑄 + 2500 Each surfboard costs $250 to make. to attain this profit.4𝑞 + 4276 Each sandstone sculpture costs $80 to make. g) The price each sculpture is sold at. l) The extent to which total revenue would change. A firm produces sandstone sculptures. and the factory in which they are manufactured has a rent of $1500. and has a demand function of the form: 𝑃𝑑 = − − 0. determine: k) The elasticity at 𝑄 = 24. h) The marginal profit function. the product rule has to be used within the product rule. c) The profit function. determine: k) The elasticity at 𝑄 = 20. 139 . if the 10% price rise went ahead. Determine: a) The total cost function b) The total revenue function. d) The break-even quantities. g) The price per surfboard to attain this profit. l) The extent to which total revenue would change. e) The quantity for maximum profit. if the 25% price rise went ahead. and its interpretation.back to the elasticity obtained. e) The quantity for maximum profit. and its interpretation. Hint: to differentiate this function. d) The break-even quantities. 3 7.5 7.Chapter 7 Multiple Variable Differentiation Differentiation with more than one variable 7.2 7.7 Additional Variables Simple Partial Differentiation Complex Partial Differentiation Second Order Partial Derivatives Application of Partial Differentiation Total Differentiation Optimisation with Many Variables Economic Applications 141 142 144 146 147 148 151 155 158 158 Chapter Seven Summary Chapter Seven Questions 140 .7 7.4 7.1 7.6 7. 7. a factory) and labour.) and labour (𝐿: people driving the tractors. soil condition etc.e. Most goods require many different inputs. For example. it is drawn on three axes): 𝑧 𝑦 𝑥 Similar to a quadratic. paper prices change up and down. paper to wrap it. 141 . but they are ignored for now. water. ink to print on the paper etc. The above multivariable function is the shape of a cup. Multivariable functions means that a dependent variable is determined by two or more other variables. an example of a multivariable function is: 𝑧 = 𝑥 2 + 𝑦 2 This is a multivariable equation (𝑧 is a function of 𝑥 and 𝑦). heat to melt it.g. cubic. constants. such as rainfall. storage sheds etc. 𝑦) This is now a three-dimensional function (i. A very common economic function is called the Cobb-Douglas Production Function which is an 𝑧 approximation for production of a good from the 𝑦 𝑥 inputs of capital equipment (e. fixing the sheds etc.4 3𝐾 0.1 additional variables Theory: multivariable or multivariate simply means more than one variable.6 Which specifies that the production of a certain good (e.). cocoa prices might be high from low rainfall in South America etc. All the previous chapters have only had a single independent variable 𝑥 determining the dependent variable 𝑦 through a certain function: 𝑦 = 𝑓 𝑥 The function could be quadratic. 𝑧 𝑦 𝑥 Theory: the Cobb-Douglas Production Function has the general form: 𝑄 = 𝐴𝐿𝛼 𝐾𝛽 where the values of 𝐴. Multivariable functions are very useful as they represent the real world much better.. a block of chocolate needs cocoa powder. 𝑦) Where 𝑧 is a dependent variable. 𝛼 and 𝛽 are all positive Mathematically. and it depends on two independent variables 𝑥 and 𝑦. Each one of these inputs has its own market. A numerical example could be: 𝑄 = 5𝐿0. etc. the function above has a minimum which can be visually determined. Obviously. 𝑄𝑡𝑜𝑛𝑛𝑒𝑠 𝑜𝑓 𝑤𝑒𝑎𝑡 ) is a function of capital (𝐾: tractors. 𝑧 is a function of 𝑥 and 𝑦. That is. Theory: the simplest multivariable functions have the form: 𝑧 = 𝑓(𝑥. these graphs are three dimensional in nature. sugar. there are other factors. Because there are now three variables. milk.g. and can also be written as: 𝑧(𝑥. Exercises: 1. List all the different variables that a baker needs to take into account when baking a cake. 2. Think of a few goods that that do not require many different types of inputs. What do they have in common? 3. Which of these are not Cobb-Douglas Production Functions? 𝑎) 𝑄 = 5𝐿0.2 𝐾0.4 𝑏) 𝑄 = 5𝐿𝐾0.4 5𝐿0.7 𝑐) 𝑄 = −0.4 𝐾 aircraft. This is the indirect effect of Boeing using more electricity. ELECTRICITY direct effect (cost of extra electricity) indirect effect (increase in metal costs) BOEING METAL MANUFACTURER When talking about partial differentiation, any possible indirect effects are ignored, and assumed not to exist. This is a big assumption, and later in this chapter it will be removed. Theory: partial differentiation is finding the rate at which the dependent variable changes when an independent variable changes assuming all other variables are held constant. That is, it is assumed that the other “independent” variables do not affect the independent variable that is changing (even though in reality, they might). Partial differentiation is written in one of two ways: 1. 𝜕𝑧 𝜕𝑥 𝑂𝑅 𝜕𝑧 𝜕𝑦 𝑑) 𝑒) 5𝐿−0.7 𝐾 −0.4 𝑄 = −3𝐿0.4 𝐾 0.6 𝑄 = 7.2 simple partial differentiation An intuitive explanation of partial differentiation is best to get your head around this concept. Intro example 1: two factories, an aircraft manufacturer (such as Boeing) and a metal manufacturing company (such as Metalex), are located close to one another; Metalex supplies Boeing with the metal sheeting to cover the aircraft. Both companies require electricity to run their large production factories (Boeing to build the planes, and Metalex to create the highest quality metal) but the electricity system cannot handle any more current (electrical flow). Say Boeing wanted to introduce electrical heating to their factory. That would mean more electricity would need to flow to Boeing, costing them more. Assuming prices of electricity are constant, this is the direct effect of increasing the use of electricity (that is, the extra cost of using more electricity for heating). What is not being taken into account is that when Boeing increases their use of electricity, this will reduce the amount of electricity available for Metalex, making the metal more expensive. Metalex must then charge Boeing more for the metal, making it more expensive to produce This is read as “the derivative of 𝑧 with respect to 𝑥 holding everything else constant”. The “curvy dees” are “deltas” and are not the same as normal 𝑑’s used in differentiation. 2. 𝑧𝑥 𝑂𝑅 𝑧𝑦 In this method of defining a partial derivative, the subscript is the variable being differentiated, and it is assumed that all other variables are constant. Intro example 2: partially differentiate 𝑧 with respect to 𝑥 for the function 𝑧 = 𝑥𝑦 + 5𝑥 − 7𝑦 The 𝑦 has to be treated as if it were a constant, even though it obviously is not: 142 𝜕𝑧 = 1𝑦 + 5 𝜕𝑥 The 1𝑦 comes from differentiating 𝑥𝑦 with respect to 𝑥 treating 𝑦 as if it were a constant; if 3𝑥 had to be differentiated, the answer would be 3 1 = 3. Similarly above, the 𝑦 is treated as if it were a 3, but in fact it is still 𝑦. An absurd example might help. Intro example 3: find the partial derivative of 𝑧 with respect to 𝑥 for the function: 𝑧 = 𝑥 𝑦 𝑤 𝑝 𝑞 𝑟 𝑒 This equation has lots of variables, all multiplying one another, but all that is relevant is the 𝑥, as everything else is held constant. 𝜕𝑧 = 𝑦 𝑤 𝑝 𝑞 𝑟 𝑒 𝜕𝑥 You may be wondering why the answer is not zero, as everything else is a constant? The reason is that these “constants” are multiplying 𝑥. If they were added to 𝑥, then the derivative would be zero. Theory: to partially differentiate a multivariable function with respect to a single variable, differentiate the function with respect to that variable while treating all other variables as constants. Example 1: find the partial derivative with respect to 𝑥 for the function 𝑧 = 𝑥 𝑦 𝑝𝑘 + 𝑝𝑓𝑑 + 3𝑥 Plan: differentiate using the normal rules, but assume everything other than 𝑥 is a constant. Solution: 𝜕𝑧 = 2𝑥𝑦 2 𝑝𝑘 + 3 𝜕𝑥 The first part of the original equation has 𝑥 2 and all the other variables 𝑦 2 𝑝𝑘 are assumed to be constant. Thus 𝑥 2 is differentiated and then multiplied by the remaining “constants”. The 2 2 second part of the original equation is 𝑝𝑓𝑑 and since there are no 𝑥’s and differentiating a constant (remember, everything other than an 𝑥 is assumed to be a constant), gives zero. The last part of the original equation is 3𝑥, so it is differentiated to give the constant 3. Example 2: find the partial derivative with respect to 𝑥 in the following equation 𝑧 = 𝑥 2 9 4 5 + 4 8 7 + 3𝑥 Plan: simplify the numbers then differentiate using the normal rules, but assume everything other than 𝑥 is a constant. Solution: simplify the numbers: 𝑧 = 180𝑥 2 + 224 + 3𝑥 Differentiate using the normal rules: 𝜕𝑧 = 2 180 𝑥 + 3 = 360𝑥 + 3 𝜕𝑥 Look back at the last two examples. The second example is identical to the first, except the variables (other than 𝑥) are replaced with actual numbers. If Example 2 was done without simplifying first, the answer would be: 𝜕𝑧 = 2𝑥 9 4 5 + 3 𝜕𝑥 Which is very similar to the answer in Example 1. It is very important you understand that the other variables are held constant when partially differentiating. Example 3: partially differentiate with respect to 𝑥 and then with respect to 𝑦 for the function 𝑧 = 𝑥 2 𝑦 − 3𝑦 3 𝑥 2 + 5𝑥 − 7𝑦 Plan: for 𝜕𝑧/𝜕𝑥 differentiate the function holding 𝑦 constant. Then for 𝜕𝑧/𝜕𝑦 differentiate the function holding 𝑥 constant. Solution: finding 𝜕𝑧/𝜕𝑥: 𝜕𝑧 = 2𝑥𝑦 − 3𝑦 3 2𝑥 + 5 𝜕𝑥 In the original function: 143 The first red section: differentiate 𝑥 2 then multiply by the rest of the red section (i.e. 𝑦) giving: 2𝑥𝑦 𝜕𝑧= 4𝑦𝑥 3 + 2𝑥 + 2 𝜕𝑦 It is vital you understand this simple partial differentiation before moving on. Exercises: 1. Find both first order partial derivatives for the functions: 𝑎) 𝑧 = 𝑥 2 𝑦 + 3𝑦 2 𝑥 + 7𝑥𝑦 𝑏) 𝑧 = 3𝑥 2 + 3𝑦 2 + 3 𝑥𝑦 7 𝑐) 𝑧 = 5𝑥𝑦 + 𝑥 2 𝑦 + 11 𝑑) 𝑧 = 2𝑥𝑦 − 2𝑥 2 𝑦 3 + 𝑥 −1 3 𝑒) 𝑧 = 2𝑥𝑦 −2 + −2 𝑥𝑦 2. Differentiate the following with respect to both 𝑥 and 𝑦 (each individually): 𝑎) 𝑧 = 3𝑥 7 𝑦 −3 + 5𝑥 + 5𝑥𝑦 − 3 𝑏) 𝑧 = −2𝑥 −2 𝑦 + 7𝑦 −2 𝑥 𝑐) 𝑧 = 15𝑥 2 𝑦 3 − 15𝑥 −2 𝑦 −3 The first blue section: differentiate 𝑥 2 then multiply by everything else in that section (i.e. −3𝑦 3 ) giving: −3𝑦 3 2𝑥 The second red section: differentiate the 𝑥, then multiply by everything else in that section (i.e. 5) giving: 5. The second blue section: there are no 𝑥’s so the derivative of a constant is zero. Put these together and simplify: 𝜕𝑧 = 2𝑥𝑦 − 6𝑦 3 𝑥 + 5 𝜕𝑥 Now for 𝜕𝑧/𝜕𝑦, differentiate the original equation holding 𝑥 as a constant: 𝜕𝑧 = 𝑥 2 − 9𝑦 2 𝑥 2 − 7 𝜕𝑦 Figure out how this last answer was obtained, one section at a time. In the above example, you may have noticed that when looking at each section, 𝑥 was isolated and differentiated, then everything else was brought back in. This is a good technique to use, but only use it where there are no division signs. Example 4: find both partial derivatives of 𝑧 = 2𝑥 𝑦 + 2𝑥𝑦 − 3𝑥 + 2𝑦 + 15 Plan: for 𝜕𝑧/𝜕𝑥 differentiate the function holding 𝑦 constant. Then for 𝜕𝑧/𝜕𝑦 differentiate the function holding 𝑥 constant. Solution: for 𝜕𝑧/𝜕𝑥: 𝜕𝑧 = 3𝑥 2 ∙ 2𝑦 2 + 2𝑦 − 3 𝜕𝑥 𝜕𝑧 = 6𝑥 2 𝑦 2 + 2𝑦 − 3 𝜕𝑥 And for 𝜕𝑧/𝜕𝑦: 𝜕𝑧 = 2𝑦 ∙ 2𝑥 3 + 2𝑥 + 2 𝜕𝑦 3 2 7.3 complex partial differentiation This section applies the more complex rules (product, chain etc.) to partial differentiation. Example 1: find both partial derivatives of 𝑧 = 𝑥 2 + 𝑦 3 3𝑥 4 − 𝑦 + 3 2 Plan: for 𝜕𝑧/𝜕𝑥 differentiate the function holding 𝑦 constant. Then for 𝜕𝑧/𝜕𝑦 differentiate the function holding 𝑥 constant. Use the product rule as the base rule, with the chain rule within it. Solution: for 𝜕𝑧/𝜕𝑥: 𝜕𝑧 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 𝜕𝑥 𝑑𝑒𝑟 1𝑠𝑡 = 2𝑥 𝑑𝑒𝑟 2𝑛𝑑 = 2 3𝑥 4 − 𝑦 + 3 1 . 3 4𝑥 3 = 24𝑥 3 3𝑥 4 − 𝑦 + 3 The 𝑑𝑒𝑟 2𝑛𝑑 was obtained by using the chain rule, and treating 𝑦 as a constant. The 𝑦 is still there as due to the chain rule, the bracketed terms are rewritten with the power reduced by one. For the solution, substitute into the base (product) rule: 144 Solution: for 𝜕𝑧/𝜕𝑥: 𝜕𝑧 𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇 − 𝑑𝑒𝑟 𝐵𝑂𝑇 × 𝑇𝑂𝑃 = 𝜕𝑥 [𝐵𝑂𝑇𝑇𝑂𝑀]2 𝑑𝑒𝑟 𝑇𝑂𝑃 = 3 𝑥 2 + 𝑦 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 1 The chain rule was used to obtain the derivative of the top. Differentiate with respect to both 𝑥 and 𝑦 (individually) using the Product Rule: 𝑎) 𝑧 = 𝑥 2 + 5𝑦 3𝑥 + 7𝑦 2 𝑏) 𝑧 = 𝑥 2 + 𝑦 𝑦 2 + 𝑥 𝑐) 𝑧 = 3𝑥𝑦 − 𝑥 2 𝑥𝑦 3 + 1 1 𝑑) 𝑧 = 3𝑥 2 + 𝑥𝑦 − 𝑦 4 𝑦 3 3 4 4 𝑒) 𝑧 = + − 2 𝑥 𝑥𝑦 𝑥 𝑥𝑦 3.𝜕𝑧 = 𝜕𝑥 2𝑥 3𝑥 4 − 𝑦 + 3 2 𝜕𝑧 𝑥 2 + 𝑦 = 𝜕𝑥 + 24𝑥 3 3𝑥 4 − 𝑦 + 3 𝑥 2 + 𝑦 3 2 6𝑥 𝑥 − 𝑦 − 𝑥 2 + 𝑦 𝑥 − 𝑦 2 Further simplify the square brackets: 𝜕𝑧 𝑥 2 + 𝑦 2 5𝑥 2 − 6𝑥𝑦 − 𝑦 = 𝜕𝑥 𝑥 − 𝑦 2 For 𝜕𝑧/𝜕𝑦: 𝜕𝑧 𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇 − 𝑑𝑒𝑟 𝐵𝑂𝑇 × 𝑇𝑂𝑃 = 𝜕𝑦 [𝐵𝑂𝑇𝑇𝑂𝑀]2 𝑑𝑒𝑟 𝑇𝑂𝑃 = 3 𝑥 2 + 𝑦 = 3 𝑥 2 + 𝑦 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = −1 2 2 And for 𝜕𝑧/𝜕𝑦: 𝜕𝑧 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡 𝜕𝑦 𝑑𝑒𝑟 1 𝑑𝑒𝑟 2 𝑠𝑡 𝑛𝑑 = 3𝑦 2 4 1 = 2 3𝑥 − 𝑦 + 3 (−1) = −2 3𝑥 4 − 𝑦 + 3 1 Substitute this back into the base function: 𝜕𝑧 = 𝜕𝑦 3𝑦 2 3𝑥 4 − 𝑦 + 3 2 + −2 3𝑥 4 − 𝑦 + 3 𝑥 2 + 𝑦 3 Substitute into the base (quotient) rule: 𝜕𝑧 3 𝑥 2 + 𝑦 = 𝜕𝑦 2 Both the solutions above can be simplified by factorisation (see Chapter 1 and 5). Exercises: 1. Example 2: find both partial derivatives of 𝑧 = 𝑥 2 + 𝑦 𝑥 − 𝑦 3 𝑥 − 𝑦 − (−1) 𝑥 2 + 𝑦 𝑥 − 𝑦 2 3 Factorise and simplify the top (do it yourself!): 𝜕𝑧 𝑥 2 + 𝑦 2 3𝑥 − 2𝑦 + 𝑥 2 = 𝜕𝑦 𝑥 − 𝑦 2 Make sure you fully understand the last two examples before trying the following exercises. Use the quotient rule as the base rule. and the chain rule within the quotient rule. take one off the power. Differentiate with respect to both 𝑥 and 𝑦 (individually) using the Quotient Rule: 𝑥 + 𝑦 𝑎) 𝑧 = 𝑥 − 𝑦 3𝑥 2 − 2𝑦 3 𝑏) 𝑧 = 𝑥 2 𝑦 2𝑥𝑦 − 2𝑥 2 𝑦 𝑐) 𝑧 = 15𝑥𝑦 + 1 4𝑥𝑦 4 𝑑) 𝑧 = 1 − 𝑥𝑦 1 + 𝑥𝑦 2 𝑒) 𝑧 = 1 − 𝑥𝑦 2 Plan: for 𝜕𝑧/𝜕𝑥 differentiate the function holding 𝑦 constant. Remember the chain rule: bring the power out front of the brackets. Then for 𝜕𝑧/𝜕𝑦 differentiate the function holding 𝑥 constant. Differentiate with respect to both 𝑥 and 𝑦 (individually) using the Chain Rule: 𝑎) 𝑧 = 𝑥 + 𝑦 3 𝑏) 𝑧 = 3𝑥 2 + 𝑦 4 𝑐) 𝑧 = 7𝑥 2 + 7𝑦 2 5 𝑑) 𝑧 = −2𝑥 −2 + 𝑦 −2 4 𝑒) 𝑧 = 3𝑥𝑦 + 2𝑥 2 𝑦 3 −3 2. then multiply by the derivative of the inside of the brackets. Substitute everything into the Quotient Rule: 𝜕𝑧 6𝑥 𝑥 2 + 𝑦 = 𝜕𝑥 Factorise 𝑥 2 + 𝑦 2 2 2 2𝑥 2 = 6𝑥 𝑥 2 + 𝑦 𝑥 − 𝑦 − (1) 𝑥 2 + 𝑦 𝑥 − 𝑦 2 3 out of the top: 145 . Differentiate with respect to 𝑦 first: 𝜕𝑧 = 7𝑦 6 𝑥 3 − 3𝑥 − 2𝑦 𝜕𝑦 Then differentiate with respect to 𝑦: 𝜕 2 𝑧 = 7(3)𝑦 6 𝑥 2 − 3 𝜕𝑦𝜕𝑥 = 21𝑥 2 𝑦 6 − 3 Notice that the two mixed partial derivatives are the same. 𝑒) 𝑧 = ln 𝑒 𝑥 𝑦 +3𝑥 𝑦 − 1 Differentiate with respect to both 𝑥 and 𝑦 (individually) using any rules: 2 𝑎) 𝑧 = 𝑒 𝑥 𝑦−1 + 𝑥𝑦 3𝑥𝑦 − 1 3 𝑏) 𝑧 = 5𝑥 2 𝑦 − 𝑥 ln 15𝑥𝑦 + 𝑒 𝑥𝑦 𝑐) 𝑑) 𝑒) 𝑧 = ln 15𝑥𝑦 + 𝑒 𝑥𝑦 + 𝑒 𝑥𝑦 𝑒 𝑦 − 𝑥 2 𝑦 𝑧 = 𝑥𝑦 − ln 𝑥𝑦 3𝑥𝑦 − 𝑥 2 4 𝑧 = 1 − 𝑥𝑦 𝑥 1 2 2 7. Theory: the second order partial derivative is simply partially differentiating the original function twice. Theory: mixed partial derivatives are found by differentiating with respect to one variable first. The second order partial derivative is denoted in one of two ways: 𝜕 2 𝑧 𝜕 2 𝑧 .4 second order partial derivatives This is a very simple concept. In theory it is the same as a normal second derivative. Solution: differentiate with respect to 𝑥: 𝜕𝑧 = 3𝑥 2 𝑦 7 − 3𝑦 + 2 𝜕𝑥 Differentiate with respect to 𝑥 again: 𝜕 2 𝑧 = 6𝑥𝑦 7 𝜕𝑥 2 For the second order partial derivative of 𝑦: 146 . no matter which way they are found. Differentiate with respect to 𝑥 first: 𝜕𝑧 = 3𝑥 2 𝑦 7 − 3𝑦 + 2 𝜕𝑥 Then differentiate with respect to 𝑦: 𝜕 2 𝑧 = 3 7 𝑥 2 𝑦 6 − 3 𝜕𝑥𝜕𝑦 = 21𝑥 2 𝑦 6 − 3 2. Differentiate with respect to both 𝑥 and 𝑦 (individually) using the 𝑒 and ln rules: 2 𝑎) 𝑧 = 𝑒 𝑥𝑦 −𝑥 𝑏) 𝑐) 𝑑) 𝑧 = 𝑒 𝑥 𝑦 −𝑦 𝑥 𝑧 = ln 𝑥𝑦 − 𝑥 2 𝑦 𝑧 = 𝑒 ln 𝑥𝑦 −1 3 3 2 2 2 𝜕𝑧 = 7𝑦 6 𝑥 3 − 3𝑥 − 2𝑦 𝜕𝑦 Differentiate this equation with respect to 𝑦 again: 𝜕 2 𝑧 = 42𝑦 5 𝑥 3 − 2 𝜕𝑦 2 This is very similar to what was learnt in Chapter 5. This is always the case. 𝑧𝑦𝑥 5. 𝜕𝑥𝜕𝑦 𝜕𝑦𝜕𝑥 𝑂𝑅 𝑧𝑥𝑦 .4. These are called mixed partial derivatives. then differentiate the function with respect to 𝑦 twice. then differentiating with respect to the other variable. however it is a good idea to always write out the theory. They are denoted in a similar way to the straight second order partial derivatives: 𝜕 2 𝑧 𝜕 2 𝑧 . 𝑧𝑥𝑦 = 𝑧𝑦𝑥 Example 1: find all second order partial derivatives for the function 𝑧 = 𝑥 3 𝑦 7 − 3𝑥𝑦 + 2𝑥 − 𝑦 2 Plan: differentiate the function with respect to 𝑥 twice. 𝑧𝑦𝑦 Example 1 cont: finding both mixed partial derivatives: 1. Theory: mixed partial derivatives are the same. however there are two more second order partial derivatives. 𝜕𝑥 2 𝜕𝑦 2 𝑂𝑅 𝑧𝑥𝑥 . and make sure that the mixed second order partial derivatives are the same. For the Cobb-Douglas Production Function 𝑄 = 𝐴𝐿𝛼 𝐾𝛽 For the straight second order partial derivatives (using the quotient and chain rules): 𝑧𝑥𝑥 1 8 𝑥 + 3 + 5 − 2𝑥 8𝑥 𝑦 = 2 1 𝑥 2 + 3 + 5 𝑦 2 𝑧𝑥𝑥 8 −8𝑥 + 3 + 40 𝑦 = 2 1 𝑥 2 + 3 + 5 𝑦 2 −5 48𝑦 𝑧𝑦𝑦 = 1 𝑥 + 3 + 5 − −3𝑦 −4 −12𝑦 −4 𝑦 2 1 𝑥 2 + 3 + 5 𝑦 2 147 . then differentiate again to find the straight second order partial derivatives. Theory: the partial derivative of the Cobb-Douglas Production Function gives the marginal product of labour (𝜕𝑄/𝜕𝐿) and the marginal product of capital (𝜕𝑄/𝜕𝐾). the greater the increase in output from an extra unit of that input.The following example has minimal narration so use scrap paper to see if you can get the same answers. The higher the value of the marginal product. Then find the mixed second order partial derivatives. Solution: for the first order partial derivatives. Determine all four second order partial derivatives for the following functions (and make sure 𝑧𝑥𝑦 = 𝑧𝑦𝑥 ): 𝑎) 𝑧 = 5𝑥𝑦 + 3𝑦 3 𝑏) 𝑧 = 12 𝑥 2 𝑦 + 𝑥𝑦 2 𝑥 + 𝑦 2 𝑥 2 + 𝑥𝑦 𝑐) 𝑧 = 𝑦 − 𝑥 𝑑) 𝑧 = 𝑒 𝑥 𝑦+4𝑥 𝑒) 𝑧 = ln 7𝑥 2 𝑦 + 4𝑦 Determine all four second order straight partial derivatives. 𝑧𝑥 = 8𝑥 1 𝑥 2 + 3 + 5 𝑦 3 1 −3𝑦 −4 4 𝑥 2 + 3 + 5 𝑦 𝑧𝑦 = 4 1 𝑥 2 + 3 + 5 𝑦 −12𝑦 −4 𝑧𝑦 = 1 𝑥 2 + 3 + 5 𝑦 𝑐) 1 1 + 𝑥 𝑥𝑦 2 𝑥 −3 𝑦 −2 + 1 7. 𝑎) 𝑧 = 𝑒 5𝑥𝑦 + 4𝑥𝑦 3 𝑏) 𝑧 = ln 𝑒 𝑥 𝑧 = 2 𝑦 2 +5𝑥𝑦 2 2.5 application of partial differentiation The Cobb-Douglas Production Function is a very common application of partial derivatives. Economic Theory: returns to scale are defined as the extent output changes when inputs are changed by a certain ratio. the ln rule is used. Example 2: find all second order partial derivates for the function 1 𝑧 = ln 𝑥 + 3 + 5 𝑦 2 4 𝑧𝑦𝑦 48𝑥 2 12 48 + 8+ 5 𝑦 𝑦 5 𝑦 = 2 1 𝑥 2 + 3 + 5 𝑦 24𝑥𝑦 −4 𝑥 2 + 1 +5 𝑦 3 2 For the mixed second order partial derivative: 𝑧𝑥𝑦 = 𝑧𝑦𝑥 = Plan: find the first order partial derivatives. with the chain rule within it: 1 4 2𝑥 𝑥 + 3 + 5 𝑦 𝑧𝑥 = 4 1 𝑥 2 + 3 + 5 𝑦 2 3 Exercises: 1. Determine the marginal productivity of labour and capital in the following Cobb-Douglas Production Functions and also the returns to scale. This is the indirect at a labour level of 46 and a capital level of 71. which reduced the amount of electricity available for the metal manufacturer Metalex. output will less than double.5 𝐾 0. and for an extra unit of capital.9 𝑔) 𝑄 = 5. if the cost of a unit of labour and a unit of capital were the same.5 𝑏) 𝑄 = 13𝐿0.2 30 𝜕𝐿 unit of labour. output would increase by approximately 21.41 ) 𝑄 = 2.7 𝐾 0.If: 1.7 when 𝐿 = 30 and 𝐾 = 46. Boeing had installed electric heaters in their plant.619 3𝑑. Since the addition of the indices is 0.5 0.64 2𝑑. the electricity example was used to demonstrate the theory of partial differentiation.25 ≈ 21.5 𝑒) 𝑄 = 0.812 𝐾 0.7 Substitute 𝐿 = 30 and 𝐾 = 46: 𝜕𝑄 = 1.118𝐿0.7 𝜕𝐿 = 1.6 𝐾 0. Add the indices to determine the returns to scale. 3.8 𝐾 0. Example 1: find the marginal productivity of labour for the function 𝑄 = 4𝐿0.2 𝐾 0. output will double. 𝜕𝐿 And for marginal productivity of capital: 𝜕𝑄 = 9𝐿0.79 𝐾 0. In this last example. Example 2: find the marginal productivities of labour and capital for the following production function 𝑄 = 12𝐿 𝐾 0. 𝐾 = 71 to find the marginal productivity of labour.06 units. which one would you increase? Why? 46 0.4 𝑐) 𝑄 = 12.75 𝜕𝐿 𝜕𝑄 = 6 46−0.5 𝐾 0. 𝑝 Output will increase by 1.2𝐿−0.5𝐿0.7 = 1. Solution: 𝜕𝑄 = 4 0. there are constant returns to scale. if inputs are doubled.417 If you had the option of increasing either labour or capital by one unit in each of the above questions. 𝑝.8𝐿0. Plan: differentiate with respect to 𝐿 then substitute in 𝐿 = 46.e. direct effect).7 ≈ 1.3 𝐿−0.75 ≈ 21. 𝛼 + 𝛽 < 1 then there are decreasing returns to scale.75 −0. Plan: partially differentiate 𝑄 with respect to 𝐿. output would increase by approximately 21. 2.73𝐿0. Similarly for 𝐾. Boeing assumed that their decision would not affect costs other than the use of electricity (i. output will more than double. 𝑎) 𝑄 = 4𝐿0.3 𝐾 0.7 𝐾 0.3 + 0. Also determine the returns to scale.619 units for an extra 7. 𝛼 + 𝛽 = 1 then there are constant returns to scale. it would be better to employ more labour.03 2𝑑.4 𝑑) 𝑄 = 75𝐿0. so will charge Boeing more per sheet of metal. 𝜕𝐾 Thus for an extra unit of labour. if inputs double.3 𝑓) 𝑄 = 100𝐿0.5 71−0. Exercises: 1. as they will add (slightly) more to output than capital.7 𝜕𝑄 = 6𝐿−0. if inputs double.64 units. Solution: for marginal productivity of labour: 148 .25 𝜕𝐾 𝜕𝑄 = 9 460.6 𝐾 0. However. Metalex will not produce as much metal.7 𝐾 0. 𝛼 + 𝛽 > 1 then there are increasing returns to scale.5 710.5 𝐾 −0. 𝑝.6 total differentiation At the beginning of the chapter. this is what total differentiation finds. then the approximation will be less accurate. Solution: the partial derivatives are: 𝜕𝑃 = 10𝐸𝑀 + 10𝐸 − 6 𝜕𝐸 𝜕𝑃 = 5𝐸 2 − 4𝑀 + 3 𝜕𝑀 The small change in 𝐸 is 𝑑𝐸 = 5 and the small change in 𝑀 is 𝑑𝑀 = −3. the variables are: total costs (𝑧). this equation is a good approximation. For small changes in either or both of the input variables. cost of electricity (𝑥) and the cost of metal (𝑦).709 = 2.674. Note: this equation finds the gradient function of a multivariable equation. As a manager. Theory: the following is a formula that allows us to move from partial differentiation to total differentiation.effect of using more electricity.470 − 332. When the changes are large.761 Production will increase by approximately two and a half million units. For the reason. the 𝑧 function (the cost 𝑃 in this example) cannot be isolated easily. It is negative because it is a decrease. Substitute the partial derivatives into the total derivative equation: 𝑑𝑃 = 𝜕𝑃 𝜕𝑃 𝑑𝐸 + 𝑑𝑀 𝜕𝐸 𝜕𝑀 It is much simpler than it seems. 𝑃 = 5𝐸 2 𝑀 + 5𝐸 2 − 2𝑀2 + 3𝑀 − 6𝐸 + 550 Plan: partially differentiate the function 𝑃. 𝐸 = 150 + 5. Adding the direct and indirect effects gives the total effect.007. To get the actual change in production. substitute the original values of 𝐸 = 150 and 𝑀 = 400: 𝑑𝑃 = 10 150 400 + 10 150 − 6 5 + 5 150 2 Partial derivative with respect to 𝑦 In the Boeing example.p. The total derivative equation is: 𝑑𝑧 = 𝜕𝑧 𝜕𝑧 𝑑𝑥 + 𝑑𝑦 𝜕𝑥 𝜕𝑦 electricity and metal into the total derivative function. This is still an approximation. when in reality. 𝑀 = 397 (these numbers come from the original plus the change in the inputs. determine the extent total production will change if electricity use increases by 5 units from 150. then substitute the partial derivatives. Knowing the total effect is better than pretending variables are constant. 𝑀 = 400 − 3). they are not. The total derivative approximation method is used because often. consult a second year text book dealing with “linear approximations and errors”. but the partial derivatives can easily be found with other 149 . better decisions are made when variables are not assumed to be constant. Partial derivative with respect to 𝑥 very small change in 𝑥 𝑑𝑃 = 10𝐸𝑀 + 10𝐸 − 6 𝑑𝐸 + 5𝐸2 − 4𝑀 + 3 𝑑𝑀 Further substitution of 𝑑𝐸 = 5 and 𝑑𝑀 = −3: 𝑑𝑃 = 10𝐸𝑀 + 10𝐸 − 6 (5) + 5𝐸 2 − 4𝑀 + 3 (−3) 𝑑𝑧 = very small change in 𝑧 𝜕𝑧 𝜕𝑧 𝑑𝑥 + 𝑑𝑦 𝜕𝑥 𝜕𝑦 very small change in 𝑦 Finally.)). Finding the difference between these two production levels gives the precise change in production (actually do it and you should get 2. the changes in electricity and metal. and the original levels of − 4 400 + 3 −3 𝑑𝑃 = 3. 𝑀 = 400 and then the production at the new input level of 𝐸 = 155. Example 1: for the following function. find the production at the original input level of 𝐸 = 150. and metal prices decrease by 3 units from 400.702million units (3d. and the current use of capital at 300 units.5 .4 −1.4 𝜕𝑄 𝜕𝑄 𝑑𝐿 + 𝑑𝐾 𝜕𝐿 𝜕𝐾 𝑑𝑄 = 28𝐿−0. 𝐾 = 700 = 5 500 𝑄 𝐿 = 502. 𝐾 = 298. 𝑑𝐿 = 4 and 𝑑𝐾 = −1. Plan: find the partial derivatives of 𝑧.6 4 −3 + 24(400)0.5 𝑑𝐾 Then substitute 𝐿 = 500. 𝑦 = 700 .96 (2𝑑.7 𝐾 −0.4 𝐾 0.7 𝐾 0.5 500 𝑑𝑄 = 1.5 ∆𝐾 = 300 = −1.6 = 28𝐿−0.3 (300)0. ) This is the approximate change in quantity from + 2. find the value of the function at the New variable level then subtract the value of the function at the Original variable level: ∆𝑧 = 𝑧 𝑥𝑁 .5 𝑄𝐾 = 2. then substitute everything into the total derivative function to solve for 𝑑𝑧. 𝐾 = 697 = 5 502 0. Also. 𝑝.4 𝐾 0.4 = 24𝐿0.6 4 + 24𝐿0.8219 0.3 𝐾 0.416 3𝑑.7 𝐾 −0.5 changes in both inputs.6 with the current use of labour at 400 units. Solution: The actual changes in 𝐿 and 𝐾 are: 1 400 = 4 100 −0.6 𝐾 0. 𝐾 = 300. 𝑦𝑂 Example 2: find the approximate change in output for the production function 𝑄 = 5𝐿0. To find the actual change in 𝑧.1863 ∆𝑄 = 𝑄 𝐿 = 502. find 𝑄𝑂 (𝐿 = 400. Partially differentiate the function. To determine the exact change.5 𝑑𝑄 = 28(400)−0.5𝐿0. 𝑦𝑁 − 𝑧 𝑥𝑂 .5 700 0.7 𝐿−0. find the difference between the values of 𝑧𝑁 𝑥 = 502. For the actual change: 700 𝑄 𝐿 = 500.6 𝜕𝐿 𝜕𝑄 = 40 0. 𝑦 = 697 and 𝑧𝑂 𝑥 = 500.5%.1863 − 5506. ) Example 3: using the following Cobb-Douglas Production Function 𝑄 = 40𝐿0. You will learn this in second year.5 into the total derivative equation: 𝑑𝑄 = 𝑑𝑄 = 3𝐿−0.3 𝐾 0. Plan: use the percentage changes in 𝐿 and 𝐾 to find the actual changes (in units).3 𝐾 0. 𝐾 = 300) then subtract it from 𝑄𝑁 𝐿 = 404. Solution: the partial derivatives are: 𝑄𝐿 = 3𝐿−0.6 0.5 𝑑𝑄 = 324.7 𝐾 −0.5 2 −0.6 0. 𝑝. find an approximation for the change in production when labour is increased by 1% and capital is reduced by 0. and 𝐾 decreases by 3 units. 𝐾 = 700 and when 𝐿 increases by 2 units.4 −1. 𝐾 = 697 − 𝑄 𝐿 = 500.6 700 697 0. then substitute everything into the total derivative equation. Theory: to get the actual change in a multivariable function. 𝐾 = 700 ∆𝑄 = 5508.6 𝐾 −0.techniques.5 100 ∆𝐿 = the partial derivatives are: 𝜕𝑄 = 40 0. if you continue with economics.5 when the original levels of the inputs are 𝐿 = 500.5𝐿0. find the exact change in 𝑧.4 𝜕𝐾 Substitute these along with 𝐿 = 400. 𝐾 = 700.6 𝐿0.7 (300)−0.5 𝑑𝐿 + 2. Do this ≈ 5506. 𝑝. 𝑑𝐿 = 2 and 𝑑𝐾 = −3: 𝑑𝑄 = 3 500 −0.5 Substitute these into the total derivative equation: 𝑑𝑄 = 𝜕𝑄 𝜕𝑄 𝑑𝐿 + 𝑑𝐾 𝜕𝐿 𝜕𝐾 ≈ 5508.6 𝐾 −0.364 (3𝑑.8219 = 1.5 150 . 5 −1 = 2. −1 = 0. 3.5 or in coordinate form 𝑥. b) An expression for the total derivative when 𝐿 = 50.7 𝐿0. 𝑦.16 2𝑑. optimisation for multivariable function has first and second order conditions. to find any stationary points. It looks like a saddle: 2 Exercises: 1. Plan: use the FOC by setting both partial derivatives equal to zero. and see if you get ∆𝑄 = 322.calculation. Similar to optimisation in Chapter 5. 𝑦 = −1. For the function 𝑧 = 0.5𝐿0. Use simultaneous equations to solve for 𝑥 and 𝑦: + 2 −1 + 2 2 − −1 −3 2 +5 7. −1. 𝑝. 𝑦 = −1.5 .5𝑦 2 + 𝑥𝑦 + 𝑥 2 − 𝑦 − 3𝑥 + 5 = 0. the 𝑧). c) The approximate change in output when 𝐿 increases by 1unit and 𝐾 increases by 2 units. what is a saddle point? Theory: a saddle point is a stationary point that is neither an overall maximum nor an overall minimum. 𝐾 = 140 c) The approximate change in output when 𝐿 decreases by 2 units and 𝐾 increases by 1 unit.71 𝐾 0. d) Find the actual change in output. The question now becomes: is this stationary point a maximum or a minimum (or something called a saddle point)? This is unknown yet.75𝐾 0. The following is a production function which approximates the production of a cabinet maker: 𝑄 = 53. This is found from the original function: 𝑧 2.3 Determine: a) An expression for the total derivative. 2. 𝜕𝑧 = 0 and 𝜕𝑥 𝜕𝑧 =0 𝜕𝑦 Then solve for 𝑥 and 𝑦 using simultaneous equations.5𝑦 2 + 𝑥𝑦 + 𝑥 2 − 𝑦 − 3𝑥 + 5 151 . d) Find the actual change in output.e. 𝐾 = 70. b) An expression for the total derivative when 𝐿 = 300. (Hint: find the output at the two different 𝐿. (Hint: find the output at the two different 𝐿.5𝑥 2 𝑦 Determine an expression for the total derivative (leaving any unknowns as variables). but firstly. 𝑧 = 2. Theory: the First Order Condition (FOC) is to set all first order partial derivatives to zero.32 Determine: a) An expression for the total derivative. Example 1: find the stationary point(s) for 𝑧 = 0.5 Thus there is a stationary point at 𝑥 = 2. 𝑧 = 2. The following production function approximates the output of a farm: 𝑄 = 4. 2. A “point” in a three dimensional graph also requires the value of the dependent variable (i. 𝐾 combinations) e) The returns to scale. 𝐾 combinations) e) The returns to scale. . then solving for 𝑥 and 𝑦.7 optimisation with many variables Partial derivatives play an important role in finding the optimal points of functions. but with some modifications. Solution: using the FOC: 𝜕𝑧 = 𝑦 + 2𝑥 − 3 = 0 𝜕𝑥 𝜕𝑧 = 𝑦 + 𝑥 − 1 = 0 𝜕𝑦 ′ : 𝑦 = −2𝑥 + 3 Substitute ′ → : −2𝑥 + 3 + 𝑥 − 1 = 0 −2𝑥 + 3 + 𝑥 − 1 = 0 𝑥 = 2 Substitute back into ′ : 𝑦 = −2 2 + 3 = −1 There is one stationary point at 𝑥 = 2.3𝑥𝑦 2 + 0. 𝐼𝑓 ∆ > 0. 34. Theory: to find out whether a stationary point is a maximum. but the following is a new rule. Theory: once the straight second order partial derivatives have been evaluated. The point is a possible minimum if: 𝜕 𝑧 𝜕 𝑧 > 0 𝐴𝑁𝐷 >0 2 𝜕𝑥 𝜕𝑦 2 That is. The point is a saddle point if the signs of the two straight first order partial derivatives do not match. a minimum or a saddle point. Solution: straight second order partial derivatives: 𝜕 2 𝑧 =2>0 𝜕𝑥 2 𝜕 2 𝑧 =1>0 𝜕𝑦 2 𝑦 𝑥 Look closely: in one direction there is a minimum. but it is not an optimal point. 2. the point is a saddle. Written simpler: ∆= 𝑓𝑥𝑥 𝑓𝑦𝑦 − 𝑓𝑥𝑦 2 Example 2 cont: confirm the nature of the point at 2. the delta test must be used: ∆= 𝜕 2 𝑧 𝜕𝑥 2 𝜕 2 𝑧 𝜕 2 𝑧 − 𝜕𝑦 2 𝜕𝑥𝜕𝑦 2 The delta test is: the two straight second partial derivatives multiplied together with the mixed partial second derivative squared. so this is a possible minimum. −1. 𝐼𝑓 ∆ < 0. Since it was thought to be a minimum in this example. subtracted. 𝑦 = −1. the point is what was suspected.5𝑦 2 + 𝑥𝑦 + 𝑥 2 − 𝑦 − 3𝑥 + 5 determine the possible nature of the stationary point at 2. and a possible minimum or possible maximum has been established. This is a stationary point. or if one is equal to zero.𝑧 Plan: evaluate the straight second order partial derivatives at 𝑥 = 2. there is a maximum. 152 . the delta test confirms this. if both second order straight partial derivatives are greater than zero. it is a possible minimum. to make sure it is a minimum or maximum. The point is a possible maximum if: 𝜕 2 𝑧 𝜕 2 𝑧 < 0 𝐴𝑁𝐷 <0 𝜕𝑥 2 𝜕𝑦 2 That is. Solution: the mixed partial second derivative: 𝜕 2 𝑧 =1 𝜕𝑥𝜕𝑦 Apply the delta test: ∆= 2 1 − 1 2 =1>0 Thus the point is what it was thought to be. 3. the Second Order Condition (SOC) is used: Find the value of all straight second order partial derivatives at the point(s) obtained from the FOC. The process used up to now has been similar to that in Chapter 5. then apply the following: 1. 2 2 Both straight second order partial derivatives are greater than zero.5 .5 . but it the other direction. ignore everything. 2. it is a possible maximum. if both second order straight partial derivatives are less than zero. Example 2: for the function from Example 1 𝑧 = 0. 7. and apply the delta test to confirm the nature. get the 𝑧 − 𝑣𝑎𝑙𝑢𝑒𝑠 for these two points (see if you get these same values): 𝑧 5. −55 5 6 𝜕 2 𝑧 𝑥 = 5.29.5 The SOC needs to be used to determine the nature of these two points: 𝜕 2 𝑧 = 2𝑥 − 3 𝜕𝑥 2 𝜕 2 𝑧 = −1 𝜕𝑦 2 Look at each of the two points individually: Point 1: 5. 𝑦 = −10 = 2 5 − 3 = 7 > 0 𝜕𝑥 2 𝜕 2 𝑧 𝑥 = 5. determine the possible nature of the stationary point. 𝑦 = −10 Point 2: 𝑥 = −3. −2. For 𝑥 = 5: For 𝑥 = −3: 𝑦 = − 5 − 5 = −10 𝑦 = − −3 − 5 = −2 =9−1=8>0 Since ∆ > 0. so this is a saddle point. the delta test is used: 𝜕 2 𝑧 ∆= 𝜕𝑥 2 𝜕 2 𝑧 𝜕 2 𝑧 − 𝜕𝑦 2 𝜕𝑥𝜕𝑦 2 The mixed second order partial derivative is: 𝜕 2 𝑧 = −1 𝜕𝑥𝜕𝑦 Put everything into the delta formula: ∆= −9 −1 − −1 2 𝑥 = 5 𝑂𝑅 𝑥 = −3 These are the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 of two stationary points. since the SOC gave a possible 153 . −2 = 29. Find the mixed second order partial derivative. 𝑦 = −2 = 2 −3 − 3 = −9 < 0 𝜕𝑥 2 𝜕 2 𝑧 𝑥 = −3. so this point is a possible maximum. the SOC determines this. 𝑦 = −2 Lastly. Solve for the variables. Solution: find the FOC: 𝜕𝑧 = 𝑥 2 − 3𝑥 − 𝑦 − 20 = 0 𝜕𝑥 𝜕𝑧 = −𝑦 − 𝑥 − 5 = 0 𝜕𝑦 Use simultaneous equations: ′ : 𝑦 = −𝑥 − 5 Substitute′ → : 𝑥 2 − 3𝑥 − (−𝑥 − 5) − 20 = 0 𝑥 2 − 3𝑥 + 𝑥 + 5 − 20 = 0 𝑥 2 − 2𝑥 − 15 = 0 Solve for 𝑥 using the Quadratic Formula: −(−2) ± (−2)2 − 4 1 (−15) 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = 2(1) 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = 2 ± 64 2 Thus the two stationary points are: Point 1: 𝑥 = 5. From their signs.Theory: the delta test does NOT tell us whether a point is a maximum or a minimum. Example 3: determine the nature of all stationary points for the function 1 1 3 𝑧 = 𝑥 3 − 𝑦 2 − 𝑥 2 − 𝑥𝑦 − 20𝑥 − 5𝑦 − 10 3 2 2 Plan: using the FOC. this confirms what was suspected from the SOC. The 𝑦 − 𝑣𝑎𝑙𝑢𝑒s are obtained from ′. Next. find the straight second order partial derivatives and evaluate them at the stationary point(s). The delta test just proves or disproves what was suspected from the SOC. Point 2: −3. find the partial derivatives and set them to zero. −10 = −55 5 6 𝑧 −3.5 𝜕 2 𝑧 𝑥 = −3. 𝑦 = −2 = −1 < 0 𝜕𝑦 2 Both signs are less than zero. −10. 𝑦 = −10 = −1 < 0 𝜕𝑦 2 The two signs do not match. To confirm this. 5 −1− 5 2 𝑒 0. 2 2 = 𝑒 −1+ 2 2 − 5 5 To find the nature of these points.5𝑥 Simplify: 2𝑥𝑒 + 𝑥 𝑒 − 0. The 𝑒 rule will have to be used.5𝑥 𝑦 (𝑒 is the number.maximum. 𝑦 = Point 2: 𝑥 = −1 + 5 . But each of these points also needs a 𝑦 − 𝑣𝑎𝑙𝑢𝑒 which is found from either of the simultaneous equations (′ is easiest as 𝑦 is already isolated): 5 𝑒 𝑦 𝑥1 = −1 − = 2 5 𝑒 = 2 0. −2. Do it and you should get: 𝑒 0.5 −1+ 5 2 ≈ 0. Determine if any of the stationary points are possible maxima/minima.5𝑥 =0 2 5 𝑒 .5𝑒 0.25𝑒 = 0 Factorise: 𝑒 𝑥 2𝑥 + 𝑥 2 − 0. set them equal to zero and solve using simultaneous equations. or 2. use the SOC: 𝜕 2 𝑧 = 2𝑒 𝑥 + 2𝑥𝑒 𝑥 + 2𝑥𝑒 𝑥 + 𝑥 2 𝑒 𝑥 − 0.25 = 0 Solve using quadratics: 𝑥 𝑥 2 𝑥 𝑥 𝑧1 −1 − 2 0.29. substitute each of these pairs of 𝑥 and 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 into the original equation and simplify. Use the delta test to make sure of their nature.25) 2(1) −2 ± 5 2 𝑃𝑜𝑖𝑛𝑡 2: −3.1734 𝑦 𝑥2 = −1 + 2 ≈ 0.5 (𝑚𝑎𝑥𝑖𝑚𝑢𝑚) Example 4: determine the nature of any stationary points for the function 𝑧 = 𝑥 2 𝑒 𝑥 + 𝑦 2 − 𝑒 0.5𝑥 𝑦 𝜕𝑥 2 𝜕 2 𝑧 =2 𝜕𝑦 2 154 .118 and 𝑥2 ≈ 0. 𝑥 2 + 2𝑥 − 0.5𝑥 𝑦 = 0 𝜕𝑥 𝜕𝑧 = 2𝑦 − 𝑒 0. 𝑒 = 0 which is impossible.118. the delta test confirms this. To find the nature of these: find the straight second order partial derivatives and evaluate them at the points found using the FOC. Plan: use the FOC to find any stationary points: find both partial derivatives. −55 5 6 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = (𝑠𝑎𝑑𝑑𝑙𝑒 𝑝𝑜𝑖𝑛𝑡) −𝑏 ± 𝑏 2 − 4𝑎𝑐 2𝑎 −2 ± 22 − 4 1 (−0.5𝑥 2 Substitute → : 2𝑥𝑒 𝑥 + 𝑥 2 𝑒 𝑥 − 0. either: 1.5𝑥 = 0 𝜕𝑦 Solve these simultaneously: ′ : 𝑦 = ′ 𝑥1 = −1 − 5/2 𝑜𝑟 𝑥2 = −1 + 5/2 This gives two stationary points at the two 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 above with approximations 𝑥1 ≈ −2.5304 So the two stationary points are: Point 1: 𝑥 = −1 − 5/2. 𝑃𝑜𝑖𝑛𝑡 1: 5. not a variable).5 −1+ 2 To obtain the 𝑧 − 𝑣𝑎𝑙𝑢𝑒𝑠.25 = 0 For the left side to be equal to zero. −10.5 −1+ 5 2 = 𝑒 −1− 2 2 + 5 5 𝑧2 −1 + 5 𝑒 .5 −1− 5 2 2 5 2 = 𝑒 0.5 −1− 5 2 2 0. 2 0.5𝑒 0.25𝑒 0. 𝑦 2 𝑒 0. Solution: the partial derivatives are: 𝜕𝑧 = 2𝑥𝑒 𝑥 + 𝑥 2 𝑒 𝑥 − 0. 118 + 2 0.5304) 𝜕 2 𝑧 ≈ 2𝑒 0.1734 𝜕 2 𝑧 ≈ 2𝑒 −2. the delta test needs to be evaluated at this point. The mixed partial derivative is: 𝜕 2 𝑧 = −0.254 < 0 𝜕 2 𝑧 =2>0 𝜕𝑦 2 The signs do not match up so this is a saddle point. then more decimal places or the exact answer would have to be used to be 100% sure that the SOC is really positive/negative.118 𝜕𝑥 2 + 0.5𝑦 − 2.118 − 0.033 > 0 This proves what was suspected from the SOC.5304 ≈ 5. and if there is a maximum. If the approximation gives a SOC result quite a bit above/below zero.118 − 0.5 −1+ 5 2 2 . 𝑎) 𝑧 = 4𝑥𝑦 − 3𝑥 2 − 5𝑦 2 + 7 𝑏) 𝑧 = 4𝑥 + 5𝑦 − 3𝑥 2 − 2𝑦 2 𝑐) 𝑧 = 𝑥 3 + 𝑦 3 + 18 𝑑) 𝑓 𝑥.118 0. 𝑦2 ≈ 0.5𝑥 𝜕𝑥𝜕𝑦 Evaluated at Point 2 (𝑥2 ≈ 0.8𝑦 2 𝑒) 𝑧 = −4𝑥 4 − 5𝑦 4 + 32𝑥 + 40𝑦 𝑓) 𝑓 𝑥. not the precise value at that point.118 𝑒 −2.8 economic applications Theory: the total revenue from two different goods is simply the sum of the revenues of the individual goods. 2 0.4𝐿 Determine any stationary points.5304 𝜕𝑥𝜕𝑦 Apply the delta test: 𝜕 2 𝑧 ∆= 𝜕𝑥 2 𝜕 2 𝑧 𝜕 2 𝑧 − 𝜕𝑦 2 𝜕𝑥𝜕𝑦 2 2 0. 0.118 + 2 −2.3𝐾 − 0. This was a very difficult example.5304 7.657 > 0 𝜕 2 𝑧 =2>0 𝜕𝑦 2 Since the signs match up and both are greater than zero. Example 1: a direct computer manufacturer/retailer sells two different types of ∆≈ 2.25𝑒 0.118 2 𝑒 0. the reason will be clear later): Point 1: 𝑥1 ≈ −2. Thus: 155 . 𝑦2 ≈ 0. and determine their nature.5304): 𝜕 2 𝑧 ≈ −0.5𝑒 0.118 Saddle point: −1 − 5 𝑒 2 . 5 0.118.118 + 2 0.118.5𝑥 + 3. if the SOC was close to zero (around 0.5(0. Since the SOC gave a possible minimum.25𝑒 0.118) ≈ −0. However.4 𝐿0.118.Evaluate each of the stationary points at each of the two partial derivatives (approximate values are used as it is very difficult to work with the exact values. the delta test proves this is true. 𝑦1 ≈ 0. Exercises: 1. 𝑦 = 2.2𝑥 2 − 1.5 −2. but look over it again and write out what is being found and why in each of the steps before trying the following exercises. then that positive/negative sign can be taken with certainty.657 2 − −0.118 𝑒 0.5 ≈ 2.5 −1− 2 2 . 𝑦 = 3𝑥𝑒 𝑥+𝑦 + 2𝑦 + 𝑥 2. 𝑒 −1+ 2 2 − 5 5 Note: the only reason approximations were used (to at least three decimal places!) when evaluating the SOC and the delta test was that the sign of the value was needed.1734 ≈ −0.118 𝑒 −2.118 𝜕𝑥 2 + 2 −2. Point 2: (𝑥2 ≈ 0. this point is a possible minimum. 𝑒 −1− 2 2 + 5 5 Minimum: −1 + 5 𝑒 .118 2 𝑒 −2.01).118 + −2.118 𝑒 0. Determine if the following functions have any stationary points.5𝑒 0. To prove that this point is a minimum. Given the profit function: 𝜋 = 𝐾 0.5 − 0. 45𝐷2 + 0.45𝐷2 − 0.3𝐿2 + 0. output should be set at 1818 laptops and 354 desktops. 𝐷) combination back into the profit function to determine the maximum profit. Substitute the (𝐿. laptops (𝐿) and desktops (𝐷).9 − −0.3𝐿𝐷 + 560 The 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 function is the sum of 𝑃 × 𝑄 for each of the two goods: 𝑇𝑅 = 1197𝐿 + 863.3𝐿 = 0 864 − 3591 + 1. determine the profit maximising level of capital and labour.3𝐿 = 0 ′ : 0.354 = 1. substitute 𝐿 = 1458 and 𝐷 = 1074 into the profit function: 𝜋 1818. the SOC must be used: 𝜋𝐿𝐿 = −0. They sell these computers at the prices: 𝑃𝐿 = $1197.10𝐷 Thus the profit function is: 𝜋 = 1197𝐿 + 864𝐷 − 0.45𝐷2 + 0. the point is a maximum).54 − 0. To determine the actual profit attainable.9 3990 − 2𝐿 − 0.3𝐷 = 1197 − 0. c) The maximum profit possible.00 𝑃𝐷 = $864. and also find the maximum profit.6𝐿 : 𝐷 = 3990 − 2𝐿 Substitute ′′ → : 864 − 0.240.3 1818 354 − 560 To find stationary points.computers.45 354 2 − 0. Solution: the 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 function is: 𝑇𝐶 = 0.45 > 0 Since the delta test is positive.09 = 0. Example 2: a company faces a demand function of the form 𝑃𝑑 = 1000 + 20 𝑄𝑑 2 − 0. use the FOC: 𝜋𝐿 = 1197 − 0. it confirms what was suspected from the SOC (i.9 < 0 Both straight second order partial derivatives are negative so this is a possible maximum.3𝐷 = 0 𝜋𝐷 = 864 − 0.3 1818 𝜋 1818.6𝐿 − 0.e.6 < 0 𝜋𝐾𝐾 = −0.3 ∆= 0. 156 .6 −0.3𝐿2 − 0.00 The company pays $560 in rent and face a variable cost function: 𝑉𝐶 = 0.9𝐷 − 0.24million.3𝐿𝐷 − 560 𝐿 = 1818 Substitute this into ′′ to give: 𝐷 = 3990 − 2 1818 = 354 To determine the nature of this stationary point.8𝐿 − 0.3𝐿𝐷 Determine: a) The profit function b) The optimal levels of 𝐿 and 𝐷 (prove this).354 = 1197 1818 + 864 354 − 0. the delta test must be used: 𝜋𝐿𝐾 = −0.3 ∆= 𝜋𝐿𝐿 𝜋𝐾𝐾 − 𝜋𝐿𝐾 2 2 ∆= −0.3𝐿 = 0 ′′ and a production function of the form: 𝑄 = −𝐿2 − 2𝐾 2 + 𝐿𝐾 + 5𝐾 + 10𝐿 If each labour unit 𝐿 costs $20 and each capital unit 𝐾 costs $30.3𝐿2 + 0.441 Thus to attain a maximum profit of approximately $1. and fixed costs are $600. Determine the profit function using: 𝜋 = 𝑇𝑅 − 𝑇𝐶 Find any stationary points using the FOC then prove their nature using the SOC and the delta test. Plan: find the 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 function using 𝑇𝐶 = 𝑉𝐶 + 𝐹𝐶 Then find the 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 function (which just comes from 𝑃 × 𝑄 for each of the two computers). To prove it. use the FOC: 𝜋𝐿 = −40𝐿 + 20𝐾 + 180 = 0 𝜋𝐾 = −80𝐾 + 20𝐿 + 70 = 0 Rearrange 𝜋𝐿 to isolate 𝐾: 𝐾 = 40𝐿 − 180 = 2𝐿 − 9 20 16 79 + 180 7 14 6 𝜋 = 987 ≈ 987. use the delta test. Solution: to find total revenue: 𝑇𝑅 = 𝑃 ∙ 𝑄 = 1000 + 20 ∙ 𝑄 𝑄 Use the SOC to determine the nature of this stationary point: 𝜋𝐿𝐿 = −40 < 0 𝜋𝐾𝐾 = −80 < 0 Since both are less than zero. and the prices for the two crops are: 𝑃𝑊 = $225/𝑡𝑜𝑛 𝑃𝐵 = $190/𝑡𝑜𝑛 With variable costs being approximated by: 𝑉𝐶 = 0. If the rent on the farm is $4. substitute in for 𝑄: 𝑇𝑅 = 1000 + 20 −𝐿 − 2𝐾 + 𝐿𝐾 + 5𝐾 + 10𝐿 𝑇𝑅 = 1000 − 20𝐿 − 40𝐾 + 20𝐿𝐾 + 100𝐾 + 200𝐿 2 2 ∆= −40 −80 − 20 = 2800 > 0 This proves that what was suspected from the SOC 2 2 is actually correct.Plan: find the 𝑇𝐶 function using the fixed costs and variable costs. Substitute into 𝜋𝐾 : −80 2𝐿 − 9 + 20𝐿 + 70 = 0 −160𝐿 + 720 + 20𝐿 + 70 = 0 𝐿 = 79 14 79 16 −9= 14 7 Substitute this into the isolated 𝐾 equation: 𝐾 = 2𝐿 − 9 = 2 Thus there is a stationary point at 𝐾 = 16 79 . b) The optimal levels of 𝐿 and 𝐾. A shipyard uses two inputs. determine: a) The profit function in terms of 𝐿 and 𝐾. 𝐿 = 7 14 157 . using the demand function and the production function. then substitute back to get the maximum profit.5𝑊 2 + 0. this stationary point is a possible maximum. The mixed second order partial derivative: 𝜋𝐿𝐾 = 20 ∆= 𝜋𝐿𝐿 𝜋𝐾𝐾 − 𝜋𝐿𝐾 2 2 𝑇𝑅 = 1000 + 20𝑄 To find 𝑇𝑅 in terms of labour and capital.5𝐿2 − 2𝐾 2 + 2𝐿𝐾 + 8𝐾 + 12𝐿 If each 𝐿 costs $25 and each 𝐾 costs $40.5𝑄𝐷 And the firm has a service production function of the form 𝑄 = −0. A farmer in a perfectly competitive industry grows wheat and barley. 𝑝. c) The maximum profit attainable. To make sure it is a maximum.6𝑊𝐵 Determine: a) The profit function b) The optimal levels of production c) The maximum profit attainable. 𝐿 and 𝐾. The SOC gave a possible maximum. and the delta test confirmed this. Then find the 𝑇𝑅 function. If the demand for their services are approximated by 2500 𝑃𝐷 = + 50 0. To find the profit at 𝐿 = 14 and 𝐾 = into the profit equation: 𝜋 = 400 − 20𝐿2 − 40𝐾 2 + 20𝐿𝐾 + 70𝐾 + 180𝐿 𝜋 = 400 − 20 79 14 2 79 16 7 Total costs come from the cost per unit costs and the fixed costs: 𝑇𝐶 = 20𝐿 + 30𝐾 + 600 The profit function is: 𝜋 = 𝑇𝑅 − 𝑇𝐶 𝜋 = 1000 − 20𝐿2 − 40𝐾 2 + 20𝐿𝐾 + 100𝐾 + 200𝐿 − 20𝐿 + 30𝐾 + 600 substitute − 40 16 7 2 + 20 79 14 16 7 + 70 𝜋 = 400 − 20𝐿2 − 40𝐾 2 + 20𝐿𝐾 + 70𝐾 + 180𝐿 To find any stationary point.86 2𝑑.000 per year. 7 Exercises: 1. Use 𝜋 = 𝑇𝑅 − 𝑇𝐶 to determine the maximum profit levels of 𝐿 and 𝐾. 2.4𝐵2 + 0. 𝑖𝑚𝑝𝑜𝑟𝑡 = $300 𝑃𝐿. Find both first order partial derivatives for: 𝑎) 𝑧 = 2𝑥𝑦 + 3𝑥 − 𝑦 + 4𝑥 2 𝑦 3 𝑏) 𝑓 𝑥.7 + 15𝑚𝑟 𝑑) 𝑧 = 18𝑥 0. determine: a) The two total revenue functions b) The total cost function c) The profit function d) The optimal sales of the two types of televisions. 𝑦 = 12𝑥 4 + 𝑥𝑒 𝑦 − 12 𝑐) 𝑝 = 15𝑟 0. 𝑦𝑂 To find an optimal point: 1. 𝛼 and 𝛽 are all positive constants. 𝑦) The Cobb-Douglas Production Function has the general form: 𝑄 = 𝐴𝐿𝛼 𝐾𝛽 where the values of 𝐴. Returns to scale are defined as the extent output changes when inputs are changed by a certain amount. 𝑂𝑅 𝑧𝑥𝑦 . 𝑦) 𝑜𝑟 𝑧(𝑥. 𝛼 + 𝛽 < 1 then there are decreasing returns to scale. To find out whether a stationary point is a maximum.4 𝑦 0. 𝐼𝑓 ∆ < 0. 𝛼 + 𝛽 = 1 then there are constant returns to scale. no matter which way they are found. Multivariable functions means that a dependent variable is determined by two or more other variables. 2.𝑖𝑚𝑝𝑜𝑟𝑡 = $500 If the domestic demand functions for each of the two types is approximated by: 2𝐿2 𝑃𝑀 = 𝑀 − + 𝐿 + 1300 𝑀 2 5𝑀 𝑃𝐿 = 𝐿 − + 2𝑀 + 1000 𝐿 3. If the cost to rent the shop is $500. A saddle point is a stationary point that is neither an overall maximum nor an overall minimum. to find any stationary points.3. 𝛼 + 𝛽 > 1 then there are increasing returns to scale. 𝑦 = 3𝑥 − 4𝑦 2 𝑥 2 − 𝑦 3 𝑔) 𝑧 = 1 − 𝑥 − 𝑦 12𝑥𝑦 2 − 𝑦 ) 𝑧 = 1 − 𝑥𝑦 𝑖) 𝑧 = ln 1 − 𝑥 2 𝑦 2 + 𝑥 𝑓) 5 𝑥𝑦 158 . e) The two domestic prices of the televisions. 𝜕 2 𝑧 𝜕 2 𝑧 . 𝑧𝑦𝑥 𝜕𝑥𝜕𝑦 𝜕𝑦𝜕𝑥 Mixed partial derivatives are the same. The total revenue from two different goods is simply the sum of the revenues of the individual goods.8 𝑥 2 𝑦 4 − 3𝑥𝑦 + 𝑥 2 𝑒) 𝑧 = 𝑥 − 𝑦 𝑥 + 𝑦 4 𝑓 𝑥. differentiate the function with respect to that variable while treating all other variables as constants. For the Cobb-Douglas Production Function 𝑄 = 𝐴𝐿𝛼 𝐾𝛽 1. the greater the increase in output from an extra unit of that input.000.3 𝑚0. then differentiating with respect to the other variable. 𝜕 2 𝑧 𝜕 2 𝑧 . chapter seven questions 1. It looks like a saddle. chapter seven summary A firm imports two types of televisions from Malaysia [plasma (𝑀) and LED (𝐿)] at a cost of: 𝑃𝑀. ignore everything. The second order partial derivative is simply partially differentiating the original function twice. Use the delta test to make sure the point is what is suspected: 2 𝜕 2 𝑧 𝜕 2 𝑧 𝜕 2 𝑧 ∆= − 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑥𝜕𝑦 𝐼𝑓 ∆ > 0. the Second Order Condition (SOC) is used: a) The point is a possible maximum if: 𝜕 2 𝑧 𝜕 2 𝑧 < 0 𝐴𝑁𝐷 <0 𝜕𝑥 2 𝜕𝑦 2 b) The point is a possible minimum if: 𝜕 2 𝑧 𝜕 2 𝑧 > 0 𝐴𝑁𝐷 >0 2 𝜕𝑥 𝜕𝑦 2 c) The point is a saddle point if the signs of the two straight first order partial derivatives do not match. or if one is equal to zero. 𝑧𝑦 𝜕𝑥 𝜕𝑦 To partially differentiate a multivariable function with respect to a single variable. 𝑧𝑦𝑦 𝜕𝑥 2 𝜕𝑦 2 Mixed partial derivatives are found by differentiating with respect to one variable first. 𝑂𝑅 𝑧𝑥 . The higher the value of the marginal product. 𝑂𝑅 𝑧𝑥𝑥 . 𝑧 = 𝑓(𝑥. a minimum or a saddle point. The First Order Condition (FOC) is: 𝜕𝑧 𝜕𝑧 = 0 𝐴𝑁𝐷 =0 𝜕𝑥 𝜕𝑦 Then solve for 𝑥 and 𝑦 using simultaneous equations. the point is what was suspected. Partial differentiation is finding the rate at which the dependent variable changes when an independent variable changes assuming all other variables are held constant. the point is a saddle. 2. 𝜕𝑧 𝜕𝑧 . 𝑦𝑁 − 𝑧 𝑥𝑂 . f) The maximum profit attainable. The total derivative equation is: 𝜕𝑧 𝜕𝑧 𝑑𝑧 = 𝑑𝑥 + 𝑑𝑦 𝜕𝑥 𝜕𝑦 To find the actual change in a function: ∆𝑧 = 𝑧 𝑥𝑁 . find any stationary points and determine their nature: 𝑎) 𝑧 = 2. f) The maximum attainable profit. Is maximum profit a possibility? Use the total derivative function to find the approximate change in 𝑧 for the function 𝜋 = 𝐾 0. d) The price at the maximum (if the maximum exists). Also determine the exact change in 𝜋. 𝑦 = ln 2𝑥 2 + 𝑦 3 ) 𝑧 = ln 13 − 𝑥𝑦 + 𝑥 2 + 15 𝑥 − 𝑦 2 For the following functions. determine: a) The total revenue function in terms of 𝐿 and 𝐾. e) The quantity sold at the maximum. d) The number of 𝑅 and 𝐴 that need to be sold to maximise profit. b) the exact change in output. d) The levels of 𝐿 and 𝐾 for maximum profit.3𝑘 2 𝑔 − 4𝑔𝑘 + 15 when 𝑔 is decreased by 0.e.2 from 154 and 𝐾 is increased by 0.01 Determine the level of utilisation of the two inputs at maximum profit. Approximate the change in the objective function 𝑅 𝑅 𝑔.1 from 310. determine: a) The two revenue functions. televisions (𝑇) and radios (𝑅).5 Find a) an expression for the approximate change in output if 𝐿 is decreased by 0.4 𝐿0. and compare the two answers.4𝐾 when 𝐿 is increased by 0. 𝑘 = 0.4𝑥 2 − 1. 11. 𝑗) 𝑧 = 𝑒 4𝑥−𝑦 𝑥 3 + 4𝑥𝑦 − 3 𝑘) 𝑧 = 𝑥 2 𝑦 2 − 1 𝑒 4𝑥𝑦 9𝑥𝑦 𝑙) 𝑧 = + ln 15 − 𝑥 + 15 ln 4 − 2 𝑥𝑦 − 1 1 − ln 1 + 𝑥𝑦 𝑚) 𝑧 = 1 − 𝑒 𝑥𝑦 Find all four second order partial derivatives for: 𝑎) 𝑧 = 𝑥 2 + 𝑦 2 − 3𝑥𝑦 − 𝑥 + 4𝑦 + 12 𝑏) 𝑧 = 2𝑥𝑦 2 + 𝑦𝑥 2 − 3 𝑐) 𝑧 = 𝑥 3 𝑦 4 + 2𝑥𝑦 − 4𝑥 − 6𝑦 + 8 𝑑) 𝑓 𝑥.7 𝐿0. b) The profit function. Given the profit function: 𝜋 = 𝐾 0. 𝑒) 𝑧 = −2𝑥 4 − 3𝑦 4 + 10𝑥 + 20𝑦 𝑓) 𝑓 𝑥. b) The total cost function. and also the maximum profit.2𝑥𝑦 − 3. If the company has a demand function for the final product as: 1 𝑃𝑑 = + 20 𝑄 The firm also has a production function of the form: 𝑄 = 𝐾 0. g) Do these prices make sense? 3. e) The maximum profit attainable. and if fixed costs are $550. 𝑦 = 2𝑥𝑦 2 4 + 18𝑥𝑦 2 − 13𝑥𝑦 𝑒) 𝑧 = 𝑒 2𝑥 2 +𝑦 2 𝑄 = 8𝐾 0.2𝑥𝑦 2 4. 5.2𝑥 + 2.02𝐾𝐿 − 0. 8.2. A firm imports two similar products from Brazil at fixed costs: refrigerators (𝑅) cost $650 and air conditioners (𝐴) cost $350. 10.2% (i. 9.5𝐿𝐾 + 8𝐿 + 12𝐾 If labour costs $50 and capital costs $35. c) The profit function. A factory producing portable music players has a demand function of the form 1100 𝑃𝐷 = + 45 𝑄𝐷 with a production function of the form: 𝑄 = −2𝐿2 − 3𝐾 2 + 2. given 𝐿 was originally 200 and 𝐾 was 180.4𝑦 − 2. 12. 7.3𝐿 − 0. only in terms of 𝐿 and 𝐾).5 − 0. and sells them on the domestic market.5 𝐿0. For the production function 159 . Also determine the exact change in 𝑅. c) The profit function. 𝑓) 𝑧 = 𝑒 𝑥 +𝑦 − 2𝑥𝑦 2 𝑔) 𝑓 𝑥. 𝑦 = 1.4𝐿 Determine any stationary points and their nature. c) The levels of 𝑇 and 𝑅 for profit maximisation. and the approximate cost function is 𝑇𝐶 = 3𝑇 2 + 4𝑅2 + 400 + 𝑅𝑇 The two goods sell on the domestic market at the following prices: 𝑃𝑇 = 300 𝑃𝑅 = 150 Determine: a) The total revenue function. given the original levels of 𝑔 and 𝑘 being 78 and 104 respectively. 6. Given the domestic demand for refrigerators is: 4𝐴2 𝑃𝑅 = 𝑅 − + 𝐴 + 1400 𝑅 And the domestic demand for air conditioners is: 6𝑅2 𝑃𝐴 = 𝐴 − + 3𝑅 + 1200 𝐴 With fixed costs amounting to $10. The costs to the firm are related to the amount they import. A profit function has the form 𝜋 = 4𝐾 + 7𝐿 − 12𝐾 2 − 14𝐿2 − 0.3𝑥 2 − 5.9. b) The total cost function in terms of 𝐿 and 𝐾.5 and 𝑘 is increased by 0.1𝑔2 𝑘 + 0. 𝑦 = 5𝑦𝑒 𝑥−𝑦 + 3𝑦 − 8𝑥 A pharmaceutical company uses capital equipment and labour as inputs into the manufacture of paracetamol. A firm imports two goods from China.4 + 0.4 And the cost function is approximated by: 𝑇𝐶 = 5𝐾 + 10𝐿 + 20 Determine in terms of 𝐾 and 𝐿: a) The total revenue function b) The profit function c) When profit is at a maximum.4% and 𝐾 is increased by 0.5𝑦 2 + 1 𝑏) 𝑧 = 2𝑥 + 2𝑦 − 3𝑥 2 − 3𝑦 2 + 2𝑥𝑦 𝑐) 𝑧 = 8𝑥 3 + 3𝑦 3 − 𝑥𝑦 + 500 𝑑) 𝑓 𝑥. e) The domestic prices of the two goods. d) The maximum profit attainable.000.8𝑦 2 + 1.4 𝐿0.3𝐾 − 0. 5 8.6 8.4 8.2 8.3 8.7 Index Numbers and Averages Series and Sums Simple Interest Compound Interest Annual Interest Rates Net Present Value Internal Rate of Return 161 163 167 168 170 172 174 177 177 Chapter Eight Summary Chapter Eight Questions 160 .Chapter 8 Financial Mathematics Mathematics and money 8.1 8. determine the value of the index in 2009. which is irrelevant.4 157.1 index numbers and averages In this context.9 164.20 × 100 731.20. You will have to know how to manipulate an index to extract useful information. The base year is given an index value of 100.g. the Consumer Price Index provides a value of the price level of a basket of goods in a certain year in terms of that same basket of goods in a given base year (e.53 Plan: use the percentage change formula %∆2002→2003 = Solution: substitute the index values: %∆2002→2003 = %∆2002→2003 = 6. Example 1: an index is to be constructed with the year 2000 as the base year (with a value of 100). usually you will only have access to an index and not the raw data. index numbers represent values in terms of a base value.17 This is not a difficult idea. Plan: use the index formula 𝐼𝑡 = 𝑃𝑡 × 100 𝑃0 1091.3 119. which in most cases.8.14% If the difference between the two years was taken. in finance. except that it is always to a base year. determine the percentage change in crude oil prices between 2002 and 2003 𝑌𝑒𝑎𝑟 2000 2001 2002 2003 𝐶𝑟𝑢𝑑𝑒 𝑂𝑖𝑙 𝐼𝑛𝑑𝑒𝑥 100 105. If gold prices in 2000 were $731. Theory: to obtain the percentage difference between two time periods in an index. Example 2: given the following data.1 Solution: substitute all the known values: 𝐼2009 = 𝐼2009 = 149.6 112. Theory: to find the indexed value of a given commodity. use the formula: 𝐼𝑛𝑑𝑒𝑥𝑡 = 𝑉𝑎𝑙𝑢𝑒𝑡 × 100 𝑉𝑎𝑙𝑢𝑒0 You might be thinking why not just take the two indices from one another as they are already in percentage form.53 and in 2009 were $1091.8 168.9%. However.3 × 100 112.3 This is very similar to finding a percentage.2 𝐼2003 − 𝐼2002 × 100 𝐼2002 119. the year 2000). but this is in terms of the price in 2000. Theory: in a finance sense. index numbers are not related to indices learnt in Chapter 1. determine the percentage change in prices between: a) 2005 and 2006 b) 2008 and 2009 𝑌𝑒𝑎𝑟 2005 2006 2007 2008 2009 𝐶𝑃𝐼 149. For example. This is correct but the difference between two index entries gives the percentage difference between the two years in terms of the base year. Example 3: given the following Consumer Price Index values. is useless.1 154. use the formula: %∆𝑡→𝑡+1 = 𝐼𝑡+1 − 𝐼𝑡 × 100 𝐼𝑡 %∆𝑡→𝑡+1 = Plan: use the percentage change formula 𝐼𝑡+1 − 𝐼𝑡 × 100 𝐼𝑡 161 . the value would be 6.2 − 112. 88% from some base amount.88. the CPI grew by 4. where the arithmetic average is finding only the average percentage relative to the first year.3 − 123.88% p.Solution: a) Substitute 𝐼𝑡+1 = 154. it is 104. so will not be explained in depth.3 Plan: use the percentage change formula: %∆𝑡→𝑡+𝑛 = 𝐼𝑡+𝑛 − 𝐼𝑡 × 100 𝐼𝑡 𝐴𝑟𝑖𝑡𝑚𝑒𝑡𝑖𝑐 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = For the geometric average: 𝐺𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 = 4 21 +1 −1 100 = 1.3 and 𝐼𝑡 = 123.69% 2𝑑.a. Solution: for the arithmetic average: 21 = 5. rather than the first year. The concept of the geometric average in this last example. given the following index entries for the two years: 𝐼2005 = 123.2. The average growth rate in gold prices over these four years can be found in one of two ways.1 − 164.00% 2𝑑.2 %∆= × 100 = 14. Solution: substitute 𝐼𝑡+𝑛 = 141. At the end of the fourth year. The concept of the arithmetic average in this last example is the simplest kind of average.4. say 100. however.1 relative to the most recent year.048809 Thus a geometric average of about 4. During the second year. the CPI grows another 4. at the end of the second year. So after one year. Plan: apply the two formulae for averages.10).88% relative to the previous year (i. there was growth of 14.2 𝐼2009 = 141. It assumes that the percentage change is compounded. 123. Theory: the arithmetic average is the traditional method of finding an average: 𝐴𝐴 = 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑦𝑒𝑎𝑟𝑠 The geometric average is a concept similar to compound interest (see later in Chapter).8 = 2. the CPI grows again at 4. 𝑝.4 − 149.1 and 𝐼𝑡 = 164. and solve: 141. The geometric average is similar to compound interest. 1.e.25% 4 = 3. Example 4: determine the percentage change in the growth of gold prices from 2005 to 2009. it is now growing from a base of 104. 𝑝. 𝑝.1 × 100 149. In the first year. Compare the geometric average to the arithmetic average. Example 5: Find the arithmetic and geometric average percentage change per year in the CPI given the percentage change over four years was 21%. and this gives.2 In the example above.8 %∆2008→2009 = × 100 164. the value of the CPI is 121. The geometric average is more realistic as it assumes growth is 162 .88. or 21% higher over the four years. however. will be. b) Substitute 𝐼𝑡+1 = 168.8 168. but in reverse.55% 2𝑑. thus to reverse the compounding: 𝐺𝐴 = 𝑛 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒 +1 −1 100 With 𝑛 being the number of years. it has a value of 115.1 %∆2005→2006 = 154.69% over the four years from 2005 to 2009. so at the end of the third year.4 and 𝐼𝑡 = 149.048809 − 1 = 0. a value of 110. During the third year.88%. c) The geometric average percentage change per annum from 1994 to 1999.5 1965 15.1 1962 − 1963 − 1964 14.7 1969 17. 38. Plan: use the percentage change formula: 𝐼𝑡+1 − 𝐼𝑡 %∆𝑡→𝑡+1 = × 100 𝐼𝑡 Then use the arithmetic average formula: 𝐴𝐴 = 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑦𝑒𝑎𝑟𝑠 𝐶𝑃𝐼 109.03% 5 For the geometric average over the five years: 𝐺𝐴 = 5 10. 8.1 1966 15. 16. you will see a common feature: 𝟓. 𝑌𝑒𝑎𝑟 1993 1994 1995 1996 1997 1998 1999 Determine a) the percentage change in the CPI from 1994 to 1999. 𝑝. 4. 45 2. f) The geometric average from 1961 to 1964. 𝑝. 25. g) The arithmetic average from 1960 to 1969. 8. 163 . 23. 31. 50. A geometric series has a constant ratio between terms.2 series and sums Series are extremely common in business (especially accounting and investments) and simply put. they are the repetition of a certain mathematical function. d) The percentage change from 1966 to 1968. e) The arithmetic average from 1961 to 1964.1 Determine: a) The percentage change from 1960 to 1961.1 122. 8.5 111.6 %∆= × 100 = 10.0195 4𝑑.9 Exercises: 1. 35. 10.6 For the arithmetic average over the five years: 𝐴𝐴 = 10. 2. 4. h) The geometric average from 1960 to 1969. 41 +6 Solution: for the percentage change: 122.8 1961 14.75 Theory: An arithmetic series has a constant difference between terms.6 − 119. 12. 29. 6. c) The percentage change from 1960 to 1969. 32. 64 100. 17. b) The percentage change from 1961 to 1964.5.13 100 +1 −1 = 0.8 − 121. 111.13% 2𝑑. 10.9 − 111. 17. Given the following data table: 𝑌𝑒𝑎𝑟 𝐶𝑃𝐼 1960 13. 24. Looking at each of the following arithmetic progressions. Thus a percentage growth per year of approximately 1. Each individual entry of a series is important but so is the sum of the series. 12 3.13% = 2.Example 6: the consumer price index table is shown below with some entries missing (which is common in data sets).95%. b) The arithmetic average percentage change per annum from 1994 to 1999. Each of the following is a series. 6. 11.5 1967 − 1968 16. try find the then the geometric average formula: 𝐺𝐴 = 𝑛 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒 100 +1 −1 pattern. Theory: a series is list of numbers with a constant pattern. 5. then apply the formula: 𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑 Solution: the first term is 𝑎 = 3. 𝑛 the term you want to find. 26. 88 … So 𝑟 = −3.5.5 (−10) 𝑑 = 7. 9. 35 … Plan: determine the first term and the difference between each term. 18. 35 … 𝑑 = 9. so to find out how much revenue the company will earn over the whole year requires summing a series. Substitute into the formula: 𝑆6 = 6 2 ∙ 4 + 5 ∙ 5 = 99 2 Use your calculator to actually sum up the sequence and makes sure 99 is correct. 26.5.5. 17. 470 +0. 490. 19.5 7. the revenue of a company might grow at a certain amount each month. and 𝑛 = 27. 94.5. Substitute into the equation: 𝑇6 = 8 + 6 − 1 ∙ 9 = 8 + 5 ∙ 9 = 53 Add two more 9’s to 35 to see if it is correct. the difference between all the numbers is: 9 9 9 8. then apply the formula 𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑 Solution: the initial number is 𝑎 = 8. 10. 14. 91. then substitute into the formula: 𝑛 𝑆𝑛 = 2𝑎 + 𝑛 − 1 𝑑 2 Solution: the first term is 𝑎 = 4. Substituting this into the equation: 𝑇13 = 100 + 13 − 1 ∙ −3 = 64 Summing a sequence is important in many business disciplines. 2. 25.5 7. and 𝑛 = 13. 3. Example 2: find the 27𝑡 term of the following progression: 3. 97. the difference is +5 (make sure of this) and there are 𝑛 = 6 terms. and 𝑑 the difference between terms in a sequence. the difference is: −3 −3 −3 −3 100. 33 … Plan: determine the first term and the difference between terms. 3.5 3.5.𝟏. Example 1: find the 6𝑡 term of the following arithmetic progression: 8. Theory: the sum of an arithmetic series is found using the formula: 𝑆𝑛 = 𝑛 2𝑎 + 𝑛 − 1 𝑑 2 Each of these sequences has an initial number (bolded) and also a constant addition/subtraction between terms (in brackets). 33 … Example 4: find the sum of the sequence: 4. 24. 94. 2. and the 6𝑡 term is 𝑛 = 6. 97. 25. 17. 480. the difference is: 7. 4 𝟓𝟎𝟎.5. 1. 91.5 = 198 Example 3: find the 13𝑡 term of the sequence: 100. Example 5: find the sum of the first 15 terms of the sequence 164 .5 7. For example. Theory: the 𝑛𝑡 term of a sequence is given by the formula: 𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑 Where 𝑎 is the initial number. 88 … Solution: the first term is 𝑎 = 100.5. Substitute this into the formula: 𝑇27 = 3 + 27 − 1 ∙ 7. 10. 18. 29 Plan: determine the values of 𝑎 and 𝑑. 158. 𝑑 and 𝑛 then substitute into the summation formula. 18. … Plan: find and substitute the constant ratio 𝑟. Above. a luxury car manufacturer. 64. 162 Plan: determine the values of 𝑎. 35. Substitute this into the formula: 𝑇8 = 2 ∙ 37 = 4374 Sometimes. Theory: for two terms with 𝑝 terms missing between them. 54. divide one term by the one before it: 162 54 18 6 = = = =3 54 18 6 2 So 𝑟 = 3. 𝑟 and 𝑛 then substitute into 𝑇𝑛 = 𝑎𝑟 𝑛−1 Solution: 𝑎 = 2. the ratio will need to be determined.5 4 Example 6: find the 8𝑡 term of the geometric progression: 𝑎 is 256 and 𝑛 is 12: 𝑇12 = 256 0. Substitute into the summation formula: 𝑆15 = 15 400 + 14 −21 2 = 795 2.5 = 0.192. the following three progressions are all geometric: 𝟑.4.0. divide a term by the one before it: 𝑟 = 𝑇𝑛 𝑇𝑛−1 54 = 2 3 27 = 3 Example 8: find the 12𝑡 term of: 256.12.179.137.12005 𝟏𝟎𝟐𝟒. but consecutive terms will be unavailable. 6. To find the ratio.64.25 (𝑟 = 2) (𝑟 = 7) 𝑟 = 0.1. Solution: the later term is 54. and the earlier term is 2. ∎. Theory: the 𝑛𝑡 term of a geometric progression is found using the formula: 𝑇𝑛 = 𝑎𝑟 𝑛−1 Where 𝑎 is the first term. 256. If Bentley manufactured 316 cars this year. 𝑑 = 10.200. increases production by ten cars each year. ∎. 54. so: 𝑟 = 3 Remember that a geometric progression has a constant ratio between terms. and 𝑛 = 8. the ratio is determined by: 𝑟 = 𝑝 +1 Example 6: Bentley.245. 𝑡 = 20.384 𝟓. … Plan: determine the values of 𝑎.48. The two terms are separated by 2 terms. 𝑟 the ratio between terms and 𝑛 the term being found. 6.16. relative to the previous year.96. ∎.125 12−1 165 . To find the ratio.1715. … Plan: use the ratio formula above. applying the sum formula: 𝑠20 20 = 2 ∙ 316 + 19 ∙ 10 = 8220 2 𝑙𝑎𝑡𝑒𝑟 𝑡𝑒𝑟𝑚 𝑒𝑎𝑟𝑙𝑖𝑒𝑟 𝑡𝑒𝑟𝑚 Moving on to geometric progressions. Solution: 𝑎 = 200. 𝑑 = −21 and 𝑛 = 15. all three progressions have an initial term (in bold) and a constant ratio between each term (in brackets). what is the total number of cars Bentley will make in two decades (including this year)? Solution: 𝑎 = 316.24. the first term 𝑎 and 𝑛 into the formula: 𝑇𝑛 = 𝑎𝑟 𝑛−1 Solution: the constant ratio is: 𝑟 = 2 64 = 256 2 1 = 0.25 Example 7: determine the ratio of the GP: 2. 05 = $1. a) Applying the sum formulae for 𝑛 = 15: Scheme 1: 𝑆15 = 1000 1. 32. 𝑎 and 𝑛. Plan: a) determine the sum of both series over the fifteen years using the two sum formulas. Plan: find 𝑎. 𝑎 = 1100 and 𝑑 = 45. then apply the sum formula: 𝑆𝑛 = 𝑎 𝑟 𝑛 − 1 𝑟 − 1 grows at 5% per year. b) Applying the term formulae for 𝑛 = 10: Scheme 1: 𝑇10 = 1000 1. 512 Plan: find 𝑟. since Scheme 2 increases by a constant $45 each year. 166 . the Arithmetic Progression formula is used. Determine: a) Which of these two schemes has a better overall return over the fifteen years.05 15 − 1 1. b) Which of the schemes has a higher return in the tenth year. 𝑛 = 8.2 − 1 9 = $21. then the sum formula.Scheme 1: has an initial return of $1000 which Theory: the sum of a geometric progression is found by applying the formula: 𝑎 𝑟 𝑛 − 1 𝑆𝑛 = 𝑟 − 1 Example 9: find the sum of the geometric progression 4. Similarly. the first term is 𝑎 = 400. 𝑎 = 1000. 8. and 𝑟 = 1. then apply the geometric progression term formula. 256. 64.578. For scheme 2.551. 10−1 = 2064 𝑟𝑜𝑢𝑛𝑑𝑒𝑑 𝑢𝑝 = 10. and sales are expected to increase by 20% each year for the next 9 years. Solution: since Scheme 1 increases by 5% each 4 2 8−1 = 1020 2−1 year.33 Scheme 2: 𝑇10 = 1100 + 10 − 1 45 = 1.2 10 − 1 = 1. so substitute this into the formula: 𝑆𝑛 = Example 10: if sales of computers this year are 400.05. the Geometric Progression formula is used.505. For scheme 1. Solution: the ratio 𝑟 is 1.2 (as it is a 20% increase over the previous year).225 Thus Scheme 1 is better overall. and how many computers will be sold in total over the next ten years (including the base year). 16. Scheme 2: has an initial return of $1100 which grows at $45 per year. and 𝑛 = 10 (as the base year is included). 128. 𝑟 and 𝑛.383 𝑟𝑜𝑢𝑛𝑑𝑒𝑑 𝑑𝑜𝑤𝑛 Example 11: a financial services firm can invest in one of two schemes which run for 15 years.05 − 1 Solution: the constant ratio is: 8 16 = = 2 = 𝑟 4 8 And 𝑎 = 4.56 Scheme 2: 𝑆15 = 15 2 1100 + 15 − 1 45 2 = $21. Apply the term formula: 𝑇10 = 400 1.00 Thus Scheme 1 is better in the tenth year. b) determine the tenth term of both schemes using the two term formulae.2 Apply the sum formula: 𝑆10 400 1. how many computers will be sold in the tenth year. and also the total amount of money in the bank when $1000 is saved for 4 years at 7% p.15. Then add the principal amount to determine the total amount in the account.5.9. and also the total amount you will have available. … Determine: a) The 12𝑡 term. some basic theory is required. b) The 25𝑡 term.a. 𝑃𝑡 = 𝑃0 + 𝐼𝑡 Example 2: a $4000 savings account is opened with a bank which offers an interest rate of 9.6%.093 5 = $1860 Add this amount to the principal to determine the total amount available: 𝑃5 = 𝑃0 + 𝐼5 𝑃5 = 4000 + 1860 = $5860 Exercises: 1.a. 𝑡 = 4 and 𝑟 = 0. but the total amount is the interest earned plus the principal: 𝑃4 = 𝑃0 + 𝐼4 = 1000 + 280 = 1280 Theory: the total value of an investment is the principal plus the interest earned over the given time. Using simple interest. 2. c) The sum of the first 20 terms. Given the sequence: 10. This is similar to arithmetic progressions in that a constant amount is earned each year. 𝐼4 = 1000 0. c) The sum of the first 19 terms.5 years.07. … Determine: a) The 15𝑡 term.1. Project 2: 4000 in the first year. determine: a) the interest earned over five years. b) The sum of the first 10 terms. with a minimum period of 5 years. Example 1: find the interest earned.3 simple interest When working with interest rate calculations. the amount of money which the interest rate applies to does not change.29..Exercises: 1.75.7. Plan: use the simple interest formula. so the extra money earned from interest each year is constant. b) Which project has a larger return in the 5𝑡 year.3% p.11.22. Determine a) which project has a larger overall income. 8. A bank offers an interest rate of 4. Theory: the simple interest formula is: 𝐼𝑡 = 𝑃0 ∙ 𝑟 ∙ 𝑡 In words. and this increase by 300 every subsequent year. Using simple interest. 3. The rate must be in decimal form. The interest rate is the rate at which this money increases in value per time period. b) the total amount available after five years.33. Theory: The “principal” is the initial amount of money invested.5. Plan: use the simple interest formula to calculate the interest over the five years. In simple interest. Solution: the principal 𝑃0 = 1000. and this increases by 9% every subsequent year.07 4 = 280 This is the interest earned. determine the amount of interest you will have earned over 2. and you invest $100. 167 . Given the sequence: 3. Solution: to determine the interest earned: 𝐼𝑡 = 𝑃0 ∙ 𝑟 ∙ 𝑡 𝐼5 = 4000 0. the interest earned after 𝑡 years from today (𝐼𝑡 ) is equal to the principal (𝑃0 ) times the per annum interest rate (𝑟) times the number of years it has been invested for (𝑡). A firm can invest money into one of two projects each which runs for 12 years and each which has a return of: Project 1: 3500 in the first year.20. 05. 𝑟 = 0.20 = 1440 = 1440 The left side can be simplified to: Year 3: the initial amount in the third year is the amount at the end of the year before: 1440 1.20 = 1728 This can be re-written as: 1000 1. The value of this amount at the end of the second year will be: 1200 1.469million The interest earned is the total value take the principal: 𝐼5 = 𝑉5 − 𝑉0 𝐼5 = 1. Example 1: find the value of $2000 invested at an interest rate of 5% for 10 years. and also the interest earned over this time. each a year in length: Year 1: at the end of the year. Determine the value of the investment after 5 years. How much will you have at the end of the three years? Solution: break the three year period into three shorter periods. The interest rate must be written as a decimal. the rate was to the power of 3.20 2 Theory: the value of an investment after 𝑡 years is given by: 𝑉𝑡 = 𝑉0 1 + 𝑟 𝑡 Where 𝑉𝑡 is the value of the investment after 𝑡 years.4 compound interest Simple interest is not realistic in most situations.05 𝑉10 = 3257.469million 168 .a. Plan: use the compound interest rate formula to find the total value of the investment. the amount in the bank will be: 1000 1.05 10 1. 𝑟 = 0.79 It should become evident that compound interest is similar in theory to a geometric progression and also a geometric average. Example 2: a finance firm invests $1million for five years in a Dubai construction project. 𝑉0 is the initial investment amount. Intro example 1: you win $1000 from a lottery and decide to save it until you finish university. Then to find the interest earned.20 comes from the original principal (the 1) plus the interest rate (0. which is the new principal.20 investment is 1728.20 1000 1.08 and 𝑡 = 5: 𝑉𝑡 = 𝑉0 1 + 𝑟 𝑡 5 10 = 2000 1.08 𝑉5 = $1. must be written as a decimal.a. 𝑟 and 𝑡.20 = 1728 1000 1. Plan: determine the values of 𝑉0 .8. 𝑡 = 10 into the compounding formula: 𝑉10 = 2000 1 + 0. then substitute into the compound formula.469 − 1 = $0. Solution: substituting 𝑉0 = 2000.20). compounded annually. This amount is 1200. This example was to show what is meant by compounding the principal each year. the value of the 1000 𝑉5 = 𝑉0 1 + 0. The interest rate at which you save it is 20% p.20 1. Year 2: at the beginning of this year. 𝑟. and 𝑟 is the interest rate. Solution: substitute 𝑉0 = 1. take the principal away from the total value of the investment. The interest rate.20 1.20 = 1200 The 1. which has an expected return of 8% p. the initial amount available is the final amount of the previous year. Notice that to find the value after three years.20 = 1440 The 1200 can be rewritten in the way it was originally found: 1000 1. 3 = 1728 After three years. 169 . compounding is occurring. always round UP to the nearest given time period.40 Example 4: find the number of years (to the nearest month) when an investment of $1000 will reach $2000 at an interest rate of 9% p. for three years.0075 4𝑡 0.0075 Rearrange and solve: 4𝑡 = log 2 log 1. Plan: use the compound formula (𝑚 times per year) to determine which of the investments will give the largest return after five years. Rather. Bank 2: 15.a. the time the investment lasts is 𝑡 = 3. The two banks offer the following returns: Bank 1: 15% p.a. they set an annual interest rate but compound monthly. per time period. Therefore the answer is 23 years and 1 quarter. Substitute into the formula: 0. compounded weekly. Example 3: find the value of a $5000 investment if it is invested at 6% p. Rearrange to isolate 𝑡 to find the solution. 𝑚. substitute into the compound formula: 𝑉𝑡 = 𝑉0 1 + 𝑟 𝑚 𝑚𝑡 2000 = 1000 1 + Simplify: 2 = 1. = 5000 1. and it is compounded 𝑚 = 12 times per year.09. Theory: for an annual interest rate 𝑟 compounding 𝑚 times per year gives the formula: 𝑟 𝑉𝑡 = 𝑉0 1 + 𝑚 𝑚𝑡 and 𝑟 = 0.0075 This is simpler than it looks. the yearly interest rate is 6% (or 𝑟 = 0. 𝑉0 = 1000. compounded semi-annually.a. compounded monthly.0075 𝑡 = ≈ 23. as with normal compounding. Plan: find the values of 𝑉0 .005 36 Example 5: a company can save $4000 in one of two banks for a five year period. Determine which of the projects is a better investment. compounded every quarter. 𝑚 = 4 Theory: the interest earned from compound interest is the total value of the investment subtract the principal amount: 𝐼𝑡 = 𝑉𝑡 − 𝑉0 Compounding every year is not what many banks do with savings accounts. the investment would not exceed the given amount. Solution: the initial amount is $5000 (𝑉0 = 5000). 𝑡. 12 3 log 2 log 1.19𝑦𝑒𝑎𝑟𝑠 4 Rounding up to the nearest quarter as that is what the question is asking. Plan: use the compound formula (𝑚 times per year) and substitute all known variables.a. If it was to be rounded down. 𝑡 and 𝑟.06). Theory: whenever finding when an investment will exceed a certain amount.06 𝑉3 = 5000 1 + 12 𝑉3 = 5983.Solution: having 𝑉𝑡 = 2000.09 12 4𝑡 Take the log of both sides: log 2 = 4𝑡 log 1. but then determine the number of times.5% p. Plan: use the continuous compounding formula.155. the closer it is to being continuously compounded.25% p. 𝑡 = 5 is substituted into the continuous compound formula: 𝑉𝑡 = 𝑉0 𝑒 𝑟𝑡 𝑉5 = 4000𝑒 0. to find 𝑉𝑡 . If the term deposit is for 10 years. determine: a) The value of the term deposit account after ten years.437. Determine the amount you would have to pay back after two years. Substitute this and all other known variables into the compound interest formula. A disreputable agent loans you $10. A conservative company decides to save a cashflow of $12.391 in a term deposit account which has a return of 8.10∙1 = 4420.000. assuming you made no repayments.86 Bank 2: 𝑟 = 0. Find the value of the investment using the compound interest formula. compounded weekly.155 2 2 5 5 52 𝑉𝑡 = 4000 1 + 0.a. Bank 1 has a better actual return due to the number of compounding periods per year. substitute all the known variables into the exponential formula. 2.a.300 for four years. A bank offers an interest rate of 4. 𝑉0 = 4000 and 𝑡 = 5: 𝑉𝑡 = 4000 1 + 𝑉𝑡 = $8.0465.600. Solution: 𝑉0 = 4000. compounded every minute of every day.000 savings account which is compounded every second.15.5% p.a. Then. 𝑝. 𝑚 = 52. compounded continuously.68363 Now use the continuous formula above: 𝑉𝑡 = 4000𝑒 0. 170 .75% p. and is compounded monthly.10 525600 525600 ∙1 𝑉𝑡 ≈ 4420. then use the exponential formula. if the savings account is left for 5 years at 4.a. 𝑟 = 0.87 Notice that just because Bank 2 offers a higher advertised interest rate.68367 The two are very close. 𝑉0 = 4000. Theory: the more times an investment gets compounded.458.. 𝑚 = 2. then set this as 𝑚. The investment has an initial value of $4000. Example 6: an investment offers an interest rate of 10%. in which case the number 𝑒 is used: 𝑉𝑡 = 𝑉0 𝑒 𝑟𝑡 Which is identical to exponential growth from Chapter 4. and you decide to save $4. and the continuous formula can be used as an accurate approximation to very short time period compounding.5 annual interest rates Example 5 from the previous section is a good indicator that the advertised interest rate is not always the realised interest rate.00 2𝑑.a. 3. and is invested for one year. b) The interest earned over the ten years. b) The total value of the savings account. determine the amount you would owe after a three year loan. Example 7: find the value of a $4. Compare the answers.Solution: Bank 1: 𝑟 = 0.0465 ∙5 𝑉5 = 5047. If the interest rate is 7. 𝑟 = 0. Determine: a) The interest earned after four years.10. then: 0. Solution: number of minutes in a year=minutes×hours×days = 60 ∙ 24 ∙ 365 = 525. 𝑉0 = 4000 and 𝑡 = 5: 0. Plan: find the number of minutes in a year.2% p.15 𝑉𝑡 = 4000 1 + 52 𝑉𝑡 ≈ $8. You have purchased new car worth $12. to find 𝑉𝑡 . Exercises: 1. 8.500 at 5. 4.65% p.. 9% and 𝑚 = 52: 𝑟𝑟𝑒𝑎 𝑙𝑖𝑠𝑒𝑑 = 1 + 0. Plan: apply the realised interest rate formula.a. compounded once every 4 months. 2.a.0485(4 𝑑.186%. Determine the realised interest rate. the value of the total investment would be: 𝑉1 = 𝑉0 + 𝐼1 = 100 + 5 = 105 It is obvious that with compound interest. Thus the realised interest rate is 4. 𝑝. Two banks are in competition.a. The −1 gets rid of the original amount.05 𝑉1 = 100 1 + 2 𝑉1 = 105. This makes a big difference with large sums of money. compounded semi-annually (every half year). Determine the realised annual interest rate. 𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 1 + Solution: Bank 1: given 𝑟 = 0.07186 5𝑑. compounded monthly.07139 5𝑑. This company has two savings accounts at two different banks. and just leaves the realised interest rate. and is compounded monthly. Thus a realised rate of approximately 7. Solution: 𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 1 + 0.06% rather than the advertised 5%.06%) is larger than the advertised 5%. 𝑟 = 0. compounded once every six months.025 2 𝑟 𝑚 𝑚 −1 −1 𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 0. Find the value of the investment after one year.45% p. The offered interest rates are: Bank 1: 7% compounded quarterly Bank 2: 6.0475 12 12 𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 0.9% compounded weekly Which is a better investment? Plan: use the realised interest rate formula to determine the realised interest rates of the two banks. which bank is a better choice? − 1 ≈ 0.07 and 𝑚 = 4: 𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 0.a. the realised interest rate (5.06 From this. Theory: the realised interest rate is found using: 𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 𝑟 = 1+ 𝑚 𝑚 2∙1 customers tomorrow morning. 𝑝. one with an advertised interest rate of 3. If you are borrowing money. Exercises: 1.05: 0. 𝑡 = 1. $100 is invested for a year. 3.139%. the precise amount is yet unknown. 𝑚 = 2.85%. An investment promises a return of 6. Example 1: find the realised annual interest rate for an investment where the advertised interest rate is 4.24% p.07 = 1+ 4 4 = 100 1.1% p.069 52 52 −1 −1 This is very similar to the compound interest formula except that there is no 𝑡 nor are there any amounts (𝑉’s). Example 2: a company has a cash inflow from its Intro example 1: for an advertised interest rate of 5%. Thus a realised rate of approximately 7.Theory: the annual interest rate is lower than the realised interest rate if the principal is compounded more than once per year. we can also see that the realised annual interest rate is 5. Bank 2: given 𝑟 = 6. and the other with an interest rate of 5. compounded monthly.45% p. ) 171 . Solution: 𝑉0 = 100. Bank 1 is a better option. 𝑝. such as home loans. Given the realised rates. If this was simple interest. which is higher than the advertised rate.75%. A bank advertises an interest rate of 5. Example 1: find the net present value of $5000 in three years’ time at an interest rate of 4.25%. how much is that investment worth in a year’s time? Solution: 𝑉1 = 1000 1 + 0. $1000 today has a higher value than $1000 some time in the future. Intro example 1: for an investment of $1000 at an interest rate of 10% per annum.6 net present value Theory: Net Present Value is a way of finding the value of income in the future.25% p. Solution: discount each of the cash-inflows: Rearranging to isolate 𝑉0 gives: 𝑡 = 𝑉0 Substitute in to get: 𝑉0 = 2000 1 + 0. 𝑁𝑃𝑉 = 𝑛 This means that getting $1000 today is the same as getting $1100 dollars in one years time. 𝑟 = 0. The reason is that if that $1000 dollars was invested at the given interest rate. except backwards.08 This means that $5000 in three years’ time has the same value as $4413.16 Cash-inflow 1: 𝑁𝑃𝑉1 = Cash-inflow 2: 𝑁𝑃𝑉2 = 2400 1 + 0. compounded at the end of each year. then sum them to find the overall NPV.48 The NPV of both cash-inflows is: 172 . compounded only once. the value of the $1000 would grow (through earning interest). 𝑡 = 3: 𝑁𝑃𝑉 = 5000 1 + 0.0425 3 = 5000 1.1 1 Theory: the formula for the net present value of a single future cash flow is: 𝑁𝑃𝑉 = 𝑉𝑡 1 + 𝑟 𝑡 The 𝑁𝑃𝑉 is the same as 𝑉0 from the rearranged annual compound formula. That is.a.8.16 today has the same value as $2000 in one year. in terms of today’s dollars. 𝑟 = 0. This means that $1869.16 would need to be invested today to have $2000 in one year’s time.0525 7 So $1869. Intro example 2: if you need to repay a loan worth $2000 in one year’s time. Plan: find the NPV of each of the two cash-inflows individually. 4000 1 + 0. 𝑡 = 1: 𝑉𝑡 = 𝑉0 1 + 𝑟 𝑉𝑡 1 + 𝑟 𝑡 𝑁𝑃𝑉 𝑛 The overall NPV is simply the sum of all 𝑛 individual net present values.a. how much do you need to invest today at 7% p.79 ≈ 1677.04253 𝑁𝑃𝑉 = 4413.08 today. the NPV of all the cash flows is the sum of the NPVs of the individual cash flows.0525 3 ≈ 3430. at the end of the year? Solution: 𝑉0 is unknown but 𝑉1 is known (= 2000).07 1 = 1869. Plan: apply the NPV formula. It is similar to the concept of compounding. Solution: 𝑉3 = 5000.0425. Example 2: find the NPV of the two cash-flows: Cash-inflow 1: $4000 in 3 years Cash-inflow 2: $2400 in 7 years The interest rate is assumed constant at 5.07. Theory: if there are multiple cash-flows at = 1100 different years. 26 Theory: any project with a positive net present value is accepted. They want you to find out the NPV of this project. Example 3: find the Net Present Value of the project which has the following cash flows: 𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 1 2 3 $983.a. Solution: Year 0: 𝑁𝑃𝑉0 = −983 Year 1: 𝑁𝑃𝑉1 = Year 2: 𝑁𝑃𝑉2 = Year 3: 𝑁𝑃𝑉3 = 399. Assume the interest rate is 4.1 532.30 $532. 𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $6000 0 1 0 $4000 2 0 $1000 3 0 $1500 2. Your supervisor gives you the following data on a project introducing a new advertising campaign.48 𝑁𝑃𝑉 = $5108. then sum them. Plan: find the NPV of the individual cash-flows.0625 𝑁𝑃𝑉2 = $442. then sum them to find the overall NPV. as it is profitable. determine the NPV of the project and if it is worthwhile. Plan: find the NPV of each of the cash flows. 173 .91 = $207.5%.71 Year 2: 500 𝑁𝑃𝑉2 = 1 + 0.1 1 = 243 2 = 440 = 300 Thus the overall NPV is: 𝑁𝑃𝑉 = 𝑁𝑃𝑉0 + 𝑁𝑃𝑉1 + 𝑁𝑃𝑉2 + 𝑁𝑃𝑉3 = −983 + 243 + 440 + 300 = 0 Thus the NPV of the project is zero.30 Assuming the interest rate is a constant 10%.40 $399.00 0 0 0 0 $267.3 1 + 0.1% p. Given the following cash-flow data. the NPV is: 𝑁𝑃𝑉0 = −4000 Year 1: 4000 𝑁𝑃𝑉1 = 1 + 0. Since it is an outflow. Exercises: 1. as the same return could be obtained by saving the initial cash outflow in a bank at an interest rate of 10%. Example 4: determine the Net Present Value for the following investment opportunity: 2 1 267.0625 𝑁𝑃𝑉1 = $3764.4 1 + 0. and if it is worthwhile to go ahead with it.71 + 442.𝑁𝑃𝑉 = 𝑁𝑃𝑉1 + 𝑁𝑃𝑉2 = 3430.91 Thus the overall NPV is the sum of all the individual NPV: 𝑁𝑃𝑉 = 𝑁𝑃𝑉0 + 𝑁𝑃𝑉1 + 𝑁𝑃𝑉2 = −4000 + 3764.79 + 1677.62 Since the overall NPV is positive. Solution: Year 0: the NPV of a cash-flow at year 0 is the value of the cash-flow. this project is worthwhile. Assume that the bank is offering an interest rate of 6.1 3 0 1 2 4000 0 0 0 4000 500 Assume that the interest rate is 6.25%. so there is no point going ahead with the project.3 1 + 0. 000 0 1 0 $3000 2 0 $4000 3 0 $5000 4 $2000 $2500 5 0 $3000 Given the following data about cash-flows from customers and suppliers. and then attempt to isolate 𝑖. so change the form of the other fractions: 10000 = 3000 1 + 𝑖 1 + 𝑖 3 3000 1 + 𝑖 2 + 4000 1 + 𝑖 5000 + 3 1 + 𝑖 1 + 𝑖 3 Simplifying: 2 10000 = + 4000 1 + 𝑖 + 5000 1 + 𝑖 3 174 . and this is unrelated to the market rate.1667 6 Thus the rate of return of the project is 16. The common denominator is 1 + 𝑖 3 . in your own words. to isolate 𝑖. there was a NPV of zero. 𝑌𝑒𝑎𝑟 0 1 2 3 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 $10. This is the definition of internal rate of return. the internal return of the project is unrelated to the market rate. use the letter 𝑖: 𝑁𝑃𝑉 = −3000 + Set this equal to zero: 0 = −3000 + 3500 1 + 𝑖 1 3500 1 + 𝑖 1 Then to be able to add the fractions on the right. so there is no way the NPV can be found. Then solve for 𝑖: 3000 = 3500 1 + 𝑖 3000 1 + 𝑖 = 3500 3000 + 3000𝑖 = 3500 𝑖 = 1 = 0. However. Intro example 2: find an expression for the internal rate of return of the following project. Also. the ability to isolate 𝑖 becomes more and more difficult.7 internal rate of return In the last example of the previous section.000 0 0 0 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $3000 $4000 $5000 8. Intro example 1: determine the internal rate of return of the following project: 𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 Solution: 𝑁𝑃𝑉 = −10000 + 3000 4000 5000 + + 1 2 1 + 𝑖 1 + 𝑖 1 + 𝑖 3 Set this equal to zero to find an expression for the internal rate of return: 0 = −10000 + 3000 4000 5000 + + 1 + 𝑖 1 1 + 𝑖 2 1 + 𝑖 3 Now. When the number of years increases. To find the IRR of the project. a common denominator is needed. the −10000 will be first moved to the other side: 10000 = 3000 4000 5000 + + 1 2 1 + 𝑖 1 + 𝑖 1 + 𝑖 3 0 $3000 0 1 0 $3500 Solution: the market rate has not been given. determine the NPV of all the cash-flows. Theory: the internal rate of return is the interest rate that would give a net present value of zero. and an interest rate of 7. the meaning of net present value.67%.3.25%. 4. The internal rate of return is the rate of return of the project. 𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $1500 $2000 1 $2500 $1000 2 0 $3000 3 0 $800 4 $4000 $1000 Describe. 𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $10. use the NPV formula and set the overall NPV equal to zero. and is unrelated to the market interest rate. instead of using a number for the interest rate. since the 3% gave a positive NPV. If it is positive. The thought process is: Try a random (low) interest rate and find the NPV. 𝑁𝑃𝑉1 = 303. Theory: STAGE 2: having two interest rates (relatively close to one another) with one giving a positive NPV and the other giving a negative NPV. use another (higher) interest rate to get a negative NPV.03 3 It does not matter which interest rate-NPV combination you define with a subscript as 1 or 2. but for part two of this approximations. This above example is an exact method of finding the internal rate of return. Example 1: find two interest rates. to find one with a negative NPV. one which gives a positive NPV. given the following investment project: 𝑌𝑒𝑎𝑟 0 1 2 3 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 $8000 0 0 0 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $5000 $2000 $2000 Solution: use a low interest rate to begin with. and one which gives a negative NPV. as the formula works either way. there is a much easier method which gives an approximate answer.Rearranging and expanding the brackets would take a long time. Plan: apply the IRR formula.02 Both 3% and 5% gave a positive NPV. However. so in this case. one with a positive NPV and one with a negative NPV. 175 . lower the interest rate further. find the approximate IRR: 𝑖1 = 5%. for 5 years.64 𝑁𝑃𝑉2 = −299. This method is a linear approximation of a curve. but a program can be used.64 This is still a positive NPV. 𝑖2 = 10%. This theory is demonstrated in the following example.10 3 5000 2000 2000 + + 1 2 1.05 1. apply the formula: 𝐼𝑅𝑅 ≈ 𝑖2 𝑁𝑃𝑉1 − 𝑖1 𝑁𝑃𝑉2 𝑁𝑃𝑉1 − 𝑁𝑃𝑉2 5000 2000 2000 + + 1 2 1. Try this for yourself and see if you can get: 0 = 10000𝑖 3 + 27000𝑖 2 + 20000𝑖 − 2000 Solving this is difficult. as the number of years increases to 4. However. The closer the positive and negative NPV’s are to being zero. 𝑁𝑃𝑉3% = −8000 + 𝑁𝑃𝑉3% = $569. to get a negative NPV.03 1. the better the approximation. Now. If it does not. so it gets more and more difficult. and in the end. say 2%. so increase the interest rate to say 10%: 𝑁𝑃𝑉10% = −8000 + 𝑁𝑃𝑉10% = −299.02 Plan: find the NPV of the project using a relatively low interest rate. the 5% has a NPV closer to zero. Example 1 cont: given the following pairs of NPV and 𝑖. the exact solution will involve solving something with 𝑖 4 which is difficult. then increase the interest rate by.03 1.10 1.05 1. This step is a trial-and-error method. This can be anything you choose. use another (lower) interest rate to find a positive NPV. say 5%: 𝑁𝑃𝑉5% = −8000 + 𝑁𝑃𝑉5% = $303. the solution will involve 𝑖 5 . so let’s use 3%. a cubic function would need to be solved. we use a higher interest rate. the interest rate that gives a NPV closest to zero is used.10 1.05 3 5000 2000 2000 + + 1 2 1. Theory: the method of finding the approximate internal rate of return is split into two stages: STAGE 1: find two interest rates.84 The low interest rate should give a positive NPV. If the initial (low) interest rate had a negative NPV. than the 3%. If the market rate is 9%. a higher interest rate will give a negative NPV. say. 𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $7. the project is worthwhile.12 654. Conversely. with appropriate subscripts: 𝑖1 = 0.02 303.053 1.10 303. 654. so it is best to rewrite the known variables. Exercises: 1. 5%: 𝑁𝑃𝑉5% = −7000 + 3000 2500 1500 1500 + + + 1.12 1. An internal rate of return higher than the market interest rate means the project is worthwhile.61%. determine if the following project is worthwhile using the IRR method.Solution: substitute this into the formula: 𝐼𝑅𝑅 ≈ 𝐼𝑅𝑅 ≈ 0.53 − 0.05 1.13 1.0961 Thus the internal rate of return is 9.14 𝑁𝑃𝑉10% = −55. find an approximation for the internal rate of return. Solution: finding the NPV at.64 − −299. Given a market interest rate of 8.054 𝑁𝑃𝑉5% = 654. describe what internal rate of return measures.05 −55.12 Substitute this into the formula: 𝐼𝑅𝑅 ≈ 0. then if that gives a positive NPV. as the NPV of the 10% is closer to zero (i.052 1. and compare it to the market rate.53).e. Apply the results to the IRR formula.0752 602.52%.31 ≈ 0.66 Since this is a positive NPV. apply the IRR formula: 𝑖2 𝑁𝑃𝑉1 − 𝑖1 𝑁𝑃𝑉2 𝑁𝑃𝑉1 − 𝑁𝑃𝑉2 Thus the internal rate of return of this project is approximately 7. −55. Note that the IRR is closer to 10% than to 5%.12 Having two interest rates.12 𝐼𝑅𝑅 ≈ 0. and will be closer to the interest rate which has its NPV closer to zero. If the first interest rate gives a negative NPV. use another higher interest rate to find a negative NPV. In your own words.1 1.e.05 −299. 𝑌𝑒𝑎𝑟 0 1 2 3 4 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 $7000 0 0 0 0 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $3000 $2500 $1500 $1500 𝐼𝑅𝑅 ≈ Many students screw up with the substitution. use a lower interest rate to find a positive NPV. For the following data. Example 2: determine the internal rate of return for the project shown below.64 − 0.53 𝑁𝑃𝑉2 = −55.10. one which gives a positive NPV and one which gives a negative NPV. 𝑁𝑃𝑉1 = 654. So try 10%: 𝑁𝑃𝑉10% = −7000 + 3000 2500 1500 1500 + + + 1.02 45.10 654.5%.12) than the NPV of the 5% (i. 𝑖2 = 0. Since the IRR is greater than the market rate of 9%.53 176 . an internal rate of return lower than the market interest rate means the project is not worthwhile.430 0 1 0 $7800 2 0 $5100 3 0 $1100 3. Theory: the IRR will always be between the two interest rates which gave the positive and negative NPV’s. 𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $12. is the project worthwhile? Plan: use a low interest rate to find the NPV.53 − −55.05.000 0 1 0 $3000 2 0 $2500 3 0 $2800 2. The realised interest rate is found using: 𝑟 𝑚 𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 1 + −1 𝑚 Net Present Value is a way of finding the value of income in the future. chapter eight questions 1.2 1987 − Determine: a) The percentage change from 1980 to 1981. the value of an investment after 𝑡 years is given by: 𝑉𝑡 = 𝑉0 1 + 𝑟 𝑡 The interest earned from compound interest is: Any project with a positive net present value is accepted. b) The percentage change from 1982 to 1984. the NPV of all the cash flows is the sum of the NPVs of the individual cash flows. A geometric series has a constant ratio between terms. apply the formula: 𝑖2 𝑁𝑃𝑉1 − 𝑖1 𝑁𝑃𝑉2 𝐼𝑅𝑅 ≈ 𝑁𝑃𝑉1 − 𝑁𝑃𝑉2 The IRR will always be between the two interest rates which gave the positive and negative NPV’s. d) The geometric average from 1982 to 1984. one which gives a positive NPV.1 107. To find an approximation to the IRR: STAGE 1: find two interest rates. STAGE 2: having these two interest rates. The 𝑛𝑡 term of a sequence is given by: 𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑 The sum of an arithmetic series is given by: 𝑛 𝑆𝑛 = 2𝑎 + 𝑛 − 1 𝑑 2 The 𝑛𝑡 term of a geometric progression is found using the formula: 𝑇𝑛 = 𝑎𝑟 𝑛 −1 To find the ratio. Given the following Consumer Price Index table: 𝑌𝑒𝑎𝑟 1980 1981 1982 1983 1984 1985 1986 𝐶𝑃𝐼 89. The formula for the NPV of a single future cash flow is: 𝑉𝑡 𝑁𝑃𝑉 = 1 + 𝑟 𝑡 If there are multiple cash-flows at different years. The percentage difference between two time periods in an index is found using: 𝐼𝑡+1 − 𝐼𝑡 %∆𝑡→𝑡+1 = × 100 𝐼𝑡 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒 𝐴𝑟𝑖𝑡𝑚𝑒𝑡𝑖𝑐 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑦𝑒𝑎𝑟𝑠 𝐺𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 𝑛 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒 +1 −1 100 A series is list of numbers with a constant pattern. Each individual entry of a series is important but so is the sum of the series.chapter eight summary Index numbers represent values in terms of a base value: 𝑉𝑎𝑙𝑢𝑒𝑡 𝐼𝑛𝑑𝑒𝑥𝑡 = × 100 𝑉𝑎𝑙𝑢𝑒0 The base year is given an index value of 100. in terms of today’s dollars. e) If the geometric average from 1982 to the missing value of 1987 is 3. An arithmetic series has a constant difference between terms. always round UP to the nearest given time period. compounding 𝑚 times per year: 𝑟 𝑚𝑡 𝑉𝑡 = 𝑉0 1 + 𝑚 Whenever finding when an investment will exceed a certain amount. An internal rate of return lower than the market interest rate means the project is not worthwhile.9 92. and one which gives a negative NPV. the ratio is found by: 𝑟 = 𝑝 +1 𝐼𝑡 = 𝑉𝑡 − 𝑉0 For an annual interest rate 𝑟.4 93. The internal rate of return (IRR) is the interest rate that would give a net present value of zero. divide a term by the one before it: 𝑇𝑛 𝑟 = 𝑇𝑛 −1 When 𝑝 terms are missing. c) The arithmetic average from 1982 to 1984. and is unrelated to the market interest rate. This step is a trial-and-error method. For continuous compounding (or very short time period compounding) the 𝑒 formula is used: 𝑉𝑡 = 𝑉0 𝑒 𝑟𝑡 The annual interest rate is lower than the realised interest rate if the principal is compounded more than once per year. An internal rate of return higher than the market interest rate means the project is worthwhile. determine the 177 .5 − 100 103. 𝑁𝑃𝑉 = 𝑛 𝑁𝑃𝑉 𝑛 𝑙𝑎𝑡𝑒𝑟 𝑡𝑒𝑟𝑚 𝑒𝑎𝑟𝑙𝑖𝑒𝑟 𝑡𝑒𝑟𝑚 The sum of a geometric progression is given by: 𝑎 𝑟 𝑛 − 1 𝑆𝑛 = 𝑟 − 1 The simple interest formula is: 𝐼𝑡 = 𝑃0 ∙ 𝑟 ∙ 𝑡 The total value of an investment is: 𝑃𝑡 = 𝑃0 + 𝐼𝑡 For compound interest.32%. and will be closer to the interest rate which has its NPV closer to zero. as it is profitable. 13.1%p.100 For a market rate of 4%.457 in the lottery. Given the following investment project: 𝑌𝑒𝑎𝑟 0 1 2 3 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 $60. you withdraw $5.a.1% p. d) The geometric average from 1971 to 1973.176. An investment firm is offered one of two investments. A student needs to have $5000 to pay his university fees in three years time. 14. Having won $11. compounded monthly. compounded monthly. and deposits $20.500 $9. 12. compounded weekly Determine which of the banks is the better deal. Investment 2: $4350 in the first year. Given the following series: 1331. compounded semi-annually.75. b) The 27𝑡 term.3 133.000. c) Which of the investments has a higher overall return.5. If the prevailing interest rate is set at 6. given that after 3years.25. A firm is looking to borrow money from one of three banks: Universal Bank: 8.000. Given the following series: 3. determine: a) The NPV b) The internal rate of return c) If the project is worthwhile. (compounded annually). but the son takes out the interest at the beginning of each year.a.000.65% p.6%p. determine how much the student needs to save today to be able to pay the university bill.2. each which has the following returns: Investment 1: $6000 in the first year.a. Mark lends David $1430 for an airfare home. f) The geometric average from 1971 to 1977. missing index entry. b) Which of the two investments has a higher return up to the fifth year. 9. Hint: find 𝐼𝑡→𝑡+𝑛 then substitute into the geometric average formula. 6. 4.a.3. compounded yearly Bank B: 5% p. Determine: a) The value of the account in 5 years. the firm withdrew $2000. d) The value of the account after 6 years. A father opens a savings account for his son.484. The deal is that David pays 3% simple interest per month.6.000 given that after 2 years. the average percentage change is negative. If the interest rate is 5% p.6.1 8. and this increases by 13% every year after. determine: a) The NPV and the IRR b) If the project is worthwhile.0 154.000 0 0 0 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $48. c) The arithmetic average from 1971 to 1973. c) When the value of the account will first exceed $20. b) The 21𝑠𝑡 term. d) The sum of the first 16 terms.000. The gold index has the following historic data: 𝑌𝑒𝑎𝑟 1971 1972 1973 1974 1975 1976 1977 𝐺𝑜𝑙𝑑 𝐼𝑛𝑑𝑒𝑥 150. however if the time period is extended slightly from 1971 to 1977. determine: a) The interest Mark has earned over the four months.000 0 0 0 $4000 0 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $5000 $5000 $2000 $2500 $4000 For a market rate of 9%. … Determine: a) The 15𝑡 term.200 $21. 10. Determine: a) The percentage change from 1971 to 1973. A firm wanting to save a future cash-flow has the option of two bank accounts: Bank A: 5. c) When the value of the account will first exceed $10. determine: a) How much money the son has taken out after 7 years. b) The percentage change from 1971 to 1977.4. b) The value of the account after 5. c) The sum of the first 9 terms. e) The arithmetic average from 1971 to 1977. compounded quarterly Sun Bank: 8%p.64.2 168. 178 . you save the total amount at 7. Determine: a) The value of the account after 4 months.5. 3.2% compounded semi-annually Determine which of the banks is the better deal.3 163. b) How much money the son has taken out after 11 years. g) Explain how the average percentage change over the years 1971 to 1973 is positive. b) When the value of the account will first exceed $20. for the four months it takes to earn the money.75. … Determine: a) The 13𝑡 term. Assuming David repays Mark in a lump-sum.a. 11. compounded weekly Bank of Vark: 8. b) The total amount David has to repay Mark after the four months. determine: a) Which of the two investments has a higher return in the fifth year.a. A firm saves $5600 at an interest rate of 4.5years.a.1% p.a. 7. If the investment is for 8 years. 5.4 172. Given the following potential project: 𝑌𝑒𝑎𝑟 0 1 2 3 4 5 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 $12. d) The sum of the first 16 terms.0 142. c) The sum of the first 11 terms. and this increases by $250 every year after. 179 . 𝐵. −9 b) 6. 𝐴.89 3111 which is elastic. 𝐶. Question 12: 8 a) 𝑄 = 2 𝑡𝑜𝑢𝑠𝑎𝑛𝑑𝑠.15 c) −1.723𝑚𝑖𝑙𝑙𝑖𝑜𝑛 3592 𝑃 = ≈ 23.25𝑥 + 16 2 e) 𝑥 2 4𝑥 2 − 1 f) 𝑥 2 𝑥 h) 9 Question 3: 5 a) b) c) e) f) g) 6 19 24 25 12 23 65 5 12 187 210 68 5 𝑜𝑟 2 7 22 1 12 d) − h) 𝑜𝑟 3 21 21 Question 4: 25 a) − b) c) 11 42 30 3+𝑥 3𝑥 h) – 25 Question 6: a) 34 b) 24 c) 9 d) −33𝑥 e) −186 f) −38 g) −64 h) −2𝑥 + 2𝑦 + 6 Question 7: a) 𝑥 2 + 4𝑥 + 4 b) 𝑥 2 − 6𝑥 + 9 c) 4𝑥 2 + 4𝑥 + 1 d) 1 − 2𝑥 + 𝑥 2 e) 𝑥 2 + 8𝑥 + 5 f) −4𝑥 2 + 7𝑥 − 1 Question 4: 6 a) 𝑦 = 𝑥 − 6 5 b) 𝑦 = −3𝑥 − 3 Question 5: a) 𝑦 = 6𝑥 − 3 b) 𝑦 = 5𝑥 − 9 4 c) 𝑦 = 𝑥 3 d) 𝑦 = −𝑥 + 6 5 1 e) 𝑦 = − 𝑥 − 8 4 4 Question 6: a) 𝑦 = 2𝑥 − 1 b) 𝑦 = 3𝑥 + 2 c) 𝑦 = 𝑥 d) 𝑦 = −2𝑥 − 12 e) 𝑦 = −2. 898 c) 𝜀𝑠 = − ≈ −14. the supply curve is backward bending in this case. 𝐹.125 g) 𝑦 7/6 𝑥 𝑦 2 g) 3𝑥 2 − 6𝑥 + 4 h) – 𝑥 2 − 2𝑥 − 4 Question 8: a) 𝑥 = 5 b) 𝑥 = 2 c) 𝑥 = 1 11 d) 𝑥 = e) f) g) i) 𝑥 = 𝑥 = 3 13 2 1 21 9 10 𝑦 3 𝑥 = − 𝑥 = 27 36 13 40 h) 𝑥 = 3 − 𝑦 +1 1 3 d) 4𝑥 𝑥 + 1 = 0 𝑥 = 0 𝑜𝑟 𝑥 = −1 e) 𝑥 2 𝑥 − 1 = 0 𝑥 = 0 𝑜𝑟 𝑥 = 1 f) 𝑥 3 3𝑥 − 5 = 0 5 𝑥 = 0 𝑜𝑟 𝑥 = 3 Question 11: a) 4 𝑥 2 + 3𝑥 + 3 b) 6 2 − 𝑥 1 c) 7𝑥 2 + 2 𝑥 = 0 𝑜𝑟 𝑥 = d) – − 2.72 61 meaning this good is elastic in supply. 𝑃 = 80 Question 9: 𝑄 = 5. 𝐸 b) 𝐴. 𝐹.25 Question 8: 𝑄 = 18. as |𝜀𝑑 | > 1. Note. 2 3𝑦 −9 𝑥 −5𝑥 4 + 3𝑥 2 − 5𝑥 + 3 Question 12: a) 𝑥 = 3 𝑜𝑟 𝑥 = 2 b) 𝑥 = −7 𝑜𝑟 𝑥 = 4 c) 𝑥 = 2 𝑜𝑟 𝑥 = 3 1 15 d) 𝑥 = − 𝑜𝑟 𝑥 = − 3 6 e) 𝑥 = 0 𝑜𝑟 𝑥 = 2 f) 𝑥 = 1 Question 13: a) 3 b) 4 c) 1 d) 15 e) 0.5 1 c) 4 8 d) 3𝑥 2 e) 243𝑥 4 𝑦 8 f) 1024𝑥 2 𝑦 0. 𝐶. 𝐹.5 8 d) − e) f) g) 21 49 6 1 d) − e) 9 5 6 f) 12𝑥 g) 7𝑥 3𝑦+2𝑥 h) 3𝑥𝑦 10 25𝑥+12 15 Question 5: a) 𝑥 3 𝑦 6 b) 216𝑥 9 𝑦1. Question 13: c) a) 𝑄 = 732 155 3391 ≈ 4. 𝐶.6 Question 7: a) −6.17 155 8980 b) 𝜀𝑑 = − ≈ −2.5𝑥 − 6. as 𝜀𝑠 > 1.61 300 3391 b) 𝜀𝑑 = − ≈ −49. −36 7 7 6𝑥 9/2 𝑦 0.12 1 5 d) 33 .chapter one solutions Question 1: a) 15 b) −4 c) 50 d) −6 Question 2: 1 a) b) 12 c) 1. 𝐵. 𝐷.22 which 1054 means this good is elastic in supply. 𝐸 g) 𝑦 = 2 5 𝑥−12 𝜀𝑠 = ≈ 3. 𝑃 = 22 Question 10: 62 𝜀𝑑 = − meaning this 27 good is elastic in demand. 𝐸 c) 𝐴. Question 11: 25.72 68 meaning this good is very elastic in demand.1 f) 1 chapter two solutions Question 1: a) 𝑦 = 𝑥 − 1 𝑥 3 b) 𝑦 = − + 2 2 c) 𝑦 = 𝑥 + 1 d) 𝑦 = 2𝑥 − 4 5 14𝑥+3 e) 𝑦 = 𝑜𝑟 7 70𝑥 + 15 𝑦 = 7 1 f) 𝑦 = 2𝑥 − 𝑜𝑟 3 5𝑥 − 60 𝑦 = 𝑜𝑟 3 5𝑥 𝑦 = − 20 3 h) 𝑦 = 5𝑥 − 12 5𝑥 i) 𝑦 = 9 − 3 Question 2: a) 𝐷.8 129 𝜀𝑠 = = meaning this 21 105 good is elastic in supply.125 j) 𝑥 = 4 Question 9: a) 2 𝑥 + 𝑥𝑦 + 1 b) 2𝑥 𝑥 + 3𝑦 − 4𝑦 2 c) 𝑦 13𝑥 − 12𝑦 + 1 d) 3𝑥 2 1 + 3𝑥 + 4𝑥 3 e) 𝑃𝑄 4 + 𝑟 − 𝑥𝑦 1 f) 2𝑥 2 + 2𝑥 4 + 1 6 g) 5𝑥 3𝑥 2 + 5 − 10𝑥 3 1 1 1 h) 1 + − 2 𝑜𝑟 𝑥 𝑥 𝑥 1 2 𝑥 + 𝑥 − 1 𝑥 3 Question 10: a) 𝑥 𝑥 − 1 = 0 𝑥 = 0 𝑜𝑟 𝑥 = 1 b) 𝑥 𝑥 + 3 = 0 𝑥 = 0 𝑜𝑟 𝑥 = −3 c) 2𝑥 1 − 3𝑥 = 0 e) 9. 𝐵.4. 𝐷. 30 182 𝑃 = 22 ≈ 22. 180 . 𝑆 = 40 𝑎𝑛𝑑 𝐿 = 94 f) −8.6 a) 1.5.1.7 Question 3: c) 1.3.2.4.8.1 e) 5.4.5. 𝑏 = −4.2 d) 13.75.1. 𝑏 = 1.011 𝑦 = 3 2𝑥 :red log 8 b) 𝑥 = 1.4𝑦 = 3 5 19 j) −2.chapter three solutions h) −0.0.041 1+log 2 120 36± 396 𝑦 = 4 log 𝑥: black 25−ln 3 ln e) 𝑥 = i) 𝑥 = − +1 2 d) 𝑡 = 55 = 26.1.4 b) 52 g) −0. 𝐼 = 180.21 𝑎𝑛𝑑 𝑄 = 8.8.91.25. 𝑐 = 6 𝑥−2 4± 24 d) Two solutions c) 𝑎 = 1.5 a) 𝑀: 2853 = 7𝑃 + 19𝑆 + 14𝐿 c) 3.7 Question 2: e) −280 c) 2.25 b) 𝑥 + 𝑦 = 3 l) 4.25.4 3𝑥 + 𝑦 = −15 3 m) 7. −1.1. 𝑏 = 0.1 d) 3.8.4.3.75.7.2.4. −1 f) −2294 d) −1.3.6 b) 𝑃 = 111.19 h) 2.4.1.5. −3 a) −11 f) 20.5.5 𝑎𝑛𝑑 𝑄 = 8.2 e) 0 2𝑥 + 𝑦 = −17 k) 2.01𝑦 18 𝑦 = log 𝑥 : blue ln 6 0.3 Question 9: = 1000 e) −1.2 c) 1.2.1.4.9 c) −𝑥 + 17𝑦 = 14 h) − 7 Question 5: −3𝑥 + 2𝑦 = 1 34 i) − a) 𝑃 = 45. 𝐷 Question 11: a) 𝑥 = Question 6: ln 2 2 Not Functions: 𝐵.2 d) Unique solution = 2608.2.0.7 e) Infinite solutions Question 15: c) Infinite solutions f) No solutions 𝐶 = 800 − 800𝑠 .3 3𝑎 − 3𝑏 − 𝑐 = 1 d) 56 b) 3. 𝐶.2. 𝐸 18 𝑦 = + 5: red ln 4 a) Two solutions 𝑥−5 Question 2: b) 𝑥 = 1 ln 3 b) Two solutions 𝑦 = − 6: blue ln 1 𝑥+4 a) 𝑎 = −5. −3 b) 2. −2.2.79 11𝑥 − 1.1 a) .5. 𝑐 = −4 d) 𝑥 = Question 12: 2 12 d) 𝑎 = 1.75𝑏 g) 𝑥 = − −𝑥 𝑦 = 2 + 2: black 182 +log 8 100 c) 𝑥 = 1 𝑜𝑟 𝑥 = −3 ln 8 Question 10: c) 𝑡1000 = 55 ≈ 14.1.4. 𝐼 = 800𝑠 𝑎𝑛𝑑 𝑌 d) 1.2.2 Question 11: Question 7: c) 3.43 f) 𝑎 − 𝑏 + 𝑐 = −4 b) −30 Question 6: 𝑎 − 𝑏 − 𝑐 = −1 c) 42 a) −1.1.9 Question 12: b) 2. −1.5.01𝑦 2 4 ln 2 0.4𝑧 = 0 3 a) 8 e) 𝑃 = 7. 𝑐 = 3 c) 𝑥 = +1 1 c) No solutions ln 4 𝑦 = − − 4: black b) 𝑎 = −1. −3 6 18 18 b) 2. −3 f) 3.03 f) 𝑁𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 181 .5.5 315 d) 11𝑥 − 1.2 f) −6.75 b) Infinite solutions Question 14: Question 4: c) Unique solution 𝐶 = 1827.75.75 1 13 17 a) 2.6 𝐼: 2873 = 9𝑃 + 21𝑆 + 11𝐿 d) 2.5. 3 . 𝑏 = −2.58𝑦 h) 𝑥 = d) 𝑥 = −4 0. 𝑐 = −2 e) Two solutions 2 ln 3+ln 2 a) 𝑥 = e) 𝑥 = 𝑜𝑟 f) One solution log 4 ln 2+3 ln 3 Question 3: 13 ln 18 2 + 27𝑥 − 54: black Question 7: 𝑦 = −3𝑥 b) 𝑥 = − 𝑥 = 2 log 3 3 ln 54 𝑦 = −𝑥 2 + 18𝑥 − 45: blue 𝑦 = −3𝑥 − 3: blue 7 3 c) 𝑥 = −3 f) No solutions 𝑦 = 2𝑥 − 3: red log 2 𝑦 = 𝑥 2 − 6𝑥: red ln 14 2 = −𝑥 3 + 5𝑥 2 g) 𝑥 = 𝑦 = 𝑥 2 − 25𝑥 + 150: green 𝑦 = 𝑥 3 + 12𝑥 2 ++ 48𝑥 − 252:grey d) 𝑥 = 14 ln 2 𝑦 11𝑥 − 168: black log 2−4 Question 4: Question 14: Question 8: e) 𝑥 = 4 14 −3± 33 𝑥 18 ln 𝑦 = 1.12 Question 8: Question 13: g) −1.75 a) Unique solution 𝐶 = 4860.70 b) 1.4.5 𝑜𝑟 𝑥 = −1.5 d) 8 a) −3𝑥 + 𝑦 = −1 j) −0.18. −11.5 chapter four questions −1± 69 Question 1: 𝑦 = 2 log 𝑥 − 2 + 5: red Question 13: g) 𝑥 = ln 15 2 Functions: 𝐴.3 𝑇: 3594 = 4𝑃 + 20𝑆 + 25𝐿 e) 13.4 e) 2.5 : blue a) 𝑥 = f) 𝑥 = a) 𝑡140 = 12 ≈ 14.6 g) 3. 𝐼 = 540 𝑎𝑛𝑑 𝑌 = 6000 h) 3.12 a) 3.5 b) 1.5𝑦 = 3 b) 𝑃 = 46 𝑎𝑛𝑑 𝑄 = 8 11 22 k) 0 e) 3𝑥 − 3𝑦 − 𝑧 = −4 c) 𝑃 = 46 𝑎𝑛𝑑 𝑄 = 19 l) 0 −15𝑥 + 4𝑦 = 13 1 d) 𝑃 = 7 𝑎𝑛𝑑 𝑄 = 1 Question 10: −𝑥 − 𝑦 + 1.8.5 c) −10 Question 1: i) −0.1. −1 g) −2.1.5 b) 𝐺𝐷𝑃10 = 𝑈𝑆$828. −2.25 d) 0.78 𝑎𝑛𝑑 𝑌 a) 3. 918𝑚𝑖𝑙𝑙𝑖𝑜𝑛 b) c) d) e) f) g) 𝑦 ′ 𝑦 ′ 𝑦 ′ 𝑦 ′ 𝑦 ′ 𝑦 ′ ′ d) 13𝑒𝑚𝑝𝑙𝑜𝑦𝑒𝑒𝑠 = 5𝑒 5𝑥−1 = −4𝑒 1−4𝑥 2 = 2𝑥𝑒 𝑥 −1 2 = 13 − 2𝑥 𝑒 13𝑥−𝑥 4 −3 = 2 4𝑥 3 − 3𝑥 −4 𝑒 𝑥 −1+𝑥 48 = − 2 𝑒 4/𝑥 𝑥 2 h) 𝑦 = − 7 Question 8: 1 a) 𝑦 ′ = b) 𝑦 ′ = c) 𝑦 ′ = d) 𝑦 ′ = e) 𝑦 = ′ 𝑥 2 𝑥 10 − 3 𝑥 4 − 1 2 𝑒 𝑥 −3 −𝑥 2 5𝑥−1 −3 2𝑥−1 𝑥 2 −𝑥 4 3 − +1 3 𝑥 4 𝑥 −3 +𝑥 i) 𝑦 ′ = − 2 − 3 + 4 𝑥 𝑥 𝑥 Question 4: a) 𝑦 ′ = 2 𝑥 + 1 b) 𝑦 ′ = 15 3𝑥 − 4 4 c) 𝑦 ′ = 12 𝑥 + 5 5 d) 𝑦 ′ = −144 1 − 4𝑥 2 e) 𝑦 ′ = 300 2𝑥 + 1 𝑥 2 + 𝑥 3 48 f) 𝑦 ′ = − 4𝑥 − 1 3 5 Question 5: a) 𝑦 ′ = 3𝑥 2 15𝑥 + 2 + 15𝑥 3 b) 𝑦 ′ = 3𝑥 2 𝑥 9 + 1 + 9𝑥 8 𝑥 3 − 1 c) 𝑦 ′ = 60𝑥 3 1 − 𝑥 3 − 45𝑥 6 𝑜𝑟 𝑦 ′ = 60𝑥 3 − 105𝑥 6 d) −28𝑥 −8 𝑥 2 + 𝑥 4 + 2𝑥 + 4𝑥 3 4𝑥 −7 1 e) 𝑦 ′ = −𝑥 2 + 4𝑥 3 𝑥 15 + 15𝑥 14 − 𝑥 3 + 𝑥 4 f) 𝑦 ′ = 1. 𝐸.05 𝑜𝑟 c) 𝑡 = ≈ 9. 𝐺. 𝐷.Question 15: a) 𝑝 5 = 1.05 ln 1.39𝑦 ln 1.03+ln 1.16 b) 15.89 c) $17.5𝑥 −7 + 2 −1 − 𝑥 −3 + 3𝑥 3 −4 𝑥 −6 −4 −6 𝑥 3 1 + 2𝑥 2 3 f) 𝑜𝑟 4 4 − 4+ 𝑥 ′ 𝑦 = −3 3 𝑥 + 𝑥 2 2𝑥+3𝑥 −4 −𝑥 −2 ′ 𝑦 = 2 −3 1 𝑥 −𝑥 + ′ 0.04 8+4𝑥 −3 𝑥 g) 𝑦 = 8𝑥−2𝑥 −2 Question 9: a) 𝑦 ′ = 12𝑥 2 2𝑥 3 − 1 3𝑥 + 5 ′ 3 3 𝑦 b) 𝑦 ′ 𝑦 ′ c) 𝑦 ′ = 2𝑥 − 1 3𝑥 + 5 54𝑥 + 60𝑥 − 9 = 5𝑒 −5𝑥−1 𝑥 − 1 4 + 4 𝑥 − 1 3 1 − 𝑒 −5𝑥−1 𝑜𝑟 = 𝑥 − 1 3 5𝑥𝑒 −5𝑥−1 − 9𝑒 −5𝑥−1 + 4 = 0.457 b) 𝑡 = ln 𝑡 = ln 1. 𝐹 Question 2: a) 𝑓 ′ = 18 b) 𝑓 ′ = 2𝑥 c) 𝑓 ′ = 2𝑥 + 7 d) 𝑓 ′ = 3𝑥 2 + 8𝑥 5 e) 𝑓 ′ = − 2 f) 𝑓 ′ = − 3 𝑥 Question 3: a) 𝑦 ′ = 6𝑥 b) 𝑦 ′ = 30𝑥 + 1 c) 𝑦 ′ = 2𝑥 + 3𝑥 2 51 d) 𝑦 ′ = 𝑥 2 + 10𝑥 2 e) 𝑦 ′ = −32𝑥 −5 + 3𝑥 2 f) 𝑦 ′ = −24𝑥 −3 − 3𝑥 −4 + 3𝑥 2 2 g) 𝑦 ′ = − 3 h) 𝑦 ′ = − 𝑥 10 𝑥 2 Question 16: a) 472.0815 61 + ln 1.170. 𝐶. 𝐵.05 42 ln 1.525 ≈ 5.0815 ln 2.16𝑦 chapter five solutions Question 1: 𝐻. 𝐴.6𝑥 2 + 1 −4𝑥𝑒 𝑥 2 2 + 9 3𝑥 + 5 3 2 2𝑥 3 − 1 2 2 𝑜𝑟 d) 𝑦 ′ = e) f) g) 𝑦 ′ = 𝑦 = ′ 1+𝑒 𝑥 180 2 2 𝑥 4 12−𝑥 −3 −5𝑥 4 1+ln 𝑥 𝑥 5 −1 2 3𝑥−1 + 1 𝑥 𝑥 5 −1 2 3𝑥−1 𝑦 ′ = 10𝑥𝑒 + 15𝑥 𝑒 𝑦 ′ = 5𝑥𝑒 3𝑥−1 2 + 3𝑥 ′ 2 𝑜𝑟 + 2𝑥 −6 h) 𝑦 ′ = − 2 2𝑥 𝑒 𝑥 𝑒 𝑥 2𝑥 −3 −𝑥 4 + 6𝑥 −4 +4𝑥 3 𝑒 𝑥 +1 3 2𝑥 −3 −𝑥 4 2 g) 𝑦 ′ = −36𝑥 −7 𝑥 6 − 6𝑥 + 6 6𝑥 5 − 6 𝑥 Question 6: 2 a) 𝑦 ′ = − 2 b) 𝑦 = c) 𝑦 = ′ ′ ′ 𝑥−1 4 1−2𝑥 2 3𝑥 2 −4𝑥+3 1−𝑥 2 2 4𝑥 i) 𝑦 = 42𝑥 2𝑥 3 − 1 𝑒 𝑥 +1 2 6 1 + ln 𝑒 𝑥 + 1 2 + 2𝑥 3 − 1 7 6 𝑜𝑟 21𝑥 1 + ln 𝑒 𝑥 + 1 2 𝑦 ′ = 2𝑥 2𝑥 3 − 1 d) 𝑦 = − e) f) g) 𝑦 ′ = 𝑦 ′ = 𝑦 ′ = ′ 𝑥 2 −1 2 −130 𝑥 −3 +39𝑥 −2 +1 5−𝑥 2 −𝑥 4 −4𝑥+6𝑥 2 𝑥 2 −2 2 −39𝑥 −4 −3𝑥 2 +13𝑥 −8 −7𝑥 −2 1+𝑥 −4 2 2 +1 2 +1−18𝑥 3 𝑥 2 +1 2 +3𝑥 2 +6𝑥 𝑥 2 +1 3 6𝑥 𝑥 1−3𝑥 2 2 + 𝑒 𝑥 2𝑥 3 − 1 𝑒 𝑥 + 1 5 𝑥 𝑥 5 +𝑥 −7 2 2 2 j) 𝑦 ′ = 7𝑥 6 7𝑒 7𝑥−1 +4𝑥 3 ln 𝑥 5 +𝑥 −7 − −7𝑥 −8 𝑒 7𝑥−1 +𝑥 4 ln + h) 𝑦 = Question 7: a) 𝑦 ′ = 2𝑒 2𝑥 𝑥 7 −1 Question 10: a) 𝑦 ′ = 4𝑥 3 + 45𝑥 2 𝑦 ′′ = 12𝑥 2 + 90𝑥 72𝑥 3 − 182 . 25𝑄 b) 𝜋 = 10 + 0. 14 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡 6 9 3 1. Question 8: a) 𝐴𝜋 = −𝑄 + 20 𝑀𝜋 = −2𝑄 + 20 c) 122468 14. 𝑝. 𝑝.312 3𝑑. 𝑝.708 3𝑑.719 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 6 183 .25𝑥 −0. − 1 − 2.16𝑥 7 + 8 𝑥 2 − 1 3 + 48𝑥 2 𝑥 2 − 1 2 2 d) 𝑦 ′ = 𝑒 2 −4𝑥 3 + 2𝑥𝑒 𝑥 −1 1 − 𝑥 − 𝑒 2 1 − 𝑥 2 + 𝑒𝑥2−1 𝑜𝑟 𝑦 ′ = 𝑒 2 −4𝑥 3 + 2𝑥𝑒 𝑥 −1 + 4𝑥 4 − 2𝑥 2 𝑒 𝑥 −1 − 1 2 + 𝑥 2 − 𝑒 𝑥 −1 2 2 𝑦 ′′ = 𝑒 2 −12𝑥 2 + 2𝑒 𝑥 −1 + 4𝑥 2 𝑒 𝑥 −1 + 16𝑥 3 2 2 − 6𝑥𝑒 𝑥 −1 − 4𝑥 3 𝑒 𝑥 −1 + 2𝑥 e) 𝑦 ′ = 𝑦 ′′ = 3𝑥 2 −1−72𝑥 3𝑥 2 +1 2 2 2 2 Question 11: 0. −4 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 b) 2. 𝐹 Global minimum: 𝐴 Global maximum: 𝐵 Question 2: a) 2𝑥 + 3 = 0 b) 45𝑥 2 − 4 1 − 𝑥 3 = 0 2 2 c) 24𝑥𝑒 1−𝑥 − 24𝑥 3 𝑒 1−𝑥 = 0 𝑜𝑟 2 24𝑥𝑒 1−𝑥 1 − 𝑥 2 = 0 d) 4 𝑥 − 1 𝑥 3 − 1 2 + 4 𝑥 3 − 1 𝑥 − 1 4 𝑥 − 1 𝑥 3 − 1 𝑥 3 + 𝑥 − 2 = 0 2 e) 2 = 0 𝑥+1 c) d) 2 = 0 𝑜𝑟 e) Question 3: a) −1.500 Question 6: a) 𝑇𝑅 = 350𝑄 b) 𝜋 = 350𝑄 − 0.0 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 1 4 − .01𝑄+4 ln 125 −4 c) 𝑄 = ≈ 82. 13 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡 3 27 −3. 6 + 4 2 𝑒 −1− 2 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡 2 − 2.2𝑄2 − 40 𝑄𝐵𝐸 = 875 ± 𝑜𝑟 0. 𝑝. −27. 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑒 2 + 2.− 2 No inflection point Question 5: a) 𝑄 = 5 𝑎𝑛𝑑 𝑄 = 95 b) 𝑄 = 50 c) 𝜋 = 202. f) 𝑦 ′ = 𝑦 ′′ = 10 5𝑥 2 + 𝑥 − 10𝑥 + 1 2 5𝑥 2 + 𝑥 2 chapter six solutions Question 1: Local minima: 𝐶.0968 4𝑑. 𝐸 Local maxima: 𝐷. 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑒 e) 1 2 4 − 244 . −2𝑒 2 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡 0.27𝑥 8 + 8𝑥 𝑥 2 − 1 3 𝑦 ′ ′ = 2. 𝑝.5 + 19 1 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 − 1 .1𝑥 +2 60−3𝑥 2 + 15 6𝑥 − 72 3𝑥 2 + 1 10𝑥+1 5𝑥 2 +𝑥 − 12𝑥 3𝑥 2 + 1 3𝑥 2 − 1 − 72𝑥 3𝑥 2 + 1 4 𝑀𝐶 ′ 30 = −0. 83 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡 3 27 4+ 244 b) .5 2 1 2 𝑒 3. 6 − 4 2 𝑒 −1+ 2 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡 3 23 8 16 f) .8314 0.5 1 2 𝑒 3.0 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 4 5.127 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 6 1 13 .01 d) 𝑃 = 0.b) 𝑦 ′ = −68𝑥 −5 + 𝑥 −1 𝑦 ′′ = 340𝑥 −6 − 𝑥 −2 c) 𝑦 ′ = 0. e) 𝜋 = 5. −52 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 −5.082 3𝑑. 𝑝. −2𝑒 4 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 d) 3.5 + 19 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 f) 1 + 2.4 𝑄𝐵𝐸 ≈ 0 𝑜𝑟 𝑄𝐵𝐸 ≈ 1750 d) 𝑄𝑚𝑎𝑥 = 875 e) 𝜋 = 153. 1 14.25 + 4𝑥 𝑓 ′ 10 = 0.0 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 4 2.371 3𝑑.75 + 4 𝑓 ′ = 𝑥 0. 28 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 3 9 10 2 − . 𝑓 ′ 35 = 0. −𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 2. 𝑀𝐶 ′ 60 = −0.085 Question 7: a) 𝑇𝑅 = 10 + 0.193 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 c) 1.2𝑒 0. Question 14: 𝑑𝑇𝐶 a) 𝑀𝐶 = = 3𝑄2 − 30𝑄 + 75 b) 𝑄 = 5 c) 𝑀𝐶 ′ 15 = 60 Question 15: 𝑑𝑇𝐶 30𝑥 a) 𝑀𝐶 = = 2 b) 𝑀𝐶 ′ = 0.0282 4𝑑.25𝑄 − 0. 4+2 2 2 4−2 2 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 Question 4: a) No stationary points 1 25 .1𝑥 2 +2 2 𝑑𝑄 𝑑𝑄 0. 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 . 41. 00 269 𝜀𝑑 = −1 which is elastic in demand.473 3𝑑.5𝑄 + 2250 − j) 𝑄 = 30003 ≈ 208. Question 17: a) 𝑇𝐶 = 2500𝑄 + 1500 b) 𝑇𝑅 = −0.1𝑞 + 2.1𝑒 5−𝑄 f) 𝐴𝜋 = 𝑒 −𝑞+1 𝑀𝜋 = 𝑒 −𝑞+1 1 − 𝑞 Question 9: 𝑄 = 0 𝑜𝑟 𝑄 = 12.45 2𝑑.02𝑥 ln 𝑥 + 1 + e) 𝐴𝜋 = −0.4 𝑦 3.5𝑄2 + 100𝑄 1 b) 𝑄 = 0 𝑜𝑟 𝑄 = 33 c) 𝑄 = 22 9 d) 𝜋𝑚𝑎𝑥 = 8230.5𝑄2 + 2250𝑄 − 1500 d) 𝑄𝐵𝐸 ≈ 1 𝑎𝑛𝑑 𝑄𝐵𝐸 ≈ 4499 e) 𝑄𝑚𝑎𝑥 = 2250 f) 𝜋𝑚𝑎𝑥 = 2. 400 2 2 𝑄 2 3 1 ≈ 0.b) 𝐴𝐶 = 0.5𝑟 0. 2 Question 11: a) 𝐴𝜋 = −1. Question 14: 3 100 ln 6 + 18 𝜀𝑠 = ≈ 3.01 k) 𝜀𝑑 = −4 ∴ 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 l) %∆𝑇𝑅 = −30% Question 18: a) 𝑇𝐶 = 80𝑄 + 200 b) 𝑇𝑅 = −0.5𝑄2 + 2500𝑄 c) 𝜋 = −0.01𝑥 ln 𝑥 + 1 𝑀𝑅 = 1. 𝑝. so total revenue will increase.003.96 + 180 𝜀𝑑 = − ≈ −15.4𝑄2 + 4196𝑄 − 200 d) 𝑄𝐵𝐸 ≈ 0 𝑎𝑛𝑑 𝑄𝐵𝐸 ≈ 10489 e) 𝑄𝑚𝑎𝑥 = 5245 f) 𝜋𝑚𝑎𝑥 = 11.750 g) 𝑃 = 1375 h) 𝑀𝜋 = −𝑄 + 2250 1500 i) 𝐴𝜋 = −0.5 Question 10: 5 3 𝐿 = ≈ 1.810 g) 𝑃 = 2178 h) 𝑀𝜋 = −0. 20 Question 13: 𝑒 1.252 107 33 %∆𝑄 = −6 % 107 74 %∆𝑇𝑅 = 18 % 107 Since demand is inelastic.5 𝑄+1 𝐴𝑅 = 0.2𝑞 + 2.01𝑥 2 𝑥 + 1 𝑀𝜋 = −0.5 d) 𝐴𝐶 = 0. 𝑝.7 + 15𝑚 𝑝𝑚 = 10.5𝑄0.2𝑥 1.5 + 𝑀𝐶 = 0.96 Since 𝜀𝑑 > 1 the good is elastic at this price.69 1.5𝑟 −0.8 − 3𝑦 + 2𝑥 𝑧𝑦 = 86. 2 e) 𝑄 = 16 3 Question 12: 𝐿 = 4 2 %∆𝑇𝑅 = −0.3 𝑚−0.842 3𝑑.68𝑒 1. the increase in price will outweigh the decrease in quantity sold.3 + 15𝑟 d) 𝑧𝑥 = 43. Question 15: chapter seven solutions Question 1: a) 𝑧𝑥 = 2𝑦 + 3 + 8𝑥𝑦 3 𝑧𝑦 = 2𝑥 − 1 + 12𝑥 2 𝑦 2 b) 𝑓𝑥 = 48𝑥 3 + 𝑒 𝑦 𝑓𝑦 = 𝑥𝑒 𝑦 c) 𝑝𝑟 = 4.5 + c) 𝑀𝐶 = 0.5𝑄2 + 50𝑄 𝑀𝜋 = −4.4𝑄 + 4196 − j) k) l) 𝑄 = 5003 ≈ 63.25 so 𝑇𝑅 will decrease if the owner Question 16: 27 𝜀𝑑 = − ≈ 0.1𝑒 𝑄−5 + 0.84 154 Since 𝜀𝑠 > 1 this good must be elastic in supply.4𝑄2 + 4276 c) 𝜋 = −0.4𝑥 2.5 𝑄 10 𝑞 0.7 𝑚0.1 𝑒 𝑄 −5 +𝑒 5−𝑄 +10 𝑄 0. 𝑝.8𝑄 + 4196 200 i) 𝐴𝜋 = −0.5 𝑄0.4 𝑦 4.529.8 − 3𝑥 e) 𝑧𝑥 = 𝑥 + 𝑦 4 + 4 𝑥 + 𝑦 3 𝑥 − 𝑦 𝑜𝑟 𝑧𝑥 = 𝑥 + 𝑦 3 5𝑥 − 3𝑦 𝑧𝑦 = − 𝑥 + 𝑦 4 + 4 𝑥 + 𝑦 3 𝑥 − 𝑦 𝑜𝑟 𝑧𝑦 = 𝑥 + 𝑦 3 3𝑥 − 5𝑦 f) 𝑓𝑥 = 𝑦 3𝑥 − 4𝑦 2 4 18𝑥 − 4𝑦 2 𝑓𝑦 = 𝑥 3𝑥 − 4𝑦 2 4 −44𝑦 2 + 3𝑥 g) 𝑧𝑥 = 𝑧𝑦 = 2𝑥−𝑥 2 −2𝑥𝑦 −𝑦 3 1−𝑥−𝑦 2 2 2 𝑄 %∆𝑇𝑅 = −16 13 16 % h) 𝑧𝑥 = 11𝑦 2 1−𝑥𝑦 2 24𝑥𝑦 − 1 − 12𝑥 2 𝑦 2 𝑧𝑦 = 1 − 𝑥𝑦 2 i) 𝑧𝑥 = 𝑧𝑦 = j) k) l) 𝑧𝑥 𝑧𝑦 𝑧𝑥 𝑧𝑦 −2𝑥𝑦 2 +1 −2𝑦𝑥 2 1 − 𝑥 2 𝑦 2 + 𝑥 = 4𝑒 4𝑥−𝑦 𝑥 3 + 4𝑥𝑦 + 3𝑥 2 + 4𝑦 𝑒 4𝑥−𝑦 = −𝑒 4𝑥−𝑦 𝑥 3 + 4𝑥𝑦 + 4𝑥𝑒 4𝑥−𝑦 = 2𝑥𝑦 2 𝑒 4𝑥𝑦 + 4𝑦𝑒 4𝑥𝑦 𝑥 2 𝑦 2 − 1 = 2𝑦𝑥 2 𝑒 4𝑥𝑦 + 4𝑥𝑒 4𝑥𝑦 𝑥 2 𝑦 2 − 1 9𝑦 𝑥𝑦 −1−𝑥 𝑥𝑦 −1 2 1−𝑥 2 𝑦 2 +𝑥 𝑧𝑥 = 𝑧𝑦 = − 1 15−𝑥 9𝑥 𝑥𝑦 − 1 − 9𝑦 𝑥𝑦 − 1 2 − 𝑦 1+𝑥𝑦 −3𝑦 − 3𝑦 𝑥 + 2𝑦 3 + 𝑥 2 1 − 𝑥 − 𝑦 2 m) 𝑧𝑥 = 1−𝑒 𝑥𝑦 +𝑦𝑒 𝑥𝑦 1−ln 1+𝑥𝑦 1−𝑒 𝑥𝑦 2 184 . 𝑓𝑦𝑦 = 896𝑥 4 𝑦 5 + 36𝑥. Delta test confirms this. 2𝑥 2 +𝑦 2 2𝑥 2 +𝑦 Question 7: e) 𝑧𝑥𝑥 = 4𝑒 + 16𝑥 𝑒 𝑑𝑅 = 2421.418% d) 7.6 b) 𝜋 = 20𝐾 𝐿 − 5𝐾 − 50𝐿 − 399 good to make a big profit on another. 𝑧𝑦𝑦 = − + 30 d) 𝜋𝑚𝑎𝑥 = 7738. and delta test c) Two stationary points: 𝑃𝑜𝑖𝑛𝑡 1: 0.476% c) 7. c) Maximum profit at 𝐿 = 10. 𝑓𝑦𝑦 = 2𝑥 2 + 𝑦 3 2 Question 10: 2 12𝑥𝑦 a) 𝑇𝑅 = 300𝑇 + 150𝑅 𝑓𝑥𝑦 = 𝑓𝑦𝑥 = − 2𝑥 2 + 𝑦 3 2 b) 𝜋 = 300𝑇 + 150𝑅 − 3𝑇 2 − 4𝑅2 − 𝑅𝑇 − 400 26+2𝑥𝑦 −2𝑥 2 −𝑦 2 41 36 h) 𝑧𝑥𝑥 = + 30 c) 𝑇 = 47 𝑎𝑛𝑑 𝑅 = 12 with the SOC indicating a 13−𝑥𝑦 +𝑥 2 2 47 47 𝑥 2 maximum. .64 however the actual change in 𝑅 is 2𝑥 2 +𝑦 𝑧𝑦𝑦 = 𝑒 2409.04 8 8 f) 𝜋𝑚𝑎𝑥 = 168. 𝑥 2 +𝑦 2 Question 9: 𝑧𝑥𝑦 = 𝑧𝑦𝑥 = 4𝑥𝑦𝑒 − 4𝑦 1 1 −8𝑥 2 +4𝑦 3 Maximum profit is obtained at 𝐿 = . 5184 5184 Question 12: is a saddle point. 639 639 indicates a possible maximum. then it 0. 2 +𝑦 𝑧𝑥𝑦 = 𝑧𝑦𝑥 = 4𝑥𝑒 2𝑥 Question 8: 2 2 2 2 f) 𝑧𝑥𝑥 = 2𝑒 𝑥 +𝑦 + 4𝑥 2 𝑒 𝑥 +𝑦 a) 𝑑𝑄 = −0. −0. Delta test confirms this.5 𝐾 0. Delta test confirms this. The profit is 7. c) 𝜋 = 750𝑅 + 850𝐴 − 5𝑅2 − 3𝐴2 + 4𝐴𝑅 − 10000 6 4 3 5 3 5 d) 𝑅 = 179 𝑎𝑛𝑑 𝐴 = 261 with the SOC giving a 11 11 e) Stationary point at . SOC gives a possible b) 𝑇𝐶 = 650𝑅 + 350𝐴 + 10.1 e) −1. Delta test confirms this.0064𝐿0. SOC possible maximum. Delta test confirms this.96 𝑎𝑛𝑑 𝑃𝑅 𝑑𝑜𝑚 𝑒𝑠𝑡𝑖𝑐 = 319.09 f) Stationary point at . SOC gives a possible c) 𝜋 = 550 − 90𝐿2 − 135𝐾 2 + 112. 𝑧𝑥𝑦 = 𝑧𝑦𝑥 = 12𝑥 2 𝑦 3 + 2 Question 6: d) 𝑓𝑥𝑥 = 192𝑥 2 𝑦 8 . SOC gives 581 319 d) 𝐿 = 3 𝑎𝑛𝑑 𝐾 = 3 for maximum profit.5 0.8601.08 e) 𝑄 = 13. That is take a small loss on one 0.48 2𝑑. The 2 4 12𝑦𝑥 − 3𝑦 maximum profit attainable is 1.714% b) 6. and the delta test proves this.897% − 1 − 𝑒 𝑥𝑦 + 𝑥𝑒 𝑥𝑦 1 − ln 1 + 𝑥𝑦 𝑥 1 + 𝑥𝑦 185 . e) 𝑃𝐴 𝑑𝑜𝑚𝑒𝑠𝑡𝑖𝑐 = 1259. SOC says the second point is a a) 𝑇𝑅𝑅 = 𝑅2 − 4𝐴2 + 𝐴𝑅 + 1400𝑅 possible minimum.80.2144.198 3𝑑.952% b) −11. 24 . 𝑑𝑧 ≈ 1.385% c) 3. 𝑧𝑦𝑦 = 12𝑥 3 𝑦 2 . delta test confirms this.0563. SOC says the first point e) 𝜋𝑚𝑎𝑥 = 2039.409.486.5 0. and the delta test b) 𝑧𝑥𝑥 = 2𝑦. and the delta test confirming this.5𝐿𝐾 + 310𝐿 + maximum.7583 4𝑑. SOC provides a possible maximum.016𝐾 0. 𝑧𝑦𝑦 = 2.0 .357% d) 3. 𝑧𝑦𝑦 = 4𝑥. 𝑝.0 2 confirms this.6 a) 𝑇𝑅 = 1 + 20𝐾 𝐿 does make sense. . SOC gives possible 4 3 possible maximum.25 .4 𝑥 2 +𝑦 2 2 𝑥 2 +𝑦 2 𝑧𝑦𝑦 = 2𝑒 + 4𝑦 𝑒 − 4𝑥 b) Exact change in output is −1.75. 𝑝. 𝑧𝑥𝑦 = 𝑧𝑦𝑥 = −3 The SOC gives a possible maximum.4 𝐿0.105% e) 110. The SOC g) 𝑓𝑥𝑥 = 2 3 2 4 6 2𝑥 +𝑦 indicates a maximum.107 𝑧𝑦 = 1 − 𝑒 𝑥𝑦 2 Question 5: Question 2: There is a stationary point at 𝐾 = 12. 𝐾 = .5 + 0.7579 and the actual change in 𝑧 is 𝑓𝑥𝑦 = 𝑓𝑦𝑥 = 512𝑥 3 𝑦 7 + 36𝑦 − 13 1. refrigerator is greater than the sales price. 3 1 1 3 𝑃𝑜𝑖𝑛𝑡 2: . 𝐿 = 12. a) 𝑧𝑥𝑥 = 2. c) 𝑧𝑥𝑥 = 6𝑥𝑦 4 . . 𝑧𝑥𝑦 = 𝑧𝑦𝑥 = 4𝑦 + 2𝑥 confirms this. maximum. chapter eight solutions Question 1: Question 2: a) 2.5𝐿𝐾 + 360𝐿 + 2 2 13 − 𝑥𝑦 + 𝑥 540𝐾 Question 3: b) 𝑇𝐶 = 50𝐿 + 35𝐾 + 550 a) Stationary point at 0. and the delta test proving this.000 11 11 maximum.781% a) 14. 505𝐾 b) Stationary point at −1.30 13 − 𝑥𝑦 + 𝑥 2 2 2 Question 11: −13 + 𝑥 𝑧𝑥𝑦 = 𝑧𝑦𝑥 = − 30 a) 𝑇𝑅 = 1100 − 90𝐿2 − 135𝐾 2 + 112. SOC says it is an inflection 5 5 g) The prices don’t make sense as the cost of the point.0874 4𝑑. 𝑝. 𝑝. 𝑇𝑅𝐴 = 𝐴2 − 6𝑅2 + 3𝐴𝑅 + 1200𝐴 5 9 d) Stationary point at . . 𝐾 = 26.214.d) 𝑃 = 20. however Question 4: due to the interdependence of the two goods. 43 c) 12𝑦 6 𝑚𝑜𝑛𝑡𝑠 d) 5099.60 Question 7: a) 6000 b) 10.349%).31 b) 7228.54 ≈ 2091.1% offered by Bank 1.57 Question 5: a) Investment 2 (7092. Question 3: a) 12 b) 18 c) 54 d) 138 Question 4: a) 𝑇15 = 1331 4 14 11 4 26 11 b) 𝑇21 = 1331 c) d) 𝑆11 = 𝑆16 = 1331 4 11 −1 11 7 − 11 4 16 1331 −1 11 7 − 11 ≈ 2091. Question 13: a) 𝑁𝑃𝑉 = 14.189.313.5% c) The project is worthwhile as 𝑁𝑃𝑉 > 0 and 𝐼𝑅𝑅 > 𝑚𝑎𝑟𝑘𝑒𝑡 𝑟𝑎𝑡𝑒 Question 14: a) 𝑁𝑃𝑉 = −122.494. c) Investment 2 has a return of 55.99 and 𝐼𝑅𝑅 ≈ 8.18) up to the fifth year. Question 12: Bank of Vark has the best effective rate of 8.98 b) 𝐼𝑅𝑅 ≈ 20.322%) and Universal Bank (8.67 b) 7𝑦 11𝑚𝑜𝑛𝑡𝑠 c) 14𝑦 8𝑚𝑜𝑛𝑡𝑠 Question 10: 4114.500) is better than investment 2 (28.35 Question 9: a) 16322.994% g) A big drop in price in between.368%.09 whilst investment 1 has a return of 55.60 b) 1601.98 Question 11: Bank 2 is a better deal at the effective rate 5. 186 .f) −1.56) is better than investment 1 (7000) in the fifth year. as more could be earned from the market interest rate.125% is better than the 5.000.51% b) The project is not worthwhile. which is better than Sun Bank (8. b) Investment 1 (32.000 Question 8: a) 5687. Question 6: a) 171. nature (120) Optimisation single variable application (124) graphical (118) mathematical (118) multiple variable (151) H Hyperbolic functions (90) I Index numbers (161) Indices (14) Inequalities (26) Inflection point (123) Interest rates annual (170) compound (168) simple (167) Internal rate of return (174) Intersecting lines (38) P Partial differentiation Profit (126) B BIMDAS (18) Break-even (129) Q Quadratic Functions defining (75) graph (76) sketching (77) C Compound interest (168) Cubic functions (80) D Determinant (65-66) Differentiation multiple variable (140) partial applications (147. graphing (34) introduction (31) macroeconomic applications (40) main features (32) obtaining equations (36) Logarithm applications (91) functions (84) graphs (88) rules (84) T Total Revenue definition (127) elasticity (135) V Variables. defining (13) Z Zero (27) M Marginal value (130) Matrix applications (62) determinant (65-66) introduction (54) notes (61) solving 2 × 2 (55) solving 3 × 3 (59) Multiple variable differentiation (140) Multiplying fractions (7) E 𝑒 (85) Elasticity differentiation (133) interpreting (44) introduction (42) total revenue (135) Equations (21) Exponential functions (82) N Negative numbers (6) Net present value (172) Non-linear function (74) F Factorisation (25) Fractions (7-12) 187 .155) complex (144) simple (142) second order (146) rules Chain (104) 𝑒 (109) ln (110) power (103) product (105) quotient (108) single variable elasticity (133) first principles (99) introduction (97) second derivative (111) total (148) Dividing fractions (7) S Simple interest (167) Simultaneous equations introduction (50) two (50) three (52) J Jacobian Determinant (68) L Linear Equations.Index A Absolute values (26) Adding fractions (8) Annual interest rates (170) Arithmetic average (162) Arithmetic series (163) Averages arithmetic (162) geometric (162) standard (130) G Geometric average (162) Geometric series (165) Gradients function (112) negative (33) O Optimal point.