ADDITIONALMATHEMATIC SPROJECT WORK 2015 Name: Muhammad Danial Hogan Class: 5 Science 1 IC: 980508146489 Instructor's Name: Mr. Faizal 1 Contents Number Content Page 1 Acknowledgement 3 2 Objective 4 3 Introduction 5 4 Part 1 6 5 Part 2 8 6 Part 3 10 7 Further Exploration 12 8 Reflection 17 Acknowledgement 2 First and foremost, I would like to thank Allah for the energy and determination to do this project. Next, my Additional Mathematics teacher,Mr. Faizal as he gives us consistent guidance during this project work. He has been a very supportive figure throughout the whole project. I also would like to give thanks to all my friends for helping me and always supporting me to complete this project work. They have done a great job at providing different reference and sharing information with other people including me. Without them this project would never have had its conclusion. Last but not least, for their strong support, I would like to express my gratitude to my beloved parents. Also for supplying the equipments and money needed for the resources to complete this project. They have always been by my side and I hope they will still be there in the future. 3 Objectives The aims of carrying out this project work are: I. II. III. IV. V. VI. VII. To apply and adapt a variety of problem-solving strategies to solve problems To improve thinking skills To promote effective mathematical communication To use the language of mathematics to express mathematical ideas precisely To provide learning environment that stimulates and enhances effective learning To develop positive attitude towards mathematics To develop mathematical knowledge through problem solving in a way that increases students interest and confidence in the subject. 4 Introduction In mathematics, the maximum and minimum of a function,known collectively as extrema are the largest and smallest value that the function takes at a point either within a given neighbourhood (local or relative extremum) or on the function domain in its entirety (global or absolute extremum). Pierre de Fermat was one of the first mathematicians to propose a general technique (called adequality) for finding maxima and minima. To locate extreme values is the basic objective of optimization. Linear programming started when Fermat and Lagrange found calculus-based formulas for identifying optima, while Newton and Gauss proposed iterative methods for moving towards an optimum. The term "linear programming" for certain optimization cases was due to George B. Dantzig, although much of the theory had been introduced by Leonid Kantorovich in 1939. (Programming in this context does not refer to computer programming, but from the use of program by the United States military to refer to proposed training and logistics schedules, which were the problems Dantzig studied at that time.) Dantzig published the Simplex algorithm in 1947, and John von Neumann developed the theory of duality in the same year. 5 Part 1 (a) i. Mathematical optimization deals with the problem of finding numerically minimums (or maximums or zeros) of a function. In this context, the function is called cost function, or objective function, or energy.It is a branch of mathematics that focuses on problems where scarce resources need to be allocated effectively, in complex, dynamic and uncertain conditions. The program combines a solid foundation in math with special sequences of courses in economics, business, and management science. In the simplest case, an optimization problem consists of maximizing or minimizing a real function by systematically choosing input values from within an allowed set and computing the value of the function. The generalization of optimization theory and techniques to other formulations comprises a large area of applied mathematics. More generally, optimization includes finding "best available" values of some objective function given a defined domain (or a set of constraints), including a variety of different types of objective functions and different types of domains. ii. In mathematical analysis, the maxima and minima (the plural of maximum and minimum) of a function, known collectively as extrema (the plural of extremum), are the largest and smallest value of the function within the entire domain of a function (the global or absolute extrema). We say that f(x) has an absolute (or global) maximum at x=c if f(x) ≤ f (c) for every x in the domain we are working on. We say that f(x) has an absolute (or global) minimum at x=c if if f(x) ≥ f (c) for every x in the domain we are working on. iii. In mathematical analysis, the maxima and minima (the plural of maximum and minimum) of a function, known collectively as extrema (the plural of extremum), are the largest and smallest value of the function, within a given range (the local or relative extrema) We say that f(x) has a relative (or local) maximum at x=c if f(x) ≤ f (c) for every x in some open interval around. We say that f(x) has a relative (or local) minimum at x=c if f(x) ≥ f (c) for every x in some open interval around. 6 (b) 1st Method Decide whether you're going to find the maximum or minimum value. It's either one or the other, you're not going to find both. The maximum / minimum value of a quadratic function occurs at it's vertex. For y = ax2+bx+c. (c-b2/4a) gives the y-value (or the value of the function) at i's vertex. If the value of a is positive, you're going to get the minimum value because as such the parabola opens upwards (the vertex is the lowest the graph can get) and vice versa. Finally, The value of a cannot be zero. 2nd Method dy/dx = 0.2ax+b = 0, -b/2a 7 3rd step dy/dx = 2ax + b 2nd step 1st step Differentiate x with respect to y. Determine the differentiation point values in terms of dy/dx. It can be found by setting these values equal to 0 and find the corresponding values. Substitute this value of x into y to get the minimum/maximum points. Part 2 y (a) x xx x y 4x+2y = 200 2x+y = 100 y = 100-2x Area of the pen = xy A = x(100-2x) A = 100x-2x2 da/dx = 100-4x When the area of the pen is maximum, da/dx = 0 100-4x = 0 -4x = -100 x = -100/-4 x = 25 m. Maximum area of the pen = 100 -2x2 =100(25) – 2(252) =1250 m2 8 30-2h (b) h 30 v = h(30-2h)(30-2h) v = h(900-120h+4h2) v = 900h-120h2+4h3 dv/dh = 900-240+12h2 When volume is maximum, dv/dh = 0 12h2-240h+900 = 0 h2-20h+75 = 0 (h-5)(h-15) = 0 h-5 = 0 or h=5 h-15 = 0 h = 15 h must be less than 15 h = 5, v = 900(5) – 120(52) + 4(h3) v = 2000 cm3 9 Part 3 (i) Based on the constructed number of hours hours and P people. t/ hours 0 1 2 3 4 5 6 7 8 9 10 11 12 13 P/ number of people 0 241 900 1800 2700 3359 3600 3359 2700 1800 900 241 0 241 equation, a table has been where t represents the starting from 0 hours to 23 represents the number of Based on the table above, a graph is generated using Microsoft Excel application. 10 (ii) The peak hours with 3600 people in the mall is after 6 hours the mall opens 9:30 a.m. + 6 hours = 3:30 p.m. (iii) (iv) 7:30 p.m. is 10 hours after the malls opens. Based on the graph, the number of people at the mall at 7:30 p.m. is 900 people. π 6 By using the formula, P(t) = + 1800 (¿t) −1800 cos ¿ π 6 2570 = + 1800 (¿t ) −1800 cos ¿ π 6 (¿t ) = cos ¿ 2570−1800 −1800 11 π t=¿ cos-1 -0.428 6 t = 3.84 hours t = 3 hours 50 minutes 9:30 a.m. + 3 hours 50 minutes = 1:20 p.m. Further Exploration a) Linear programming is a considerable field of optimization for several reasons. Many practical problems in operations research can be expressed as linear programming problems. Certain special cases of linear programming, such as network flow problems and multicommodity flow problems are considered important enough to have generated much research on specialized algorithms for their solution. A number of algorithms for other types of optimization problems work by solving LP problems as sub-problems. 12 Historically, ideas from linear programming have inspired many of the central concepts of optimization theory, such as duality, decomposition, and the importance of convexity and its generalizations. Likewise, linear programming is heavily used in microeconomics and company management, such as planning, production, transportation, technology and other issues. Although the modern management issues are ever-changing, most companies would like to maximize profits or minimize costs with limited resources. Therefore, many issues can be characterized as linear programming problems. The example of its uses in a daily life includes – Crew Scheduling i) ii) iii) iv) Make sure that each flight is covered Meet regulations, eg: each pilot can only fly a certain amount each day Minimize costs, eg: accommodation for crews staying overnight out of towns, crews deadheading Would like a robust schedule. The airlines run on small profit margins, so saving a few percent through good scheduling can make an enormous difference in terms of profitability. They also use linear programming for yield management b) (i) (a) I. Cost : 100x + 200y ≤ 1400 II. Space : 0.6x + 0.8y ≤ 7.2 III. Volume = 0.8x + 1.2y 13 (b) I. x y −1 x+7 2 0 7 II. x y y= y= 0 9 2 6 4 5 6 4 8 3 12 1 14 0 2 7.5 4 6 6 4.5 8 3 10 1.5 12 0 −3 x+ 9 4 14 15 (ii) Maximum storage volume Method 1 – Test using corner point of Linear Programming Graph (8, 3), (0, 7), and (12, 0) Volume = 0.8x + 1.2y Coordinate 1 - (8,3) Volume = 0.8(8) + 1.2(3) Volume = 10 cubic meter Coordinate 2 - (0,7) Volume = 0.8(0) + 1.2(7) Volume = 8.4 cubic meter Coordinate 3 - (12,0) Volume = 0.8(12) + 1.2(0) Volume = 9.6 cubic meter Thus the maximum storage volume is 10 cubic meter. Method 2 – Using simultaneous equation y= −1 x+7 2 y= −3 x+ 9 -------- 2 4 ------- 1 Substitute equation 1 into 2 −3 −1 x +9= x +7 4 2 x=8 y=3 Applying the value of x and y in formula , Volume = 0.8x + 1.2y Thus, the maximum storage volume is 10 cubic metre. 16 (iii) Carbinet x 4 5 6 7 8 9 Carbinet y 6 5 4 3 3 2 Total Cost (RM) 1600 1500 1400 1300 1400 1300 (iv) If I was Aaron, I would choose the combination of 6 cabinet x and 4 cabinet y. It statisfies the term which is the ratio of number of cabinet x to cabinet y is not less than 2:3. Furthermore, I have an allocation of RM1400 for the cabinets and this combination is affordable and just perfect for me as it costs exactly RM1400. The volume of the carbinet is also large enough. 17 Reflection I’ve found a lot of information while conducting this Additional Mathematics project. I’ve learnt the uses of function in our daily lives. I never thought addmaths could be used this way! Throughout this project, I have learned that hardwork is the key to success. My hardships in doing this project has made me realized that challenges can turn into something better. As a change, I have came to understand this topic better. I have also tightened the bonds between myself, my instructor and my friends through this project. We should always offer help to anyone who asks. I also obtained a brand new thinking skill and can practice effective mathematical communication thanks to this project work. In a nutshell, I think this project teaches a lot of moral values, and also tests the students’ understanding in Additional Mathematics. Let me end this project with a modified lyrics to Jason Mraz’s song, Make it mine; Wake up everyone How can you sleep at a time like this unless you’re gonna score A plus Listen to his voice, The man who teach everything from just the tip of his thumbs Making his students understand. I’m not gonna waste these times As they are precious The success is guaranteed It’s just that I know it 18 I’m gonna make it mine Yes I’ll make it all mine. 19