Linear Programming Formulation Exercises from TextbookISM 4400, Fall 2006: Page 1 /14 SOLUTIONS TO SELECT PROBLEMS FROM CHAPTER 7 7-14 The Electrocomp Corporation manufactures two electrical products: air conditioners and large fans. The assembly process for each is similar in that both require a certain amount of wiring and drilling. Each air conditioner takes 3 hours of wiring and 2 hours of drilling. Each fan must go through 2 hours of wiring and 1 hour of drilling. During the next production period, 240 hours of wiring time are available and up to 140 hours of drilling time maybe used. Each air conditioner sold yields a profit of $25. Each fan assembled may be sold for a $15 profit. Formulate and solve this LP production mix situation to find the best combination of air conditioners and fans that yields the highest profit. Use the corner point graphical approach. Let X1 = the number of air conditioners scheduled to be produced X2 = the number of fans scheduled to be produced Maximize Subject to: 7-15 25 X31 2X1 X1 + + + 15 ≤ X22 XX ≤ 2 X1, 2 ≥ 0 X2 Optimal Solution: X1 = 40 X2 = 60 (maximize profit) (wiring capacity constraint) (drilling capacity constraint) (non-negativity constraints) Profit = $1,900 24 0 14 0 Electrocomp’s management realizes that it forgot to include two critical constraints (see Problem 7-14). In particular, management decides that to ensure an adequate supply of air conditioners for a contract, at least 20 air conditioners should be manufactured. Because Electrocomp incurred an oversupply of fans in the preceding period, management also insists that no more than 80 fans be produced during this production period. Resolve this product mix problem to find the new optimal solution. Let X1 = the number of air conditioners scheduled to be produced X2 = the number of fans scheduled to be produced Maximize Subject to: 25 X31 2X1 XX 1 + + + 15 X22 XX 2 2 ≤ 24 0 ≤ 14 02 ≥ 08 ≤ ≥ 0 0 (maximize profit) (wiring capacity constraint) (drilling capacity constraint) (a/c contract constraint) (maximum # of fans constraint) (non-negativity constraints) X X1, 2 X2 Optimal Solution: X1 = 40 X2 = 60 Profit = $1,900 1 The firm also has a stock of 3500 feet of good-quality redwood.000 for last-minute advertising in the days preceding the election. benches and picnic tables. In planning the advertising campaign.20 0 35X2 ≤ 3. each picnic table takes 6 labor hours and 35 feet of redwood. The firm has two main resources: its carpenters (labor force) and a supply of redwood for use in the furniture. How many benches and tables should Outdoor Furniture produce to obtain the largest possible profit? Use the graphical LP approach.000 people. ≥ 0 12 X2 Optimal Solution: X1 = 262.000 people. Each bench that Outdoor Furniture produces requires 4 labor hours and 10 feet of redwood.50 .Linear Programming Formulation Exercises from Textbook ISM 4400. Let X1 = the number of benches produced X2 = the number of tables produced Maximize Subject to: 9 4X1 10X1 X1 + + + 20 X62 ≤ 1. the fourth constraint would be rewritten as: X1 − X2 ≥ 0 Optimal Solution: X1 = 175 7-17 X2 = 10 Exposure = 595.000 + X2500 X2 X X 2 1 ≤ 40. but she has stipulated that at least 10 ads of each type must be used. the number of radio ads must be at least as great as the number of television ads. Each television ad costs $500 and reaches an estimated 7. for use in yards and parks.862. Completed benches will yield a profit of $9 each. the campaign manager would like to reach as many people as possible. Two types of ads will be used: radio and television. Also.50 0 XX .5 X2 = 25 (maximize profit) (labor hours constraint) (redwood capacity constraint) (non-negativity constraints) Profit = $2.200 hours of labor are available under a union agreement. 1. During the next production cycle. How many ads of each type should be used? How many people will this reach? Let X1 = the number of radio ads purchased X2 = the number of television ads purchased Maximize Subject to: 3.00 ≥ 0 1 ≥ 10 0 ≥ X ≥ 0 2 (maximize exposure) (budget constraint) (at least 10 radio ads purchased) (at least 10 television ads purchased) (# of radio ads ≥ # of television ads) (non-negativity constraints) X1. and tables will result in a profit of $20 each. Fall 2006: Page 2 /14 7-16 A candidate for mayor in a small town has allocated $40. 1 X2 For solution purposes.000 X1200 X1 X + 7.000 people The Outdoor Furniture Corporation manufactures two products. Each radio ad costs $200 and reaches an estimated 3. e.800 each.. Let X1 = the number of Alpha 4 computers scheduled for production next month X2 = the number of Beta 5 computers scheduled for production next month Maximize Subject to: 1. 2 X2 Optimal Solution: X1 = 40 X2 = 20 1 7-19 3 20 06 0 (minimize faculty salaries) (schedule at least 30 undergrad courses) at least 20 grad courses) (schedule (schedule at least 60 total courses) (non-negativity constraints) Cost = $160.500 in faculty wages. 2 ≥ 0 5 (non-negativity constraints) X2 Optimal Solution: X1 = 10 X2 = 24 Profit = $55. Determine the most profitable number of each model of minicomputer to produce during the coming month. Management insists that full employment (i.000 X2 + X X 2 ≥ ≥ ≥ ≥ 0 X1.Linear Programming Formulation Exercises from Textbook ISM 4400. and each graduate course costs $3. Fall 2006: Page 3 /14 7-18 The dean of the Western College of Business must plan the school’s course offerings for the fall semester. Alpha 4s generate $1. and Beta 5s yield $1. How many undergraduate and graduate courses should be taught in the fall so that total faculty salaries are kept to a minimum? Let X1 = the number of undergraduate courses scheduled X2 = the number of graduate courses scheduled Minimize Subject to: 2. It requires 20 labor hours to assemble each Alpha 4 computer and 25 labor hours to assemble each Beta 5 model. Each undergraduate course taught costs the college an average of $2. working 160 hours each per month. on its assembly line. Faculty contracts also dictate that at least 60 courses be offered in total. all 160 hours of time) be maintained for each worker during next month’s operations.000.200 profit per unit.200 X1 20 X1X + 1.500 X1 X 1 X + 3.200 . the Alpha 4 and the Beta 5. 5 workers x 160 X2 01 (make hours) at least 10 Alpha 4 computers) ≥ 0 X ≥ 1 (make at least 15 Beta 5 computers) 1 X1. The firm employs five technicians. MSA wants to see at least 10 Alpha 4s and at least 15 Beta 5s produced during the production period.000 MSA Computer Corporation manufactures two models of minicomputers. Student demands make it necessary to offer at least 30 undergraduate and 20 graduate courses in the term.800 (maximize profit) + X2 25 = 80 (full employment. The total risk of the portfolio is found by multiplying the risk of each stock by the dollars invested in that stock.33 (which equates to an average risk of 283.000 X2 = $30.000). A risk index on a scale of 1–10 (with 10 being the most risky) is assigned to each of the two stocks.33/50. 1 X2 Optimal Solution: X1 = $16.000 Return = $4.X2 ≥ 0 (non-negativity constraints) X2 Optimal Solution: X1 = $20.00 (limit investment) 22 ≤ 0 0 (average risk cannot exceed 6) X1. How much should be invested in each stock? What is the average risk for this investment? What is the estimated return for this investment? Let X1 = the number of dollars invested in petrochemical stocks X2 = the number of dollars invested in utility stocks Maximize Subject to: 7-21 . Although a long-range goal is to get the highest possible return. Fall 2006: Page 4 /14 7-20 A winner of the Texas Lotto has decided to invest $50.000).000/50.67 + . suppose the investor has changed his attitude about the investment and wishes to give greater emphasis to the risk of the investment.666.000 (9 x $20.67 X2 = $33. Formulate this as an LP problem and find the optimal solution. .333.09 x 33. The total return would be $4000 (. 2 X02X . Referring to the Texas Lotto situation in Problem 7-20. How much should be invested in each stock? What is the average risk for this investment? What is the estimated return for this investment? Let X1 = the number of dollars invested in petrochemical stocks X2 = the number of dollars invested in utility stocks Minimize Subject to: 9 XX 1 .000 + 4 x $30. Now the investor wishes to minimize the risk of the investment as long as a return of at least 8% is generated.000 = 6). some consideration is given to the risk involved with the stocks.333. which yields an average risk of 6 (300.Linear Programming Formulation Exercises from Textbook ISM 4400.000 per year in the stock market. 1 04X + + − 4 XX 2 ≤ 50.00 ≥ 0 0 ≥ 0 (minimize total risk) (limit on total investment) (average return must be at least 8%) (non-negativity constraints) .000 = 5.67). 12XX 31 X1 + + − . which just happens to be a return of exactly 8% ($4000/$50.333.33).200 The total risk is 300. (maximize return on on total investment) 06XX ≤ 50. The following table provides a summary of the return and risk: Stock Petrochemical Estimated Return 12 6% % Utility Risk Index 9 4 The investor would like to maximize the return on the investment.666.33 Total risk = 283.333.12 x 16. but the average risk index of the investment should not be higher than 6. Under consideration are stocks for a petrochemical firm and a public utility. 67 1.000 in the next three years. Fall 2006: Page 5 /14 7-24 The stock brokerage firm of Blank. and Weinberger has analyzed and recommended two stocks to an investors’ club of college professors. (2) an appreciation of at least $5.818.359 X2 = $1. and dividend rates. and (3) a dividend income of at least $200 per year.177.00 (appreciation in next three X. per dollar invested Dividend rate Power .5 4% 8% potential Each member of the club has an investment goal of (1) an appreciation of no less than $720 in the short term. Leibowitz.24 1.2 0 20 (dividend years) ≥ income per year) 0 X08X . 1 36X 1. intermediate growth. Minimize Subject to: X .67 X.Linear Programming Formulation Exercises from Textbook ISM 4400. 2 ≥ 72 (appreciation in the short term) 24X ≥ 0 1.18 . ≥ 0 (non-negativity constraints) 1 X2 Optimal Solution: X1 = $1.50 5. What is the smallest investment that a professor can make to meet these three goals? Let X1 = the number of dollars invested in Louisiana Gas and Power X2 = the number of dollars invested in Trimex Insulation Co. The professors were interested in factors such as short term growth. These data on each stock are as follows: Stock Louisiana Gas and Trimex Insulation Factor Short term growth potential.18 Total investment = $3.1 04X + + + + X (minimize total investment) . per dollar invested Intermediate growth potential (over next three years).36 Company . 90XX 101 X1 12 X1 + + + + . How many pounds of beef and grain should be included in each pound of dog food? What is the cost and vitamin content of the final product? Let X1 = the number of pounds of beef in each pound of dog food X2 = the number of pounds of grain in each pound of dog food Minimize Subject to: .X2 ≥ 0 0 (non-negativity constraints) X2 Optimal Solution: X1 = . A pound of grain contains 6 units of Vitamin 1 and 9 units of Vitamin 2.7-25 Woofer Pet Foods produces a low-calorie dog food for overweight dogs. Each pound of beef costs $0.75 X2 = .25 Cost = $. Formulate this as an LP problem to minimize the cost of the dog food.60. This product is made from beef products and grain.90.825 . A pound of beef contains 10 units of Vitamin 1 and 12 units of Vitamin 2. (minimize cost per pound of dog food) 60XX = 1 (total weight should be one pound) 62 ≥ 9 (at least 9 units of vitamin 1 in a pound) 9X2 ≥ 1 (at least 10 units of vitamin 2 in a pound) X1. and each pound of grain costs $0. A pound of the dog food must contain at least 9 units of Vitamin 1 and 10 units of Vitamin 2. The table below contains all relevant information concerning production times per cabinet produced and production capacities for each operation per day. 71 2 Net Revenue per Cabinet 2 ($) 28 5 Let X1 = the number of French Provincial cabinets produced each day X2 = the number of Danish Modern cabinets produced each day Maximize Subject to: 28 X31 1. 2 ≤ 12 (finishing hours available) 75X ≥ 56 (contract requirement on F. X ≥ 1 cabinets) X1.M.P. 1 75XX + + + + 25 (maximize revenue) X22 ≤ 36 (carpentry hours available) XX 0 ≤ 20 (painting hours available) 2 0 . along with net revenue per unit produced. (b) Solve using an LP software program or spreadsheet. painting.930 . (a) Formulate as an LP problem.5X1 X . and finishing. Carpentr y (Hours/Cabine 3 2 3 6 Cabinet Style French Provincial Danish Modern Dept.SOLUTIONS TO SELECT PROBLEMS FROM CHAPTER 8 8-1 (Production problem) Winkler Furniture manufactures two different types of china cabinets: a French Provincial model and a Danish Modern model. Each cabinet produced must go through three departments: carpentry.1 2 0 Finishin g (Hours/Cabine . capacity (hrs) Paintin g (Hours/Cabine 1 . The firm has a contract with an Indiana distributor to produce a minimum of 300 of each cabinet per week (or 60 cabinets per day). Owner Bob Winkler would like to determine a product mix to maximize his daily revenue. 2 ≥ 0 0 (non-negativity constraints) X2 Optimal Solution: X1 = 60 X2 = 90 Revenue = $3. 7. cabinets) 06 (contract requirement on D. aerospace firms. - . + X1. -. aerospace.300 .00 0 0 ≥ 0 ≤ 0 X2. Subject to these restraints. + . + X + . United Aerospace Corp. she requests that the firm select whatever stocks and bonds they believe are well rated.8 4. X3. ≤ ≥ X 250. (b) Solve this problem. . X5 ≥ (total funds available) (municipal bond restriction) (electronics. prepare a list of high-quality stocks and bonds and their corresponding rates of return. (c) No more than 50% of the amount invested in municipal bonds should be placed in a highrisk.3 6.9 8. Inc. - .8-2 (Investment decision problem) The Heinlein and Krarnpf Brokerage firm has just been instructed by one of its clients to invest $250. + X + . (b) At least 40% of the funds should be placed in a combination of electronic firms. Projected Rate of Return (%) 5. X4. The client has a good deal of trust in the investment house. but within the following guidelines: (a) Municipal bonds should constitute at least 20% of the investment. - . aware of these guidelines. In particular. + .000 X2 = $0 X3 = $0 X4 = $175. Let X1 = dollars invested in Los Angeles municipal bonds X2 = dollars invested in Thompson Electronics X3 = dollars invested in United Aerospace X4 = dollars invested in Palmer Drugs X5 = dollars invested in Happy Days Nursing Homes Maximize . + X + .8 Investment Los Angeles municipal bonds Thompson Electronics.000 X5 = $25.5 (maximize return on investment) X + - . the client’s goal is to maximize projected return on investments.4 + . The analysts at Heinlein and Krampf.000 ROI = $20. and drug manufacturers. Palmer Drugs Happy Days Nursing Homes (a) Formulate this portfolio selection problem using LP.000 for her money obtained recently through the sale of land holdings in Ohio.4 11. + X Subject to: . drugs combo) (nursing home as a percent of 0bonds) (non-negativity constraints) Optimal Solution: X1 = $50. -. high-yield nursing home stock. - . but she also has her own ideas about the distribution of the funds being invested. M. and each works an 8-hour shift. 3 P.M. (Hint: Let Xi equal the number of waiters and busboys beginning work in time period i. staff size) (period 1) (period 2) (period 3) (period 4) (period 5) (period 6) (nonnegativity) Minimize X Subject to: X 1 + X 2 + X 3 + X 4 + X 5 1 X 1 + X + X 6 + X 6 + X 6 2 X 2 + X X 3 3 + X 4 X 4 + X 5 X 5 X .M–3 P.X .4.M.M. The following table shows the minimum number of workers needed during the six periods into which the day is divided. Chang’s scheduling problem is to determine how many waiters and busboys should report for work at the start of each time period to minimize the total staff required for one day’s operation.5..M. 7 AM.3. 7 P.6.6) ≥ 3 ≥ ≥ 1 21 ≥ 69 ≥ 1 14 ≥ ≥ 0 (min.X .M–7 P. 2.M. 11 A. Chang Restaurant is open 24 hours a day..M–7 A. The famous Y. Waiters and busboys report for duty at 3AM. where i = 1. or 11 P.5.M.2. Busboys Required 3 12 16 9 11 4 Let Xi = the number workers beginning work at the start of time period i (i=1. 11 P.M–11 P.) Number of Waiters and Period 1 2 3 4 5 6 Time 3 A.. 11 AM. S..4. 3 P.M.M–11 A.X .X . 7 P.8-3 (Restaurant work scheduling problem).3.X 1 2 3 4 5 6 .. 7 A..M–3 A.M. The feed mixes available for the horses’ diet are an oat product.8-4 (Animal feed mix problem) The Battery Park Stable feeds and houses the horses used to pull tourist-filled carriages through the streets of Charleston’s historic waterfront area.14 Oat Produc t 2 .5 $0. Diet Requirement (Ingredients) A B C D E Cost/lb Let Minimize s. 5X 6 2X3 1. 2 . Feed Mix Enrich ed Grain 3 1 5 1.5X2 X. At the same time.t.09 Minera l Produc 1 . recognizes the need to set a nutritional diet for the horses in his care. 1 5XX 1 + + + + + + + . units of each ingredient per pound of feed mix. a highly enriched grain.X1 5X 3 XX 1 . and a mineral product. The stable owner. Consequently. ≥ ≥ ≥ ≥ ≥ ≤ ≥ 6 2 9 8 5 6 0 (minimize cost) (ingredient A) (ingredient B) (ingredient C) (ingredient D) (ingredient E) (maximum feed per day) (non-negativity constraints) Minimum Daily Requirement 6 2 9 8 5 .5X3 XX33 X2.2 5XX 2 + + + + + + + . X1. 5 $0.5 . X3 X3 . he would like to keep the overall daily cost of feed to a minimum. 5 $0. an ex-racehorse trainer. he determines that 6 pounds of feed per day are the most that any horse needs to function properly. and costs for the three mixes. 53 1 . The table below shows these minimum requirements. 3 XX 2 52 1. Each of these mixes contains a certain amount of five ingredients needed daily to keep the average horse healthy.17 X1 = the number pounds of oat product per horse each day X2 = the number pounds of enriched grain per horse each day X3 = the number pounds of mineral product per horse each day . In addition. the stable owner is aware that an overfed horse is a sluggish worker. 56 2 1. Formulate this problem and solve for the optimal daily mix of the three feeds. (b) Solve the problem. a television spot costs $2.04X 03X + 2. 1 X2 ≥ ≥ ≥ 0 .000 + X2. The stores are expanding their lines of do-it-yourself tools. + . The Sunday newspaper has corresponding exposure rates of 4% and 3% per ad. Let Minimize Subject to: X1 = the number of newspaper ads placed X2 = the number of TV spots purchased 925 X. . a chain of four retail stores on Chicago’s North Side. and the other is for advertising time on Chicago TV. is considering two media possibilities. Diversey Paint would like to select the least costly advertising strategy that would meet desired exposure levels. and the advertising director is interested in an exposure level of at least 40% within the city’s neighborhoods and 60% in northwest suburban areas. One plan is for a series of half.05X X03X . The cost of a half-page Tribune advertisement is $925.000. The TV viewing time under consideration has an exposure rating per spot of 5% in city homes and 3% in the northwest suburbs.4 6 (minimize cost) (city exposure) (suburb exposure) (non-negativity constraints) . (a) Formulate using LP.8-6 (Media selection problem) The advertising director for Diversey Paint and Supply.page ads in the Sunday Chicago Tribune newspaper. 1 . .25X 394X 1 + 5 + 3 + 2 6 + 4 2 3 4 3 1216X . is responsible for formulating a nutritious meal plan for students.2 3.X .T.X .1X + 4 2 + 279X 6 .8X 74X 2.2 4.2X 7 (Fat) 1. (4) at least 26 grams of protein.5X + . On a particular day.17X 1 .35 1.X . campus dietician for a small Idaho college.33 What combination and amounts of food items will provide the nutrition Roniger requires at the least total food cost? Let X1 = the number of pounds of milk per student in the evening meal X2 = the number of pounds of ground meat per student in the evening meal Etc.2X + 9X 2.2X . down to X7 = the number of pounds of potatoes per student in the evening meal ≥ ≤ 9 150 ≥ 00 4 ≤ ≥ ≤ 5 02 65 0 Minimize .58X + 358X 5 + + 394X 1 + . she feels that the following five meal-content requirements should be met: (1) between 900 and 1.15X 7 (Cal. (3) no more than 50 grams of fat.15 2.) 295X 118X 128X 3.2 Calorie s/ 295 1216 394 358 128 118 279 Food Item Milk Ground Meat Chicken Fish Beans Spinach Potatoes Table of Food Values and Costs Fat Protein Carbs (gm/lb) (gm/lb) .25 0.2 0.33X 7 + 5 + 358X + 4.X .) 22X 1 1 + + 1216X 1.500 calories.5X 81X + 7 (Iron) 14.4 14 19 0.X 1 2 3 4 5 6 7 7 7 ≥ 0 .8 7 28 1.2X + .60 2.1 2. Roniger’s food stock includes seven items that can be prepared and served for supper to meet these requirements.) 295X 118X + 3.3 3.2 14.3X + 96X 4 3 2 7 5 + 14X 6 + 8X 5 + 19X 6 + 63X X . The cost per pound for each food item and the contribution to each of the five nutritional requirements are given in the accompanying table: Iron (mg/lb) 0.8-11 (College meal selection problem) Kathy Roniger.6X S. 16 16 22 96 81 0 9 74 0 0.X .35X + 279X + 5 + + 1 + 6 + 1 + 3 (Protein) 16X (Carbs. (Cal.4X 4 + + 7X + 28X + 128X 3 2 16X 6 6 4 2 + 1. For an evening meal.17 0. and (5) no more than 50 grams of carbohydrates.2X 5 + + 83X 2. (2) at least 4 milligrams of iron.5 83 0 0.58 1.5 8 63 Cost/ Pound 0. 75 S. Quitmeyer Electronics wants to maximize its profits. variable labor costs are $15 per hour for test device 1. $12 per hour for test device 2. on which equipment? Let X1 = the number of internal modems scheduled for manufacture each week X2 = the number of external modems scheduled for manufacture each week Etc. (a) Formulate this problem as an LP model. as shown in the table the following table: Test device 1 Test device 2 Test device 3 Internal Modem 7 2 5 External Modem 3 5 1 Circuit Board 12 3 3 CD Drive 6 2 2 Hard Drive 18 15 9 Memory Board 1 17 72 The first two test devices are available 120 hours per week.80 + 191. Each of these technical products requires time. external modems. (b) Solve the problem by computer. 7 + 3 + 12 + 6 + 18 + 17 ≤ 2 + 5 + 3 + 2 + 15 + 17 ≤ 5 + 1 + 3 + 2 9 + 2 + X1. The table that follows summarizes the revenues and material costs for each product: Device Internal modem External modem Graphics circuit board CD drive Hard disk drive Memory expansion board Revenue Per Material Cost Unit Sold ($) 200 120 180 130 430 260 Per Unit ($) 35 25 40 45 170 60 In addition. X6 ≤ ≥ 0 72 00 72 00 60 00 . The market for all six computer components is vast. on three types of electronic testing equipment. X5.T. X4.95 + 135.35 + 92. and memory expansion boards. What is the best product mix? (c) What is the value of an additional minute of time per week on test device 1? Test device 2? Test device 3? Should Quitmeyer Electronics add more test device time? If so. in minutes. The third (device 3) requires more preventive maintenance and may be used only 100 hours each week. X2.8-12 (High tech production problem) Quitmeyer Electronics Incorporated manufactures the following six microcomputer peripheral devices: internal modems.50 + 249. graphics circuit boards. down to X6 = the number of mem. and $18 per hour for test device 3.. X3.50 + 82. expansion boards scheduled for mfg. each week Maximize 161. and Quitmeyer Electronics believes that it can sell as many units of each product as it can manufacture. CD drives. hard disk drives. 0 3 .09 Let X1 = the number of pounds of alloy 1 in one ton of steel X2 = the number of pounds of alloy 2 in one ton of steel Etc.0 25. 09 0.0 20.30X 1 . 0.01X 3 .05X 5 4 .0 30.7X (Mn-max) . The Japanese auto manufacturer has strict quality control standards for all of its component subcontractors and has informed Amalgamated that each frame must have the following steel content: Material Manganese Silicon Minimum Percent 2.8-15 (Material blending problem) Amalgamated Products has just received a contract to construct steel body frames for automobiles that are to be produced at the new Japanese factory in Tennessee.25X 7 . .05X 5 4 + .55X + ..12X S. The table below details these materials.10X 5 + 6 + .15X 1 + .0 1 5. 15 0. 3 6 5.0 2.05 Carbon Maximum Percent 2. down to X8 = the number of pounds of carbide 3 in one ton of steel ≥ 4 24 ≤ 68 69 ≥ ≤ ≥ ≤ ≤ ≤ ≤ ≤ = 12 01 03 05 20 01 200 00 Minimize . 12 0. . 10 0.0 12 . .07X 4 + .13X 1 + .12X 2 + .01X 3 + .12X 6 + 7 + .0 Poun ds Availab No limit 30 No0limit No limit No limit 5 0 20 0 100 Cost Per Pound ($) 0.15X 2 + . (Mn- .00 Carbo n 3( 1. 4.55X 1 + . Material Available Alloy 1 Alloy 2 Alloy 3 Iron 1 Iron 2 Carbide 1 Carbide 2 Mangane se ( 70 55 .0 .0 0 0 Carbide 3 Silico n ( 15.0 23.12X 2 2 + + . 12 13 0.3 5.26X 3 + + .09X + 8 .0 25.1 4.T.025X .0 26.35 Amalgamated mixes batches of eight different available materials to produce one ton of steel used in the body frames.0 10.10X 4 + 5 + . 07 0.7X (Si-min) + . 5 24.0 18.09X 3 + .24X . Formulate and solve the LP model that will indicate how much of each of the eight materials should be blended into a 1-ton load of steel so that Amalgamated meets its requirements while minimizing cost. 03X Alloy 2 lim.26X .18X .23X + 6 .23X 24X 6 8 (Si-max) + .X .18X 3 + + .03X .15X . X 2 + . .25X + 1 .01X .X .20X + .X . Weighs 1 ton 4 + 2 + 7 + 2 + X 3 + X 4 + X 5 + X 6 + X . X X 1 + X 8 2 Carbide 1 lim.10X .20X 7 + 4 + .03X 1 8 (C-min) .01X 4 .X .X . X 7 Carbide 3 lim.X .X 1 2 3 4 5 6 7 8 X + 7 ≥ 8 X 0 8 .25X X 6 Carbide 2 lim.025X + 5 + 2 8 (C-max) .25X .03X1 6 + .30X + 7 + .