Solving linear programming problems using the graphical method. COPYRIGHT 2006 by LAVON B. PAGE
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1 Solving linear programming problems using the graphical method
2 Example - designing a diet A dietitian wants to design a breakfast menu for certain hospital patients. The menu is to include two items A and B. Suppose that each ounce of A provides 2 units of vitamin C and 2 units of iron and each ounce of B provides 1 unit of vitamin C and 2 units of iron. Suppose the cost of A is 4 /ounce and the cost of B is 3 /ounce. If the breakfast menu must provide at least 8 units of vitamin C and 10 units of iron, how many ounces of each item should be provided in order to meet the iron and vitamin C requirements for the least cost? What will this breakfast cost?
3 x = #oz. of A y = #oz. of B vit. C: 2x + y! 8 iron: 2x + 2y! 10 x! 0, y! 0 Cost = C = 4x+3y
4 x = #oz. of A y = #oz. of B vit. c vit. C: 2x + y! 8 iron: 2x + 2y! 10 x! 0, y! 0 Cost = C = 4x+3y
5 x = #oz. of A y = #oz. of B vit. c vit. C: 2x + y! 8 iron: 2x + 2y! 10 x! 0, y! 0 Cost = C = 4x+3y iron
6 x = #oz. of A y = #oz. of B vit. c vit. C: 2x + y! 8 iron: 2x + 2y! 10 x! 0, y! 0 Cost = C = 4x+3y iron
7 x = #oz. of A y = #oz. of B vit. c vit. C: 2x + y! 8 iron: 2x + 2y! 10 x! 0, y! 0 iron Cost = C = 4x+3y The 3 blue lines are 48 = 4x+3y 36 = 4x+3y 24 = 4x+3y
8 vit. c x = #oz. of A y = #oz. of B The cost will be minimized if the strategy vit. C: followed 2x + y! is 8the one corresponding iron: 2x + 2y to! this 10corner point x! 0, y! 0 iron Cost = C = 4x+3y The 3 blue lines are 48 = 4x+3y 36 = 4x+3y 24 = 4x+3y
9 x = #oz. of A y = #oz. of B vit. c vit. C: 2x + y! 8 iron: 2x + 2y! 10 x! 0, y! 0 2x + y = 8 2x + 2y = 10 iron
10 x = #oz. of A y = #oz. of B vit. c vit. C: 2x + y! 8 iron: 2x + 2y! 10 x! 0, y! 0 2x + y = 8 2x + 2y = 10 iron Solution: x=3, y=2 C = 4x + 3y = 18
11 x = #oz. of A y = #oz. of B vit. c corner pt. C = 4x + 3y (0,8) 24 cents (5,0) 20 cents (3,2) 18 cents iron
12 Example - bicycle factories A small business makes 3-speed and 10-speed bicycles at two different factories. Factory A produces 16 3-speed and speed bikes in one day while factory B produces 12 3-speed and speed bikes daily. It costs $1000/day to operate factory A and $800/day to operate factory B. An order for 96 3-speed bikes and speed bikes has just arrived. How many days should each factory be operated in order to fill this order at a minimum cost? What is the minimum cost? x = # days factory A is operated y = # days factory B is operated
13 x = # days factory A is operated y = # days factory B is operated
14 x = # days factory A is operated y = # days factory B is operated 3-speed constraint: 16x + 12y! 96
15 x = # days factory A is operated y = # days factory B is operated 3-speed constraint: 16x + 12y! speed constraint: 20x + 20y! 140 x! 0, y! 0
16 x = # days factory A is operated y = # days factory B is operated 3-speed constraint: 16x + 12y! speed constraint: 20x + 20y! 140 x! 0, y! 0 Minimize: C = 1000x + 800y
17 x = # days factory A is operated y = # days factory B is operated 3-speed constraint: 16x + 12y! speed constraint: 20x + 20y! 140 x! 0, y! 0 Minimize: C = 1000x + 800y 3-speed
18 x = # days factory A is operated y = # days factory B is operated 3-speed constraint: 16x + 12y! speed constraint: 20x + 20y! 140 x! 0, y! 0 Minimize: C = 1000x + 800y 3-speed 10-speed
19 x = # days factory A is operated y = # days factory B is operated 3-speed constraint: 16x + 12y! speed constraint: 20x + 20y! 140 x! 0, y! 0 Minimize: C = 1000x + 800y 3-speed 10-speed
20 x = # days factory A is operated y = # days factory B is operated 3-speed constraint: 16x + 12y! speed constraint: 20x + 20y! 140 x! 0, y! 0 Minimize: C = 1000x + 800y corner pts C = 1000x + 800y 3-speed 10-speed
21 x = # days factory A is operated y = # days factory B is operated 3-speed constraint: 16x + 12y! speed constraint: 20x + 20y! 140 x! 0, y! 0 Minimize: C = 1000x + 800y corner pts C = 1000x + 800y (0,8) $ speed 10-speed
22 x = # days factory A is operated y = # days factory B is operated 3-speed constraint: 16x + 12y! speed constraint: 20x + 20y! 140 x! 0, y! 0 Minimize: C = 1000x + 800y corner pts C = 1000x + 800y (0,8) $6400 (7,0) $ speed 10-speed
23 x = # days factory A is operated y = # days factory B is operated 3-speed constraint: 16x + 12y! speed constraint: 20x + 20y! 140 x! 0, y! 0 Minimize: C = 1000x + 800y corner pts C = 1000x + 800y (0,8) $6400 (7,0) $7000 (3,4) $ speed 10-speed
24 Example - ski manufacturing Michigan Polar Products makes downhill and crosscountry skis. A pair of downhill skis requires 2 man-hours for cutting, 1 man-hour for shaping and 3 man-hours for finishing while a pair of crosscountry skis requires 2 man-hours for cutting, 2 man-hours for shaping and 1 man-hour for finishing. Each day the company has available 140 man-hours for cutting, 120 man-hours for shaping and 150 man-hours for finishing. How many pairs of each type of ski should the company manufacture each day in order to maximize profit if a pair of downhill skis yields a profit of $10 and a pair of cross-country skis yields a profit of $8?
25 x = # pairs of downhill skis y = # pairs of cross country skis cutting: 2x + 2y " 140 shaping: x + 2y " 120 finishing: 3x + y " 150 x! 0, y! 0 P = 10x + 8y
26 x = # pairs of downhill skis y = # pairs of cross country skis cutting: 2x + 2y " 140 shaping: x + 2y " 120 finishing: 3x + y " 150 x! 0, y! 0 P = 10x + 8y cutting
27 x = # pairs of downhill skis y = # pairs of cross country skis cutting: 2x + 2y " 140 shaping: x + 2y " 120 finishing: 3x + y " 150 x! 0, y! 0 P = 10x + 8y cutting shaping
28 finishing x = # pairs of downhill skis y = # pairs of cross country skis cutting: 2x + 2y " 140 shaping: x + 2y " 120 finishing: 3x + y " 150 x! 0, y! 0 P = 10x + 8y cutting shaping
29 finishing x = # pairs of downhill skis y = # pairs of cross country skis cutting: 2x + 2y " 140 shaping: x + 2y " 120 finishing: 3x + y " 150 x! 0, y! 0 P = 10x + 8y cutting shaping
30 finishing x = # pairs of downhill skis y = # pairs of cross country skis cutting: 2x + 2y " 140 shaping: x + 2y " 120 finishing: 3x + y " 150 x! 0, y! 0 P = 10x + 8y cutting shaping
31 finishing x = # pairs of downhill skis y = # pairs of cross country skis cutting: 2x + 2y " 140 shaping: x + 2y " 120 finishing: 3x + y " 150 x! 0, y! 0 P = 10x + 8y cutting shaping corners P = 10x + 8y (0,0) (0,60) (20,50) (40,30) (50,0) $0 $480 $600 $640 $500
32 finishing cutting: 2x + 2y " 140 shaping: x + 2y " 120 finishing: 3x + y " 150 x! 0, y! 0 cutting x = # pairs of downhill skis y = # pairs of cross country skis Make 40 pairs of downhill skis and 30 pairs of cross country skis for a profit of $640 P = 10x + 8y corners P = 10x + 8y (0,0) (0,60) (20,50) (40,30) (50,0) $0 $480 $600 $640 $500
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