Chapter 3 Section 2

1. Chapter 3 Section 2 Linear Programming I

2. Fundamental Theorem of Linear Programming • The maximum (or minimum) value of the objective function is achieved at one of the vertices

3. Exercise 21 (page 131) • Minimize the objective function 3x + 4y subject to the below constraints. • Solution: 2x + y> 10 y> – 2x + 10 x + 2y> 14 y> – ½ x + 7 x> 0 x> 0 y> 0 y> 0

4. Graph of the Inequalities x = 0 I y = – 2x + 10 II y = – ½ x + 7 III y = 0

5. Finding the Vertices (by hand) Vertex IVertex IIVertex III x = 0 y = – 2x + 10 y = – ½ x + 7 y = – 2x + 10 y = – ½ x + 7 y = 0 y = – 2(0) + 10 – 2x + 10 = – ½ x + 7 ( 0 ) = – ½ x + 7 y = 10 3/2 x = 3 ½ x = 7 x = 2 x = 14 ( 0 , 10 ) y = – 2( 2 ) + 10 ( 14 , 0 ) y = 6 ( 2 , 6 )

6. Find the Optimal Point VertexObjective Function: 3 x + 4 y ( 0 , 10 ) 3 ( 0 ) + 4 ( 10 ) = 40 ( 2 , 6 ) 3 ( 2 ) + 4 ( 6 ) = 30 ( 14 , 0 ) 3 ( 14 ) + 4 ( 0 ) = 42 The minimum value of the objective function occurs at the vertex ( 2 , 6 )

7. Exercise 33 (page 132) • Define the variables being used. Look at the question being asked! • Last sentence: “How many cans of Fruit Delight and Heavenly Punch should be produced…?” • Let x represent the number of cans of Fruit Delight produced y represent the number of cans of Heavenly Punch produced

8. The Objective Function • Last sentence: “How many cans of Fruit Delight and Heavenly Punch should be produced each week to maximize profits?” • The objective function is: Profit = 0.20 · x + 0.30 · y

9. Table

10. Table

11. Table

12. Table

13. Restrictions from the Table and Problem • Restrictions that are placed on the x and y variables: 10 x + 10 y< 9,000 3 x + 2 y< 2,400 x + 2 y< 1,400 x> 0 y> 0

14. Change From General Form to Standard Form 10 x + 10 y< 9,000 y< – x + 900 3 x + 2 y< 2,400 y< – 3/2 x + 1,200 x + 2 y< 1,400 y< – ½ x + 700 x> 0 x> 0 y> 0 y> 0

15. Graph of the System of Equations x = 0 Not to scale y = 0 y = – 1/2 x + 700 y = – 3/2 x + 1200 y = – x + 900

16. Shading for the Inequalities x = 0 Not to scale y = 0 y = – 1/2 x + 700 y = – 3/2 x + 1200 y = – x + 900

17. A Modified Graph of the Feasible Set from the Previous Slide II y = – 1/2 x + 700 III y = – x + 900 IV x = 0 Feasible Set y = – 3/2 x + 1200 V I y = 0

18. Finding the Vertices (using a calculator when possible) • Vertex I x = 0 ( 0 , 0 ) y = 0 • Vertex II x = 0 ( 0 , 700 ) y = – ½ x + 700 • Vertex III y = – ½ x + 700 ( 400 , 500 ) y = – x + 900 • Vertex IV y = – x + 900 ( 600 , 300 ) y = – 3/2 x + 1200 • Vertex V y = – 3/2 x + 1200 ( 800 , 0 ) y = 0

19. Find the Optimal Point VertexObjective Function 0.2 x + 0.3 y ( 0 , 0 ) 0.2 ( 0 ) + 0.3 ( 0 ) = 0 ( 0 , 700 ) 0.2 ( 0 ) + 0.3 ( 700 ) = 210 ( 400 , 500 ) 0.2 ( 400 ) + 0.3 ( 500 ) = 230 ( 600 , 300 ) 0.2 ( 600 ) + 0.3 ( 300 ) = 210 ( 800 , 0 ) 0.2 ( 800 ) + 0.3 ( 0 ) = 160 • ( 400 , 500 ) maximizes the objective function

20. Answer Produce 400 cans of Fruit Delight and 500 cans of Heavenly Punch to maximize profits

21. Exercise 35 (page 133) • Define the variables being used. Look at the question being asked! • Key Question: “…, what planting combination will produce the greatest total profit?” • Let x represent the number of acres of oats planted y represent the number of acres of corn planted • This now defines Column Headings

22. The Objective Function First we need to recognize these relationships • Profit = Revenue + Left over capital cash + Left over labor costs • Where • Revenue = 55 x + 125 y • Left over capital cash = 2100 – 18 x – 36 y • Left over labor cash = 2400 – 16 x – 48 y

23. The Objective Function • Now add revenue, left over capital cash, and left over labor costs together [i.e. ( 55 x + 125 y ) + (2100 – 18 x – 36 y ) + ( 2400 – 16 x – 48 y )] to get the profit and the objective function becomes: Profit = 4500 + 21 x + 41 y

24. Table

25. Table

26. Restrictions from Table and Problem • The inequalities that restrict the values of the variables: 18 x + 36 y< 2,100 16 x + 48 y< 2,400 x + y< 100 x> 0 y> 0 Why do we need this restriction?

27. Why x + y< 100? • The farmer has only 100 acres available for the two crops. The amount of oats plus the amount of corn that the farmer plants has to be 100 acres or less.

28. Convert from General Form to Standard Form • Restrictions: 18 x + 36 y< 2,100 y< – ½ x + 175/3 16 x + 48 y< 2,400 y< – 1/3 x + 50 x + y< 100 y< – x + 100 x> 0 x> 0 y> 0 y> 0

29. Graph of the System of Equations x = 0 Not to scale y = 0 y = – 1/3 x + 50 y = – x + 100 y = – ½ x + 175/3

30. Graph of the System of Inequalities x = 0 Not to scale y = 0 y = – 1/3 x + 50 y = – x + 100 y = – ½ x + 175/3

31. Modified Graph of the Feasible Set form the Previous Slide II y = – 1/3 x + 50 III y = – ½ x + 175/3 IV x = 0 Feasible Set y = – x + 100 V I y = 0

32. Finding the Vertices (using a calculator when possible) • Vertex I x = 0 ( 0 , 0 ) y = 0 • Vertex II x = 0 ( 0 , 50 ) y = – 1/3 x + 50 • Vertex III y = – 1/3 x + 50 ( 50 , 100/3 ) y = – ½ x + 175/3 • Vertex IV y = – ½ x + 175/3 ( 250/3 , 50/3) y = – x + 100 • Vertex V y = – x + 100 ( 100 , 0 ) y = 0

33. Finding the Optimal Point VertexObjective Function 4500 + 21x + 41y ( 0 , 0 ) 4500 + 21 ( 0 ) + 41 ( 0 ) = 4,500.00 ( 0 , 50 ) 4500 + 21 ( 0 ) + 41 ( 50 ) = 6,550.00 ( 50 , 100/3 ) 4500 + 21 ( 50 ) + 41 ( 100/3 ) ~ 6,916.67 ( 250/3 , 50/3 ) 4500 + 21 ( 250/3 ) + 41 ( 50/3 ) ~ 6,933.33 ( 100 , 0 ) 4500 + 21 ( 100 ) + 41 ( 0 ) = 6,600.00 • ( 250/3 , 50/3 ) = ( 83 1/3 , 16 2/3 ) maximizes the objective function

34. The Answer Plant 83-and-a-third acres of oats and 16-and-two-thirds acres of corn to maximize the profit