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### Alg II 3-4 Linear Programming

• 1. 3-4 Linear Programming Algebra II Unit 3 Linear Systems © Tentinger
• 2. Essential Understanding and Objectives ● Essential Understanding: Some real-world problems involve multiple linear relationships. Linear programming accounts for all of these linear relationships and gives the solution to the problem. ● Objectives: ● Students will be able to: ● Solving problems using linear programming ● Define constraint, linear programming, feasible region, and objective function
• 3. Iowa Core Curriculum • Algebra • A.CED.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods.
• 4. Linear Programming Businesses use linear programming to find out how to maximize profit or minimize costs. Most have constraints on what they can use or buy.
• 5. Linear Programming Linear programming is a process of finding a maximum or minimum of a function by using coordinates of the polygon formed by the graph of the constraints.
• 6. What is a constraint? A restriction... A boundary… A limitation…
• 7. What is the feasible region? The feasible region is the area of the graph in which all the constraints are met.
• 8. Objective Function • The quantity you are trying to maximize or minimize is modeled by this. • Usually this quantity is the cost or profit • Looks something like this C = ax + by, where a and b are real numbers
• 9. Vertex Principle • If there is a maximum or minimum value of the linear objective function, it occurs at one or more vertices of the feasible region. Online Example
• 10. Find the minimum and maximum value of the function f(x, y) = 3x - 2y. We are given the constraints: •y ≥ 2 • 1 ≤ x ≤5 •y ≤ x + 3
• 11. Linear Programming • Find the minimum and maximum values by graphing the inequalities and finding the vertices of the polygon formed. • Substitute the vertices into the function and find the largest and smallest values.
• 12. 8 1 ≤ x ≤5 7 6 5 4 3 y≥2 2 y≤x+3 1 1 2 3 4 5
• 13. Linear Programming •The vertices of the quadrilateral formed are: (1, 2) (1, 4) (5, 2) (5, 8) •Plug these points into the objective function f(x, y) = 3x - 2y
• 14. Linear Programming f(x, y) = 3x - 2y • f(1, 2) = 3(1) - 2(2) = 3 - 4 = -1 • f(1, 4) = 3(1) - 2(4) = 3 - 8 = -5 • f(5, 2) = 3(5) - 2(2) = 15 - 4 = 11 • f(5, 8) = 3(5) - 2(8) = 15 - 16 = -1
• 15. Linear Programming •f(1, 4) = -5 minimum •f(5, 2) = 11 maximum
• 16. Find the minimum and maximum value of the function f(x, y) = 4x + 3y We are given the constraints: • y ≥ -x + 2 1 •y ≤ x+2 4 • y ≥ 2x -5
• 17. 6 y ≥ 2x -5 5 1 y x 2 4 4 3 y ≥ -x + 2 2 1 1 2 3 4 5
• 18. Vertices f(x, y) = 4x + 3y • f(0, 2) = 4(0) + 3(2) = 6 • f(4, 3) = 4(4) + 3(3) = 25 • f( 3 , - 3 ) = 4( 3) + 3(- 3 ) = 7 1 7 1 28 3 -1 = 25 3
• 19. Linear Programming •f(0, 2) = 6 minimum •f(4, 3) = 25 maximum
• 20. Example 1 A farmer has 25 days to plant cotton and soybeans. The cotton can be planted at a rate of 9 acres per day, and the soybeans can be planted at a rate of 12 acres a day. The farmer has 275 acres available. If the profit for soybeans is \$18 per acre and the profit for cotton is \$25 per acre, how many acres of each crop should be planted to maximize profits?
• 21. Step 1: Define the variables What are the unknown values? Let c = number of acres of cotton to plant Let s = number of acres of soybeans to plant
• 22. Step 2: Write a System of Inequalities Write the constraints. What are the limitations given in the problem? The number of acres planted in cotton must c 0 be greater than or equal to 0. s 0 The number of acres planted in soybeans must be greater than or equal to 0. c s 275 The total number of acres planted must be less than or equal to 275. c s 25 The time available for planting must be less 9 12 than or equal to 25 days.
• 23. Step 3: Graph the Inequalities s c The purple area is the feasible region.
• 24. Step 4: Name the Vertices of the Feasible Region of the vertices of the Find the coordinates feasible region, the area inside the polygon. (0,275) (225,0) (0,0) (75,200)
• 25. Step 5: Write an Equation to be Maximized or Minimized p(c,s) = 25c + 18s Maximum profit = \$25 times the number of acres of cotton planted + \$18 times the number of acres of soybeans planted.
• 26. Step 6: Substitute the Coordinates into the Equation the coordinates of the vertices into the Substitute maximum profit equation. (c,s) 25c + 18s f(c,s) (0,275) 25(0) + 18(275) 4950 (225,0) 25(225) + 18(0) 5625 (0,0) 25(0) + 18(0) 0 (75,200) 25(75) + 18(200) 5475
• 27. Step 7: Find the Maximum (c,s) 25c + 18s f(c,s) (0,275) 25(0) + 18(275) 4950 (225,0) 25(225) + 18(0) 5625 (0,0) 25(0) + 18(0) 0 (75,200) 25(75) + 18(200) 5475 225 acres of cotton and 0 acres of soybeans should be planted for a maximum profit of \$5,625.
• 28. Example 2: The Bethlehem Steel Mill can convert steel into girders and rods. The mill can produce at most 100 units of steel a day. At least 20 girders and at least 60 rods are required daily by regular customers. If the profit on a girder is \$8 and the profit on a rod is \$6, how many units of each type of steel should the mill produce each day to maximize the profits?
• 29. Step 1: Define the Variables Let x = number of girders Let y = number of rods Step 2: Write a System of Inequalities x 20 At least 20 girders are required daily. y 60 At least 60 rods are required daily. The mill can produce at most 100 units x y 100 of steel a day.
• 30. Step 3: Graph the Inequalities x 20, y 60, and x y 100 100 80 60 40 20 The purple region represents the feasible region. -50 50 100
• 31. Step 4: Name the Vertices of the Feasible Region Find the coordinates of the vertices of the feasible region, the area inside the polygon. (20, 60) (20, 80) (40, 60)
• 32. Step 5: Write an Equation to be Maximized or Minimized p(x,y) = 8x + 6y Maximum profit = \$8 times the number of girders produced + \$6 times the number of rods produced
• 33. Step 6: Substitute the Coordinates into the Equation Substitute the coordinates of the vertices into the maximum profit equation. (x,y) 8x + 6y p(x,y) (20, 60) 8(20) + 6(60) 520 (20, 80) 8(20) + 6(80) 640 (40, 60) 8(40) + 6(60) 680
• 34. Step 7: Find the Maximum (x,y) 8x + 6y p(x,y) (20, 60) 8(20) + 6(60) 520 (20, 80) 8(20) + 6(80) 640 (40, 60) 8(40) + 6(60) 680 40 girders and 60 rods of steel should be produced for a maximum profit of \$680.
• 35. Homework • Pg 160-161 #10-13, 16-19 • 8 problems
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