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Section 3.4

Section 3.4. Linear Programming. y – x + y x + y 3 x – 11. 2 3. 11 3. <. 1 4. 11 4. >. >. Step 1 : Graph the constraints. 2 3. 11 3. y = – x + y = x +. To find A , solve the system .

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Section 3.4

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  1. Section 3.4 Linear Programming

  2. y – x + yx + y 3x – 11 2 3 11 3 < 1 4 11 4 > > Step 1: Graph the constraints. 2 3 11 3 y = – x + y = x + To find A, solve the system . The solution is (1, 3), so A is at (1, 3). 11 4 1 4 11 4 1 4 y = x + y = 3x – 11 To find B, solve the system . The solution is (5, 4), so B is at (5, 4). Linear Programming ALGEBRA 2 LESSON 3-4 Find the values of x and y that maximize and minimize P if P = –5x + 4y. Step 2: Find the coordinates for each vertex. 3-4

  3. 2 3 11 3 y = – x + y = 3x – 11 To find C, solve the system . The solution is (4, 1), so C is at (4, 1). Linear Programming ALGEBRA 2 LESSON 3-4 (continued) Step 3: Evaluate P at each vertex. Vertex P = –5x + 4y A(1, 3) P = –5(1) + 4(3) = 7 B(5, 4) P = –5(5) + 4(4) = –9 C(4, 1) P = –5(4) + 4(1) = –16 When x = 1 and y = 3, P has its maximum value of 7. When x = 4 and y = 1, P has its minimum value of –16. Quick Check 3-4

  4. As you come in… • Log into NavNet • Make sure Inequalz app is running • Make sure you have all four applications entered and properly graphed

  5. Define: Let x = number of tables made in a day. Let y = number of chairs made in a day. Let P = total profit. Relate: Organize the information in a table. Tables Chairs Total No. of Products xyx + y No. of Units 30 x 60 40 y 100 120 Profit 150x 65y 150x + 65y < < < < constraint objective Linear Programming ALGEBRA 2 LESSON 3-4 A furniture manufacturer can make from 30 to 60 tables a day and from 40 to 100 chairs a day. It can make at most 120 units in one day. The profit on a table is $150, and the profit on a chair is $65. How many tables and chairs should they make per day to maximize profit? How much is the maximum profit? 3-4

  6. > < > < < Step 1: Graph the constraints. Linear Programming ALGEBRA 2 LESSON 3-4 (continued) x 30 x 60 y 40 y 100 x + y 120 Write: Write the constraints. Write the objective function. P = 150x + 65y Step 2: Find the coordinates of each vertex. Vertex A(30, 90) B(60, 60) C(60, 40) D(30, 40) Step 3: Evaluate P at each vertex. P = 150x + 65y P = 150(30) + 65(90) = 10,350 P = 150(60) + 65(60) = 12,900 P = 150(60) + 65(40) = 11,600 P = 150(30) + 65(40) = 7100 The furniture manufacturer can maximize their profit by making 60 tables and 60 chairs. The maximum profit is $12,900. Quick Check 3-4

  7. < < < < > (–2, 3), (4, 3), (4, –1); maximum of 33 when x = 4 and y = 3, minimum of 5 when x = 4 and y = –1 Linear Programming ALGEBRA 2 LESSON 3-4 1. Graph the system of constraints. Name all vertices of the feasible region. Then find the values of x and y that maximize and minimize the objective function P = 2x + 7y + 4. –2 x 4 –1 y 3 y – x + 2. If the constraint on y in the system for Question 1 is changed to 1 < y < 3, how does the minimum value for the objective function change? 5 3 2 3 There is a new minimum value of 13 when x = 1 and y = 1. 3-4

  8. Pages 138–140 Exercises 1. When x = 4 and y = 2, P is maximized at 16. 2. When x = 600 and y = 0, P is maximized at 4200. 3. When x = 6 and y = 8, C is minimized at 36. 4. vertices: (0, 0), (5, 0), (5, 4), (0, 4);    maximized at (5, 4) 5. 5. (continued) vertices: (3, 5), (0, 8); minimized at (0, 8) 6. vertices: (0, 0), (5, 0), (2, 6), (0, 8); maximized at (5, 0) Linear Programming ALGEBRA 2 LESSON 3-4 3-4

  9. 7. vertices: (1, 5), (8, 5), (8, –2);    minimized at (8, –2) 8. vertices: (8, 0), (2, 3); minimized at (8, 0) 9. vertices: (2, 1), (6, 1), (6, 2), (2, 5), (3, 5); maximized at (6, 2) 10. P = 500x + 200y 10. (continued) b. 15 experienced teams, 0 training teams; none; 7500 trees c. 11 experienced teams; 8 training teams; 7100 trees 11. 70 spruce; 0 maple 12. Solving a system of linear equations is a necessary skill used to locate the vertex points. < < > > 2x + y 30 2y 16 x 0, y 0 Linear Programming ALGEBRA 2 LESSON 3-4 3-4

  10. 13. 60 samples of Type I and 0 samples of Type II 14. vertices: (0, 0), (1, 4), (0, 4.5), ,0 ; maximized when P = 6 at (1, 4) 15. vertices: (75, 20), (75, 110), 25, 86 , (25, 110); minimized when C = 633 at 25, 86 16. vertices: (0, 0), , (0, 11); maximized when P = 29 at 1 3 2 3 1 3 1 3 2 3 2 3 1 3 7 3 7 , 3 7 , 3 2 3 Linear Programming ALGEBRA 2 LESSON 3-4 3-4

  11. 17. vertices: (0, 0), (150, 0), (100, 100), (0, 200); maximized when P = 400 at (0, 200) 18. vertices: (12, 0), (0, 10), (4, 2), (1, 5); minimized when C = 80,000 at (4, 2) 19. vertices: (3, 3), (3, 10), (5, 1), (12, 1); maximized when P = 51 at (12, 1) Linear Programming ALGEBRA 2 LESSON 3-4 20. 3 trays of corn muffins and 2 trays of bran muffins 3-4

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