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SECTION 3.4 Linear Programming

Section 3.4:

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SECTION 3.4 Linear Programming

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    1. SECTION 3.4 Linear Programming WHATS IMPORTANT: -- Solve linear programming problems. -- Be able to use linear programming to solve real-life problems.

    2. Section 3.4: Linear Programming Optimization: finding the maximum or minimum value of some quantity Linear Programming: the process of optimizing a linear objective function subject to a system of linear inequalities called constraints Feasible Region: the graph of the system of constraints

    3. Section 3.4: Linear Programming If an objective function has a maximum of a minimum value, then it must occur at a vertex of the feasible region. Moreover, the objective function will have both a maximum and a minimum value if the feasible region is bounded.

    4. Section 3.4: Linear Programming Example: Find the minimum and maximum values of C = -x + 3y subject to the following constraints: x ? 2 x ? 5 y ? 0 y ? -2x + 12 Begin by graphing the lines for the constraints. (All will be solid lines.)

    5. Section 3.4: Linear Programming Each mark is 2 units. Graph x = 2 Graph x = 5 Graph y = 0 Graph y = -2x + 12

    6. Section 3.4: Linear Programming Fill in the vertices and shade in the region. As you can see, the vertices are at (2,0), (5,0), (5,2), and (2,8).

    7. Section 3.4: Linear Programming Since this is a bounded region, there will be both a minimum and a maximum. Use the points of the vertices in the equation C = -x + 3y to find them.

    8. Section 3.4: Linear Programming Put in (2,0): C = -x + 3y = -2 + 3(0) = -2 + 0 = -2 Put in (5,0): C = -x + 3y = -5 + 3(0) = -5 + 0 = -5 Put in (5,2): C = -x + 3y = -5 + 3(2) = -5 + 6 = 1 Put in (2,8): C = -x + 3y = -2 + 3(8) = -2 + 24 = 22 So, the minimum is -5 and the maximum is 22.

    9. Section 3.4: Linear Programming Example: Find the minimum and maximum values of C = x + 5y subject to the following constraints: x ? 0 y ? 2x + 2 5 ? x + y Begin by graphing the lines for the constraints. (All will be solid lines.)

    10. Section 3.4: Linear Programming Each mark is 1 unit. Graph x = 0 Graph y = 2x + 2 Graph 5 = x + y

    11. Section 3.4: Linear Programming Fill in the vertices and shade the appropriate region. The vertices are at (0,0), (0,2), (1,4), and (5,0)

    12. Section 3.4: Linear Programming Since this is an unbounded region, there will be a maximum, but not a minimum. Use the points of the vertices in the equation C = x + 5y to find the maximum.

    13. Section 3.4: Linear Programming Put in (0,0): C = x + 5y = 0 + 5(0) = 0 + 0 = 0 Put in (0,2): C = x + 5y = 0 + 5(2) = 0 + 10 = 10 Put in (1,4): C = x + 5y = 1 + 5(4) = 1 + 20 = 21 Put in (5,0): C = x + 5y = 5 + 5(0) = 5 + 0 = 5 So, the maximum is 21. No minimum.

    14. Section 3.4: Linear Programming Example: A furniture manufacturer makes chairs and sofas. The table on the next slide gives the number of packages of wood, stuffing, and fabric required for each chair or sofa. The manufacturer has room for 1300 packages of wood parts, 2000 packages of stuffing, and 800 packages of fabric. The manufacturer earns $200 per chair and $350 per sofa. How many chairs and sofas should they try to make to maximize profit?

    15. Section 3.4: Linear Programming

    16. Section 3.4: Linear Programming

    17. Section 3.4: Linear Programming Each mark is 100 units. Graph 2x + 3y = 1300 Graph 4x + 3y = 2000 Graph x + 2y = 800 And graph x = 0 and y = 0

    18. Section 3.4: Linear Programming Fill in the vertices and shade the appropriate region. The vertices are at (0,0), (0,400), (200,300), (300,200), and (500,0)

    19. Section 3.4: Linear Programming Use the vertices in the equation P = $200x + $350y to find the maximum.

    20. Section 3.4: Linear Programming At (0,0): P = 200(0) + 350(0) = 0 + (0) = $0 At (0,400): P = 200(0) + 350(400) = 0 + 1400 = $1400 At (200,300): P = 200(200) + 350(300) = 40,000 + 105,000 = $145,000 At (300,200): P = 200(300) + 350(200) = 60,000 + 70,000 = $130,000 At (500,0): P = 200(500) + 350(0) = 100,000 + 0 = $100,000 So, the maximum profit is $145,000 at (200,300). The manufacturer needs to sell 200 chairs and 300 sofas to maximize profits.

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