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NCTM Standards: 1, 2, 6, 7, 8, 9, 10

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Constraints: Limitation placed upon the variables.

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Example 1 1. Define the variables:Let x = the units of lumber produced Let y = the units of plywood produced. 2. Write the constraints:x + y 600 x 150 y 225 3. Graph the system: 4. Write the profit equation: 5. Substitute the values of the vertices: 150, 450 30(150) + 45(450)=24,750 150, 225 30(150) + 45(225)= 14,625 375, 225 30(375) + 45(225)=21,375 6. Determine whether the situation calls for a maximum value or a minimum value & answer the question. The mill wants to maximize profit, so they should produce 150 units of lumber & 450 units of plywood for a maximum profit of $24,750

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Infeasible Problem When the constraints of a linear programming problem cannot be solved simultaneously. The constraints do not define a region with any points in common. Ex: Unbounded The region formed is not a polygon. Ex: The graph has a minimum value at (5, 3), but there is no maximum value.

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Alternate Optimal Solutions When a linear programming application has two or more optimal solutions. This usually occurs when the graph of the function to be maximized or minimized is parallel to one side of the polygonal convex set.

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Example 2 1.Define the variables: 2. Write the constraints: & the profit equation: Time available for woodworking: Time available for finishing: Two obvious ones we almost miss: Let b = # bookcases Let c = # of cabinets

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3. Graph the constraints: 4. Test the values in the profit equation: (0, 0) 60(0) + 40(0) $0 (0, 7) 60(0) + 40(7) $280 (2, 4) 60(2) + 40(4) $280 (4, 0) 60(0) + 40(0) $240 5.Answer the question: The shop will make a maximum of $280 if they make 0 bookcases & 7 cabinets or 2 bookcases & 4 cabinets.

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Homework: Page 115

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30S Applied Math Mr. Knight – Killarney School Slide 1 Unit: Linear Programming Lesson 5: Problem Solving Problem Solving with Linear Programming Learning.

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