 # Copyright 2013, 2010, 2007, Pearson, Education, Inc. Section 7.6 Linear Programming.

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Copyright 2013, 2010, 2007, Pearson, Education, Inc. Section 7.6 Linear Programming

Copyright 2013, 2010, 2007, Pearson, Education, Inc. What You Will Learn Linear Programming 7.6-2

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Linear Programming In a linear programming problem, there are restrictions called constraints. Each constraint is represented as a linear inequality. The list of constraints forms a system of linear inequalities. When the system of inequalities is graphed, often a feasible region is obtained. 7.6-3

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Linear Programming The points where two or more boundaries intersect are called the vertices of the feasible region. 7.6-4

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Objective Function The objective function is the formula for the quantity K = Ax + By (or some other variable) that we want to maximize or minimize. The values we substitute for x and y determine the value of K. From the information given in the problem, we determine the real number constants A and B. 7.6-5

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Fundamental Principle of Linear Programming If the objective function, K = Ax + By, is evaluated at each point in a feasible region, the maximum and minimum values of the equation occur at vertices of the region. 7.6-6

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Solving a Linear Programming Problem 1.Determine all necessary constraints. 2.Determine the objective function. 3.Graph the constraints and determine the feasible region. 4.Determine the vertices of the feasible region. 7.6-7

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Solving a Linear Programming Problem 5.Determine the value of the objective function at each vertex. The solution is determined by the vertex that yields the maximum or minimum value of the objective function. 7.6-8

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: Modeling–Washers and Dryers, Maximizing Profit The Admiral Appliance Company makes washers and dryers. The company must manufacture at least one washer per day to ship to one of its customers. No more than 6 washers can be manufactured due to production restrictions. The number of dryers manufactured cannot exceed 7 per day. 7.6-9

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: Modeling–Washers and Dryers, Maximizing Profit Also, the number of washers manufactured cannot exceed the number of dryers manufactured per day. If the profit on each washer is \$20 and the profit on each dryer is \$30, how many of each appliance should the company make per day to maximize profits? What is the maximum profit? 7.6-10

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: Modeling–Washers and Dryers, Maximizing Profit Solution Let x = the number of washers manufactured per day y = the number of dryers manufactured per day 20x = the profit on washers 30y = the profit on dryers P = the total profit 7.6-11

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: Modeling–Washers and Dryers, Maximizing Profit Solution Constraints Number of appliances manufactured cannot be negative: x ≥ 0, y ≥ 0 At least one washer per day: x ≥ 1 No more than 6 washers per day: x ≤ 6 No more than 7 dryers per day: y ≤ 7 Washers cannot exceed dryers: x ≤ y x ≥ 0, y ≥ 0, x ≥ 1, x ≤ 6, y ≤ 7, x ≤ y 7.6-12

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: Modeling–Washers and Dryers, Maximizing Profit Solution Objective function is the profit formula P = 20x + 30y Graph all the constraints to find the feasible region and the vertices. 7.6-13

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: Modeling–Washers and Dryers, Maximizing Profit Solution The vertices are: (1, 1), (1, 7), (6, 7), and (6, 6) 7.6-14

Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: Modeling–Washers and Dryers, Maximizing Profit Solution Calculate the objective function at each vertex, P = 20x + 30y (1, 1): P = 20(1) + 30(1) = 50 (1, 7): P = 20(1) + 30(7) = 230 (6, 7): P = 20(6) + 30(7) = 330 (6, 6): P = 20(6) + 30(6) = 300 The company should manufacture 6 washers and 7 dryers. Profit is \$330. 7.6-15