Question

The Janie Gioffre Drapery Company makes three types of draperies at two different locations. At location I, it can make 10 pairs of deluxe drapes, 20 pairs of better drapes, and 13 pairs of standard drapes per day. At location II, it can make 20 pairs of deluxe, 50 pairs of better, and 6 pairs of standard per day. The company has orders for 1000 pairs of deluxe drapes, 2100 pairs of better drapes, and 600 pairs of standard drapes. If the daily costs are $650 per day at location I and $750 per day at location II, how many days should Janie schedule at each location to fill the orders at minimum cost?

Answer 1

Answer:

• 30 days at location I
• 35 days at location II

Step-by-step explanation:

Let x and y represent days of operation of Location I and Location II, respectively. Then we want to minimize the objective function ...

  650x +750y

subject to the constraints on drape production:

  10x +20y ≥ 1000 . . . . . order for deluxe drapes

  20x +50y ≥ 2100 . . . . . order for better drapes

  13x +6y ≥ 600 . . . . . . . order for standard drapes

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I find a graphical solution works well for this. The vertices of the feasible solution space are (x, y) = (0, 100), (30, 35), (80, 10), (105, 0). The vertex at which the cost is minimized is

  (x, y) = (30, 35)

This schedule will produce exactly the required numbers of deluxe and standard drapes, and 2350 pairs of better drapes, 250 more than required.

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In the attached graph, we have reversed the inequalities so that the solution space (feasible region) is white, not triple-shaded. Minimizing the objective function means choosing the vertex of the feasible region so that the line representing the objective function is as close to the origin as possible.