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.
