A playground is in the shape of a square with each side equal to 109 yards. It has skating rinks in the shape of the quadrants o
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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.
The vertical distance can be found by subtracting parabola from the given line.
Given line equation is y= x+20
Equation of parabola is y= x^2
Subtract parabola from line equation
so y =
y =
Now we take derivative to find out the maximum that is the vertex
y' = -2x + 1
Now we set derivative =0 and solve for x
0 = -2x+1
2x = 1
Now we plug in x values in
Take common denominator
So our maximum vertical distance is at