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+1 vote
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in Linear Programming by (45.0k points)
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The corner points of the feasible region determined by the following system of linear inequalities:

2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is

A. p = q

B. p = 2q

C. p = 3q

D. q = 3p

1 Answer

+1 vote
by (47.4k points)
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Best answer

Correct answer is D.

Given the vertices of the feasible region are:

O (0, 0)

A (5, 0)

B (3, 4)

C (0, 5)

Also given the objective function is Z = px + qy

Now substituting O, A, B and C in Z

As per the condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5)

Then we can equate Z values at B and C, this gives

3p + 4q = 5q

3p = 5q – 4q

3p = q

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