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Question
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
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
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Solution
(d) q = 3p
The maximum value of Z is unique.
It is given that the maximum value of Z occurs at two points (3, 4) and (0, 5).
∴ Value of Z at (3, 4) = Value of Z at (0, 5)
⇒ p(3) + q(4) = p(0) + q(5)
⇒ 3p + 4q = 5q
⇒ q = 3p
The maximum value of Z is unique.
It is given that the maximum value of Z occurs at two points (3, 4) and (0, 5).
∴ Value of Z at (3, 4) = Value of Z at (0, 5)
⇒ p(3) + q(4) = p(0) + q(5)
⇒ 3p + 4q = 5q
⇒ q = 3p
Suggest Corrections
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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