The corner points of the feasible region determined by the following system of linear inequalities: 2x+y≤10,x+3y≤15,xy≥ 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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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
Hence, the correct answer is D.

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