The corner points of the feasible region determined by the following system of linear inequalities:
$2 x+y \leq 10, x+3 y \leq 15, x y \geq 0$ are $(0,0),(5,0),(3,4)$ and $(0,5)$ Let $Z=p x+q y$, 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) $p=2 q$
(C) $p=3 q$
(D) $q=3 p$
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).
$\therefore$ Value of $Z$ at $(3,4)=$ Value of $Z$ at $(0,5)$
$\Rightarrow p(3)+q(4)=p(0)+q(5)$
$\Rightarrow 3 p+4 q=5 q$
$\Rightarrow q=3 p$
Hence, the correct answer is D.
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