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The corner points of the feasible region determined by the following system of linear inequalities:

Question:

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) $q=3 p$

Solution:

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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