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

        $2x+y \leq 10,x+3y \leq 15,x,y\geq 0$ 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

               $(A) p=q$

   $(B)p=2q$

   $(C)p=3q$

   $(D)q=3p$

Answers (1)

best_answer

The maximum value of Z is unique.

It is given that maximum value of Z occurs at two points $(3,4)\, \, and\, \, \, (0,5)$ .

$\therefore$ Value of Z at $(3,4)$ =value of Z at $(0,5)$

$\Rightarrow \, \, \, p(3)+q(4)=p(0)+q(5)$

$\Rightarrow \, \, \, 3p+4q=5q$

$\Rightarrow \, \, \, q=3p$

Hence, D is correct option.

Posted by

seema garhwal

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