Question

The corner points of the feasible region determined by the following system of linear inequalities :
$2x+y\le 10,x+3y\le 15,x,y\ge 0$ 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. $q=3p$

Answer 1

Answer: (D)

$q=3p$

Since, maximum of $Z=px+qy$ occurs at both points $(3,4)$ and $(0,5)$

Hence value of $Z=px+qy$ must be equal at both the points.

$\therefore p\times 3+q\times 4=p\times 0+q\times 5$

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

$\Rightarrow 3p=q$

So, $q=3p$ is the correct answer.