Question

Corner points of the feasible region determined by the system of linear constraints are $(0, 3), (1, 1)$ and $(3, 0)$. Let $z = px = qy$, where $p, q > 0$. Condition on $p$ and $q$ so that the minimum of $z$ occurs at $(3, 0)$ and $(1, 1)$ is

• A. $p = 2q$
• B. $p = \frac {q}{2}$
• C. $p = 3q$
• D. $p = q$

Answer 1

Answer: (B)

$p = \frac {q}{2}$

Answer 2

Given corner points are $(0,3),(1,1),(3,0)$
$z=p x+q y$
At $(3,0), z=3 p$
At $(1,1), z=p+q$
It is given that the minimum of $z$ occurs at $(3,0)$ and $(1,1)$
$\Rightarrow 3 p=p+q$
$\Rightarrow 2 p = q$
$\Rightarrow p = q / 2$