The corner points of the bounded feasible region determined by a system of linear constraints are $(0, 3), (1, 1)$ and $(3, 0)$. Let $Z = px + qy$, where $p, q > 0$. The condition on $p$ and $q$ so that the minimum of $Z$ occurs at $(3, 0)$ and $(1, 1)$ is:

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

The correct answer is Option (2) → $p = \frac{q}{2}$ ##

$Z = px + qy$ — (i)

At $(3, 0)$, $Z = 3p$ — (ii)

and at $(1, 1)$, $Z = p + q$ — (iii)

From (ii) & (iii),

$3p = p + q$

$⇒2p = q ⇒ p = \frac{q}{2}$