The corner points of the bounded feasible region determined by the system of linear constraints 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 (5, 0) and (3, 4) is

$2q=p$

The correct answer is Option (3) → $2q=p$

Require $Z(5,0)=Z(3,4)$.

$5p = 3p + 4q$

$2p = 4q$

$p = 2q$

Since $p,q>0$, $p=2q$ (equivalently $\frac{p}{q}=2$ or $p:q=2:1$) ensures $Z$ attains the same (maximum) value at $(5,0)$ and $(3,4)$.

Condition: $p=2q$.