The corner points of the bounded feasible region determined by the system of linear constraints are (0, 8), (4, 4), (12, 12), (0, 20). Let $z = px + qy$, where $p, q>0$.Condition on $p$ and $q$ so that the maximum of $z$ occurs at both the points (12, 12), (0, 20) is

$3p = 2q$

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

Corner points: $(0,8),\,(4,4),\,(12,12),\,(0,20)$.

Maximum of $z=px+qy$ must occur at both $(12,12)$ and $(0,20)$.

Compute $z$ at these two points:

$z(12,12)=12p+12q$

$z(0,20)=20q$

For maximum at both points:

$12p+12q = 20q$

$12p = 8q$

$\frac{p}{q} = \frac{8}{12} = \frac{2}{3}$

Condition: $\frac{p}{q} = \frac{2}{3}$ (with $p,q>0$)