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

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

The correct answer is Option (3) → $p=\frac{q}{2}$

for minimum to occurs at both (3, 0), (1, 1)

$Z(3, 0)=Z(1, 1)$

$3p=p+q$

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