The corner points of the feasible region determined by the following system of linear inequalities $2x+y ≤10, x+3y ≤ 15, x, y ≥0$ are (0, 0), (5, 0), (3, 4) and (0, 5). Let $z=px +qy $ where p and $q > 0. $ The condition on p and q so that the maximum of z occurs at both (3, 4) and (0, 5) is :

$p=1$ $p=2q$ $p=3q$ $q=3p$

$q=3p$

The correct answer is Option (4) → $q=3p$ $Z_{max}(3,4)=Z_{max}(0,5)$ $⇒3p+4q=5q$ $3p=q$