The corner points of a bounded feasible region determined by the following system of linear inequalities $x + 3y ≤ 60, x + y ≥10,x≤y,x≥0, y≥ 0$ are (0, 10), (5, 5), (15, 15) and (0, 20). Let $z = 2px + qy,p,q>0$. If maximum of z occurs at both (15, 15) and (0, 20), then the relation between $p$ and $q$ is

$6p = q$

The correct answer is Option (3) → $6p = q$

$\text{Max at }(15,15)\text{ and }(0,20)\Rightarrow z(15,15)=z(0,20)$

$2p(15)+q(15)=2p(0)+q(20)$

$30p+15q=20q\Rightarrow 30p=5q\Rightarrow q=6p$

${q=6p}$