The corner points of feasible region determined by the following system of linear inequalities:

$2x+y≤10, x+3y≤15, x≥0, y≥0$ are $(0, 0), (5, 0), (3, 4)$ and $(0, 5)$

Let $ z= px +qy ,$ when $p, q > 0$ then the relation between p and q, so that maximum of z occurs at both points $(3, 4)$ and $(0, 5)$ is :

$3p=q$

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

$z= px +qy$

At (3, 4): $z= 3p +4q$

At (0, 5): $z= 0p +5q=5q$

$3p+4q=5q$ (∵ At maximum, objective function (z) must yield same value)

$3p=q$