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

q = 3p

objective fn → Z = px + qy

or Z(x, y) = px + qy

so if maximum occurs at (3, 4) and (0, 5) both

$\Rightarrow Z(3,4) = Z(0,5)$

$\Rightarrow 3 p+4 q =0+5 q$

3p = q