Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let $Z=px +qy $ where, $p, q> 0.$ Condition on p and q so that minimum of Z occurs at (3, 0) and (1, 1) is :

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

Corner points: (0,3), (1,1), (3,0)

Objective function: $Z = p x + q y$, with $p,q > 0$

Evaluate $Z$ at each corner point:

$Z(0,3) = p*0 + q*3 = 3q$

$Z(1,1) = p*1 + q*1 = p + q$

$Z(3,0) = p*3 + q*0 = 3p$

For minimum at $(3,0)$ and $(1,1)$:

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

From $3p = p + q \Rightarrow q = 2p$

Check inequality: $p + q \le 3q \Rightarrow p + 2p \le 3*2p \Rightarrow 3p \le 6p$, which is true

Condition: $q = 2p$, $p > 0$, $q > 0$