Question
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 the minimum of $Z$ occurs at (3,0) and (1,1) is
So, condition of $p$ and $q$,
so that the minimum of $Z$ occurs at (3,0) and (1,1) is
$p + q = 3p \Rightarrow 2p = q$
$\therefore p = \frac{q}{2}$
- (a) $p = 2q$
- (b) $p = \frac{q}{2}$ ✓ Correct
- (c) $p = 3q$
- (d) $p = q$