The corner points of the feasible region determined by the system of linear inequalities are $(0,3), (1,1)$ and $(3,0)$. Let $Z = px + qy$ where $p, q > 0$. Find the condition on $p$ and $q$ such that the minimum of $Z$ occurs at both $(3,0)$ and $(1,1)$.
- A$p = 2q$
- B$p = \frac{q}{2}$
- C$p = 3q$
- D$p = q$
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DifficultMHT CET 2026
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