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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 $p \cdot q > 0$. Find the condition on $p$ and $q$ so that the minimum of $z$ occurs at (3,0) and (1,1).

$p = q$

$p = 2q$

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

$p = 3q$

Solution

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