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The comer points of the feasible region for an L.P.P. are (0, 3), (1, 1) and (3, 0). If objective function is $Z = px + qy, p, q > 0$ , then the condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is

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p = 2q

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

p = 3q

p = q

answer is B.

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Detailed Solution

$Z = p x + q y, p, q > 0$

$∵$ Minimum of Z occurs at (3, 0) and (1, 1)

$\begin{array}{l} \Rightarrow p \times 3 + q \times 0 = p \times 1 + q \times 1 \\ \Rightarrow 3 p = p + q \Rightarrow 2 p = q \Rightarrow p = \frac{q}{2} \end{array}$

$∴$ Option (b) is the correct answer.

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