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For an objective function Z =ax+ by where a, b > 0, the corner points of the feasible region determined by a set of constraints (linear inequalities) are (0, 20), (10, 10), (30, 30) and (0, 40). The condition on a and b such that the maximum Z occurs at both the points (30, 30) and (0, 40) is

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$b - 3 a = 0$

$a = 3 b$

$a + 2 b = 0$

$2 a - b = 0$

answer is A.

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Detailed Solution

$Z = a x + b y, a, b > 0$

Maximum of Z occurs at (30, 30) and (0, 40)

$\begin{array}{l} \Rightarrow a \times 30 + b \times 30 = a \times 0 + b \times 40 \\ \Rightarrow 30 a + 30 b = 40 b \\ \Rightarrow 10 b = 30 a \Rightarrow b = 3 a \\ \Rightarrow b - 3 a = 0 \end{array}$

$∴$ Option (a) is the correct answer.

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