MCQ

The corner points of the feasible region determined by the following system of linear inequalities:

2*x* + *y* ≤ 10, *x* + 3*y* ≤ 15, *x*, *y* ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let *Z* = px + qy, where p, q > 0. Condition on *p* and *q* so that the maximum of *Z* occurs at both (3, 4) and (0, 5) is

#### Options

p = q

p = 2q

p = 3q

q = 3p

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#### Solution

q = 3p

**The maximum value of Z is unique.

It is given that the maximum value of Z occurs at two points (3, 4) and (0, 5).

∴ Value of Z at (3, 4) = Value of Z at (0, 5)

⇒ p(3) + q(4) = p(0) + q(5)

⇒ 3p + 4q = 5q

⇒ q = 3p

Concept: Mathematical Formulation of Linear Programming Problem

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