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CUET · MATHS · PYQ PAPER 2025

The corner points of a bounded feasible region determined by the following system of linear inequalities \(x+3 y \leq 60, x+y \geq 10, x \leq y, x \geq 0, y \geq 0\) are \((0,10),(5,5),(15,15)\) and (0,20). Let \(z=2 p x+q y, p, q>0\). If maximum of z occurs at both (15, 15) and (0, 20), then the relation between p and q is

  1. A 3p = 2q + 1
  2. B 2p = 4q
  3. C 6p = q
  4. D 2p = 3q
Verified Solution

Answer & Solution

Correct Answer

(C) 6p = q

Step-by-step Solution

Detailed explanation

\(z(15, 15) = 2p(15) + q(15) = 30p + 15q\) \(z(0, 20) = 2p(0) + q(20) = 20q\) \(30p + 15q = 20q\) \(30p = 5q\) \(6p = q\)