CUET · MATHS · PYQ PAPER 2025
Consider the LPP: Minimize \(Z=x+2 y\) subject to \(2 x+y \geq 3, x+2 y \geq 6, x, y \geq 0\). The optimal feasible solution occurs at:
- A \((6,0)\) only
- B \((0,3)\) only
- C Neither \((6,0)\) nor \((0,3)\)
- D Both \((6,0)\) and \((0,3)\)
Answer & Solution
Correct Answer
(D) Both \((6,0)\) and \((0,3)\)
Step-by-step Solution
Detailed explanation
Corner points of feasible region: Intersection of \(x=0\) and \(x+2y=6 \implies (0,3)\). Intersection of \(y=0\) and \(x+2y=6 \implies (6,0)\). Evaluate \(Z=x+2y\): At \((0,3)\), \(Z = 0+2(3) = 6\). At \((6,0)\), \(Z = 6+2(0) = 6\). Both points yield the minimum value \(Z=6\).
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