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CUET · MATHS · PYQ PAPER 2025

Consider the LPP : Maximize \(z=5 x+3 y\) subject to \(3 x+5 y \leq 15,5 x+2 y \leq 10, x, y \geq 0\).
The optimal feasible solution occurs at

  1. A (2,0) only
  2. B neither \((2,0)\) nor \(\left(\frac{20}{19}, \frac{45}{19}\right)\)
  3. C (0,3) only
  4. D \(\left(\frac{20}{19}, \frac{45}{19}\right)\) only
Verified Solution

Answer & Solution

Correct Answer

(D) \(\left(\frac{20}{19}, \frac{45}{19}\right)\) only

Step-by-step Solution

Detailed explanation

\(3x+5y=15 \text{ and } 5x+2y=10 \Rightarrow x=\frac{20}{19}, y=\frac{45}{19}\) Feasible corner points: \((0,0), (2,0), (0,3), \left(\frac{20}{19}, \frac{45}{19}\right)\) \(z(0,0) = 5(0)+3(0)=0\) \(z(2,0) = 5(2)+3(0)=10\) \(z(0,3) = 5(0)+3(3)=9\)…
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