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JEE Mains · Maths · STD 12 - 12. linear programming

The maximum value of \(z\) in the following equation \(z=6 x y+y^{2},\) where \(3 x+4 y \leq 100\) and \(4 x+3 y \leq 75\) for \(x \geq 0\) and \(y \geq 0\) is \(......\)

  1. A \(904\)
  2. B \(846\)
  3. C \(952\)
  4. D \(882\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(904\)

Step-by-step Solution

Detailed explanation

\(z=6 x y+y^{2}=y(6 x+y)\) \(3 x+4 y \leq 100\) \(....(i)\) \(4 x+3 y \leq 75\) \(......(ii)\) \(x \geq 0\) \(y \geq 0\) \(z \leq \frac{75-3 y}{4}\) \(Z=y(6 x+y)\) \(Z \leq y\left(6 \cdot\left(\frac{75-3 y}{4}\right)+y\right)\)…
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