Question

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 _________.

Answer 1

$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) $
$Z \leq y\left(6 \cdot\left(\frac{75-3 y}{4}\right)+y\right)$
$Z \leq \frac{1}{2}\left(225 y -7 y ^{2}\right) \leq \frac{(225)^{2}}{2 \times 4 \times 7}$
$=\frac{50625}{56} $
$\approx 904.0178$
$\approx 904.02$
It will be attained at $y =\frac{225}{14}$