Question:
The maximum value of $\mathrm{z}$ 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
Solution:
$z=6 x y+y^{2}=y(6 x+y)$
$3 x+4 y \leq 100$...(1)
$4 x+3 y \leq 75$...(2).
$x \geq 0$
$y \geq 0$
$x \leq \frac{75-3 y}{4}$
$Z=y(6 x+y)$
$z \leq y\left(6 \cdot\left(\frac{75-3 y}{4}\right)+y\right)$
$z \leq \frac{1}{2}\left(225 y-7 y^{2}\right) \leq \frac{(225)^{2}}{2 \times 4 \times 7}$
$=\frac{50625}{56}$
$\approx 904.0178$
$\approx 904.02$
It will be attained at $\mathrm{y}=\frac{225}{14}$