Question

The maximum value of $z$ in the following equation $z=6xy+y^2$, where $3x+4y\le 100$ and $4x+3y\le 75$ for $x\ge 0$ and $y\ge 0$ is _________.

Answer 1

Answer: 904

$z=6xy+y^2=y(6x+y)$
$3x+4y\le 100$ ...(i)
$4x+3y\le 75$ ...(ii)
$x\ge 0$
$y\ge 0$
$Z\le \frac{75-3y}{4}$
$Z=y(6x+y)$
$Z\le y\left(6\cdot \left(\frac{75-3y}{4}\right)+y\right)$
$Z\le \frac{1}{2}\left(225y-7y^2\right)\le \frac{(225{)}^2}{2\times 4\times 7}$
$=\frac{50625}{56}$
$\approx 904.0178$
$\approx 904.02$
It will be attained at $y=\frac{225}{14}$