The maximum value of z in the following equation z = 6 x y + y 2 , where 3 x + 4 y ≤ 100 and 4 x + 3 y…

$z=6xy+y^2=y(6x+y)$

$3x+4y\le 100 \dots \left(\mathrm{i}\right)$

$4x+3y\le 75 \dots \left(\mathrm{ii}\right)$

$x\ge 0$

$y\ge 0$

$x\le \frac{75-3y}{4}$

Given $z=y(6x+y)$

$\Rightarrow z\le y\left(6·\left(\frac{75-3y}{4}\right)+y\right)$

$\Rightarrow z\le \frac{1}{2}\left(225y-7y^2\right)$

Now range of $\frac{1}{2}\left(225y-7y^2\right)$ is $(-\infty ,\frac{(225{)}^2}{2\times 4\times 7}]$

So, $z\le \frac{1}{2}\left(225y-7y^2\right)\le \frac{(225{)}^2}{2\times 4\times 7}$

$=\frac{50625}{56}$

$\approx 904.0178$

$\approx 904.02$

It will be attained at $y=\frac{225}{14}$