The maximum value of z in the following equation z = 6xy + y ^ 2 ,...

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

$3x + 4y \le 100 \ldots \left(i\right)$

$4x + 3y \le 75 \ldots \left(ii\right)$

$x \ge 0$

$y \ge 0$

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

Given $z = y ( 6x + y )$

$⇒ z \le y\left(6 \cdot \left(\dfrac{75 - 3y}{4}\right) + y\right)$

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

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

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

$= \dfrac{50625}{56}$

$\approx 904 . 0178$

$\approx 904 . 02$

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