The maximum value of

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 $\mathrm{y}=\frac{225}{14}$

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