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The maximum value of z in the following equation z=6xy+y^{2}, where 3x+4y\leq 100 and 3x+4y\leq 75 for x\geqslant 0 and y\geqslant 0 is ______________.
Option: 1 904
Option: 2 905
Option: 3 906
Option: 4 907

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\begin{array}{l} z =6 xy + y ^{2}= y (6 x + y )\\ 3 x+4 y \leq 100\qquad\ldots(i)\\ 4 x+3 y \leq 75 \qquad\ldots(ii)\end{array}

\begin{array}{l} x \geq 0 \\ y \geq 0 \end{array}

\begin{aligned} &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) \end{array}

For maxima and minima

\frac{\mathrm{dZ} }{\mathrm{d} x} \leq \frac{1}{2}\left(225 -14 y\right) =0

\Rightarrow y =\frac{225}{14}

Hence maximum value of Z is

\begin{aligned} 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 \end{array}

 

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Suraj Bhandari

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