$The greatest value of x^{2} y^{3} where x > 0, y > 0 and 3 x + 4 y = 5 is$

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$\frac{3}{16}$

$\frac{3}{8}$

$\frac{6}{5}$

$\frac{9}{5}$

answer is B.

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Detailed Solution

$\frac{2. \frac{3 x}{2} + 3 \frac{4 y}{3} \geq}{5} {({(\frac{3 x}{2})}^{2} {(\frac{4 y}{3})}^{3})}^{\frac{1}{5}} \Rightarrow x^{2} y^{3} \leq \frac{3}{16}$

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