Let positive real numbers $x$ and $y$ be such that $3 x + 4 y = 14$ . The maximum value of $x^{3} y^{4}$ is

4

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128

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216

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432

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Solution

The correct option is A $128$
GIven both $x$ and $y$ are positive and such that $3 x + 4 y = 14$ ,
$⟹ x^{3} y^{4} = x^{3} {(\frac{14 - 4 x}{4})}^{4} = f (x)$
Now for the maximum $f^{^{'}} (x) = 0 ⟹ 3 x^{2} {(\frac{14 - 3 x}{4})}^{4} = {3 x}^{3} {(\frac{14 - 3 x}{4})}^{3} ⟹ 4 x = 14 - 3 x ⟹ x = 2 ⟹ m a x (f (x)) = 128$

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