If $x + y = 28$ then the maximum value of $x^{3} y^{4}$ is

4^{3} {.24}^{4}

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12^{3} {.16}^{4}

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4321

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1234

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Solution

The correct option is C $12^{3} {.16}^{4}$
$x + y = 28 ⟶ (1)$
$∴ y = 28 - x$
$∴ x^{3} y^{4}$
$⟹ x^{3} (28 - x)^{4}$
$\frac{d}{d x} [x^{3} (28 - x)^{4}] = 0$
$3 x^{2} (28 - x)^{4} - 4 (28 - x)^{3} x^{3} = 0$
$3 (28 - x) = 4 x$
$∴ x = 12$
$∴ y = (28 - 12) = 16$
$∴$ Maximum value $= 12^{3} 16^{4}$

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