lf $x + y = 28$ 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 B $12^{3} {.16}^{4}$
If $x + y = k$ then maximum value of $x^{m} y^{n}$ is at $x = \frac{k m}{m + n}, y = \frac{k m}{m + n}$ where $x, y > 0$ and $m, n \geq 1$
Here $k = 28, m = 3, n = 4$
So, $x = 12, y = 16$
Hence maximum value is $12^{3} {.16}^{4}$

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Similar questions

x^{3} y^{4}

A

x + y = 12

x^{2} + y^{2}

x + y = k

k^{2}

x + y = 28

x^{3} y^{4}

x^{2} + y^{2} = 1

5 cos x + 12 cos y = 13

5 sin x + 12 sin y

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