Find two positive numbers $x$ and $y$ such that $x + y = 60$ and $x y^{3}$ is maximum.

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Solution

Let $P = {x y}^{3}$
It is given that $x + y = 60$
$\Rightarrow x = 60 - y$
$P = (60 - y) y^{3}$ [Putting value of $x$ ]
$= {60 y}^{3} - y^{4}$
$\Rightarrow \frac{d P}{d y} = {180 y}^{2} - {4 y}^{3}$
$\frac{d^{2} P}{{d y}^{2}} = 360 y - 12 y^{2}$
For maximum or minimum values of $y$ , $P$ we have
$\frac{d P}{d y} = 0$
$\Rightarrow {180 y}^{2} - {4 y}^{3} = 0$
$\Rightarrow {4 y}^{2} (45 - y) = 0$
$\Rightarrow y = 0$ $45 - y = 0$ $y = 45$
Now ${(\frac{d^{2} P}{{d y}^{2}})}_{y = 45} = 360 \times 45 - 12 {(45)}^{2}$
$= 12 \times 45 - (30 - 45)$
$= - 8100 < 0$
$P$ is maximum when $y = 45$
when $y = 45$ $x + y = 60$ $\Rightarrow x = 60 - 45$
$x = 15$
Numbers are $15$ and $45$

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