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

We have two numbers $x$ and $y$ such that
$x + y = 60$ … (i)

Let $P = x{y^3} \Rightarrow P = (60 - y){y^3}$ [from(i)]
$\Rightarrow P = 60{y^3} - {y^4} \Rightarrow \cfrac{{dP}}{{dy}} = 180{y^2} - 4{y^3}$

For maximum $P$, we must have $\cfrac{{dP}}{{dy}} = 0$

$\Rightarrow 180{y^2} - 4{y^3} = 0 \Rightarrow 4{y^2}(45 - y) = 0 \Rightarrow y = 45$
($y = 0$ is not possible for $x{y^3}$ to be maximized)

Also, $\cfrac{{{d^2}P}}{{d{y^2}}} = 360y - 12{y^2}$
and ${\left( {\cfrac{{{d^2}P}}{{d{y^2}}}} \right)_{y = 45}} = 360 \times 45 - 12 \times {(45)^2} < 0$

Therefore, $P$ is maximum when $y = 45$.

Therefore the required numbers are $x = 60 - y = 60 - 45 = 15$ and $y = 45$.