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

Given $x+y=60$

$x=60-y........\left(1\right)$

let $f\left(y\right)=60y^3-y^4$

$f^{\mathrm{\prime }}\left(y\right)=180y^2-4y^3\Rightarrow f^{\prime \prime }\left(y\right)=360y-12y^2$

Consider $f^{\mathrm{\prime }}\left(y\right)=0$

$180y^2-4y^3=0\Rightarrow 4y^2(45-y)=0$

$4y^2=045-y=0$

$y=0\text{ (Not possible) }$

$y=45;\text{ At }y=45\text{, }$

$f^{\prime \prime }\left(y\right)=360y-12y^2=360\left(45\right)-12(45{)}^2$

$=12\times 45(30-45)=12\times 45\times -15$

$=-12\times 45\times 15<0$

It has maximum value at $y=45$

Sub in $\text{ (1) }x=60-y=60-45=15$

$\therefore$the numbers are $15 \& 45$