Find two positive numbers $x$ and $y$ such that $x+y=60$ and $xy^3$ is minimum.

The correct answer is $x = 15$ and $y = 45$

Given that $x+y = 60$ ⇒ $x = 60 - y$


Let $P = xy^3$ = $(60 - y)y^3 $ = $60y^3-y^4$


∴ $\dfrac{dP}{dy}$ = $180y^2 - 4y^3$ and $\dfrac{d^2P}{dy^2}$ = $360y - 12y^2$


Now, for maxima, $\dfrac{dP}{dx} = 0$ ⇒ $180y^2 - 4y^3$ = 0 or $4y^2(45-y) = 0$

∴ Either $y = 0$ or $y=45$

At $x = \dfrac{45}{2}$, $ \left( \dfrac{d^2P}{dy^2} \right)_{x= 45}$ = $360 (45) - 12 (45)^2$ = $-8100 < 0$

∴ By second derivative test, we have a local maxima of $P$ at $y = 45$

As $x+y=60$, $x= 15$

Hence $P$ is maximum when the numbers are $15$ and $45$