The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube.
Let r be the radius of the sphere, x be the side of the cube and S be the sum of the surface area of both. Then,
$S=4 \pi r^{2}+6 x^{2}$
$\Rightarrow x=\left(\frac{S-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}$ .....(1)
Sum of volumes, $V=\frac{4}{3} \pi r^{3}+x^{3}$
$\Rightarrow V=\frac{4 \pi r^{3}}{3}+\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{3}{2}}$ [From eq. (1)]
$\Rightarrow \frac{d V}{d r}=4 \pi r^{2}-2 \pi r\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}$
For the minimum or maximum values of $V$, we must have
$\frac{d V}{d r}=0$ ....(2)
$\Rightarrow 4 \pi r^{2}-2 \pi r\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}=0$ [From eq. (2)]
$\Rightarrow 4 \pi r^{2}=2 \pi r\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}$
$\Rightarrow 4 \pi r^{2}=2 \pi r x$ [From eq. (1)]
$\Rightarrow x=2 r$
Now,
$\frac{d^{2} V}{d r^{2}}=8 \pi r-2 \pi\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}-\frac{2 \pi r}{2}\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{-\frac{1}{2}} \frac{(-8 \pi r)}{6}$
$\Rightarrow \frac{d^{2} V}{d r^{2}}=8 \pi r-2 \pi\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}+\frac{4}{3} \pi^{2} r^{2}\left[\frac{6}{\left(S-4 \pi r^{2}\right)}\right]^{\frac{1}{2}}$
$\Rightarrow \frac{d^{2} V}{d r^{2}}=8 \pi r-2 \pi x+\frac{4}{3} \pi^{2} r^{2} \frac{1}{x}=8 \pi r-4 \pi r+\frac{2}{3} \pi^{2} r$
$\Rightarrow \frac{d^{2} V}{d r^{2}}=4 \pi r+\frac{2}{3} \pi^{2} r>0$
So, volume is minimum when x = 2r.