Question

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.

Answer 1

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,
$S=4\mathrm{\pi }{r}^2+6x^2$
$\Rightarrow$$x={\left(\frac{S-4\mathrm{\pi }{r}^2}{6}\right)}^{\frac{1}{2}}$ ...(1)

Sum of volumes, V= $\frac{4}{3}\mathrm{\pi }{r}^3+x^3$

$\Rightarrow$V = $\frac{4\mathrm{\pi }{r}^3}{3}+{\left[\frac{\left(S-4\mathrm{\pi }{r}^2\right)}{6}\right]}^{\frac{3}{2}}$ [From eq. (1)]

$\Rightarrow \frac{dV}{dr}=4\mathrm{\pi }{r}^2-2\mathrm{\pi }r{\left[\frac{\left(S-4\mathrm{\pi }{r}^2\right)}{6}\right]}^{\frac{1}{2}}$

For the minimum or maximum values of V, we must have
$\frac{dV}{dr}=0$ ...(2)
$$\begin{gathered} \Rightarrow 4\mathrm{\pi }{r}^2-2\mathrm{\pi }r{\left[\frac{\left(S-4\mathrm{\pi }{r}^2\right)}{6}\right]}^{\frac{1}{2}}=0\left[\mathrm{From}\mathrm{eq}.\left(2\right)\right] \\ \Rightarrow 4\mathrm{\pi }{r}^2=2\mathrm{\pi }r{\left[\frac{\left(S-4\mathrm{\pi }{r}^2\right)}{6}\right]}^{\frac{1}{2}} \\ \Rightarrow 4\mathrm{\pi }{r}^2=2\mathrm{\pi }rx\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow x=2r \end{gathered}$$

Now,
$$\begin{gathered} \frac{d^{2}V}{d{r}^2}=8\pi r-2\pi {\left[\frac{\left(S-4\pi r^2\right)}{6}\right]}^{\frac{1}{2}}-\frac{2\pi r}{2}{\left[\frac{\left(S-4\pi r^2\right)}{6}\right]}^{-\frac{1}{2}}\frac{\left(-8\pi r\right)}{6} \\ \Rightarrow \frac{d^{2}V}{d{r}^2}=8\pi r-2\pi {\left[\frac{\left(S-4\pi r^2\right)}{6}\right]}^{\frac{1}{2}}+\frac{4}{3}{\pi }^2{r}^2{\left[\frac{6}{\left(S-4\pi r^2\right)}\right]}^{\frac{1}{2}} \\ \Rightarrow \frac{d^{2}V}{d{r}^2}=8\pi r-2\pi x+\frac{4}{3}{\pi }^2{r}^2\frac{1}{x}=8\pi r-4\pi r+\frac{2}{3}{\pi }^{2}r \\ \Rightarrow \frac{d^{2}V}{d{r}^2}=4\pi r+\frac{2}{3}{\pi }^{2}r>0 \end{gathered}$$

So, volume is minimum when x = 2r.