I the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

Let length of one edge of cube be $x$ units and radius of sphere be $r$ units.

Therefore surface area of cube $= 6{x^2}$
and surface area of sphere $= 4\pi {r^2}$
Also, $6{x^2} + 4\pi {r^2} = k$ [constant, given]

$\Rightarrow$ $6{x^2} = k - 4\pi {r^2}$

$\Rightarrow$ ${x^2} = \frac{{k - 4\pi {r^2}}}{6}$

$\Rightarrow$ $x = {\left[ {\frac{{k - 4\pi {r^2}}}{6}} \right]^{1/2}}$ ……(i)

Now, volume of cube $= {x^3}$

and volume of sphere $= \frac{4}{3}\pi {r^3}$

Let sum of volume of the cube and volume of the sphere be given by $S = {x^3} + \frac{4}{3}\pi {r^3} = {\left[ {\frac{{k - 4\pi {r^2}}}{6}} \right]^{3/2}} + \frac{4}{3}\pi {r^3}$

On differentiating both sides w.r.t. $r$, we get
$\frac{{dS}}{{dr}} = \frac{3}{2}{\left[ {\frac{{k - 4\pi {r^2}}}{6}} \right]^{1/2}} \cdot \left( {\frac{{ - 8\pi r}}{6}} \right) + \frac{{12}}{3}\pi {r^2}$

$= - 2\pi r{\left[ {\frac{{k - 4\pi {r^2}}}{6}} \right]^{1/2}} + 4\pi {r^2}$ ….(ii)

$= - 2\pi r\left[ {{{\left\{ {\frac{{k - 4\pi {r^2}}}{6}} \right\}}^{1/2}} - 2r} \right]$

Now, $\frac{{dS}}{{dr}} = 0$
$\Rightarrow$ $r = 0$ or $2r = {\left( {\frac{{k - 4\pi {r^2}}}{6}} \right)^{1/2}}$

$\Rightarrow$ $4{r^2} = \frac{{k - 4\pi {r^2}}}{6} \Rightarrow 24{r^2} = k - 4\pi {r^2}$

$\Rightarrow$ $24{r^2} + 4\pi {r^2} = k \Rightarrow {r^2}[24 + 4\pi ] = k$

Therefore,$r = 0$ or $r = \sqrt {\frac{k}{{24 + 4\pi }}} = \frac{1}{2}\sqrt {\frac{k}{{6 + \pi }}}$

We know that, $r \ne 0$

Therefore, $r = \frac{1}{2}\sqrt {\frac{k}{{6 + \pi }}}$

Again, differentiating w.r.t. $r$ in Eq. (ii), we get
$\frac{{{d^2}S}}{{d{r^2}}} = \frac{d}{{dr}}\left[ { - 2\pi r\left\{ {{{\left( {\frac{{k - 4\pi {r^2}}}{6}} \right)}^{1/2}} + 4\pi {r^2}} \right\}} \right]$

$= - 2\pi \left[ {r \cdot \frac{1}{2}{{\left( {\frac{{k - 4\pi {r^2}}}{6}} \right)}^{ - 1/2}} \cdot \left( {\frac{{ - 8\pi r}}{6}} \right) + {{\left( {\frac{{k - 4\pi {r^2}}}{6}} \right)}^{1/2}} \cdot 1} \right] + 4\pi \cdot 2r$

$= - 2\pi \left[ {r \cdot \frac{1}{{2\sqrt {\frac{{k - 4\pi {r^2}}}{6}} }} \cdot \left( {\frac{{ - 8\pi r}}{6}} \right) + \sqrt {\frac{{k - 4\pi {r^2}}}{6}} } \right] + 8\pi r$

$= - 2\pi \left[ {\frac{{ - 8\pi {r^2} + 12\left( {k - \frac{{4\pi {r^2}}}{6}} \right)}}{{12\sqrt {\frac{{k - 4\pi {r^2}}}{6}} }}} \right] + 8\pi r$

$= - 2\pi \left[ {\frac{{ - 48\pi {r^2} + 72k - 48\pi {r^2}}}{{72\sqrt {\frac{{k - 4\pi {r^2}}}{6}} }}} \right] + 8\pi r = - 2\pi \left[ {\frac{{ - 96\pi {r^2} + 72k}}{{72\sqrt {\frac{{k - 4\pi {r^2}}}{6}} }}} \right] + 8\pi r > 0$

For $r = \frac{1}{2}\sqrt {\frac{k}{{6 + \pi }}}$, then the sum of their volume is minimum.

For $r = \frac{1}{2}\sqrt {\frac{k}{{6 + \pi }}}$, $x = {\left[ {\frac{{k - 4\pi \cdot \frac{1}{4}\frac{k}{{(6 + \pi )}}}}{6}} \right]^{1/2}}$

$= {\left[ {\frac{{(6 + \pi )k - \pi k}}{{6(6 + \pi )}}} \right]^{1/2}} = {\left[ {\frac{k}{{6 + \pi }}} \right]^{1/2}} = 2r$

Since, the sum of their volume is minimum when $x = 2r$.

Therefore the ratio of an edge of cube to the diameter of the sphere is 1: 1 .