Application of Derivatives — Class 12 Maths Solution

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 .