# If the sum of the surface areas of cube and a sphere is constant,

Question:

If 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?

Solution:

sphere.

Surface area of cube = 6x2

And, surface area of the sphere = 4πr2

Now, their sum is

$6 x^{2}+4 \pi r^{2}=\mathrm{K}($ constant $) \Rightarrow r=\sqrt{\frac{\mathrm{K}-6 x^{2}}{4 \pi}}$.............(i)

Volume of the cube $=x^{3}$ and the volume of sphere $=\frac{4}{3} \pi r^{3}$

Now,

Sum of their volumes $(V)=$ Volume of cube

+ Volume of sphere

$V=x^{3}+\frac{4}{3} \pi r^{3}$

$\Rightarrow \mathrm{V}=x^{3}+\frac{4}{3} \pi \times\left(\frac{\mathrm{K}-6 x^{2}}{4 \pi}\right)^{3 / 2}$

Differentiating both sides w.r.t. $x$, we get

$\frac{d \mathrm{~V}}{d x}=3 x^{2}+\frac{4 \pi}{3} \times \frac{3}{2}\left(\mathrm{~K}-6 x^{2}\right)^{1 / 2}(-12 x) \times \frac{1}{(4 \pi)^{3 / 2}}$

$=3 x^{2}+\frac{2 \pi}{(4 \pi)^{3 / 2}} \times(-12 x)\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}$

$=3 x^{2}+\frac{1}{4 \pi^{1 / 2}} \times(-12 x)\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}$

Thus, $\frac{d \mathrm{~V}}{d x}=3 x^{2}-\frac{3 x}{\sqrt{\pi}}\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}$ $\ldots$ (ii)

For local maxima and local minima, $\frac{d V}{d x}=0$

$3 x^{2}-\frac{3 x}{\sqrt{\pi}}\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}=0$

$3 x\left[x-\frac{\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}}{\sqrt{\pi}}\right]=0$

$x \neq 0$ and $x-\frac{\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}}{\sqrt{\pi}}=0$

$\Rightarrow \quad x=\frac{\left(\mathrm{K}-6 x^{2}\right)^{1 / 2}}{\sqrt{\pi}}$

Squaring both sides, we get

$x^{2}=\frac{\mathrm{K}-6 x^{2}}{\pi} \Rightarrow \pi x^{2}=\mathrm{K}-6 x^{2}$

So, $\quad \pi x^{2}+6 x^{2}=\mathrm{K} \Rightarrow x^{2}(\pi+6)=\mathrm{K} \Rightarrow x^{2}=\frac{\mathrm{K}}{\pi+6}$

Thus, $x=\sqrt{\frac{\mathrm{K}}{\pi+6}}$

Now pulting the value of $\mathrm{K}$ in eq. (i), we get

$6 x^{2}+4 \pi r^{2}=x^{2}(\pi+6)$

$6 r^{2}+4 \pi r^{2}=\pi r^{2}+6 r^{2} \rightarrow 4 \pi r^{2}=\pi r^{2} \rightarrow 4 r^{2}=r^{2}$

$2 r=x$

$\therefore x: 2 r=1: 1$

Now differentiating eq. (ii) w.r.t $x$, we have

$\frac{d^{2} \mathrm{~V}}{d x^{2}}=6 x-\frac{3}{\sqrt{\pi}} \frac{d}{d x}\left[x\left(\mathrm{~K}-6 x^{2}\right)^{1 / 2}\right]$

$=6 x-\frac{3}{\sqrt{\pi}}\left[x \cdot \frac{1}{2 \sqrt{K-6 x^{2}}} \times(-12 x)+\left(\mathrm{K}-6 x^{2}\right)^{1 / 2} \cdot 1\right]$

$=6 x-\frac{3}{\sqrt{\pi}}\left[\frac{-6 x^{2}}{\sqrt{\mathrm{K}-6 x^{2}}}+\sqrt{\mathrm{K}-6 x^{2}}\right]$

$=6 x-\frac{3}{\sqrt{\pi}}\left[\frac{-6 x^{2}+\mathrm{K}-6 x^{2}}{\sqrt{\mathrm{K}-6 x^{2}}}\right]=6 x+\frac{3}{\sqrt{\pi}}\left[\frac{12 x^{2}-\mathrm{K}}{\sqrt{\mathrm{K}-6 x^{2}}}\right]$

Put $x=\sqrt{\frac{K}{\pi+6}}=6 \sqrt{\frac{K}{\pi+6}}+\frac{3}{\sqrt{\pi}}\left[\frac{\frac{12 K}{\pi+6}-K}{\sqrt{K-\frac{6 K}{\pi+6}}}\right]$

$=6 \sqrt{\frac{\mathrm{K}}{\pi+6}}+\frac{3}{\sqrt{\pi}}\left[\frac{12 \mathrm{~K}-\pi \mathrm{K}-6 \mathrm{~K}}{\sqrt{\frac{\pi \mathrm{K}+6 \mathrm{~K}-6 \mathrm{~K}}{\pi+6}}}\right]$

$=6 \sqrt{\frac{\mathrm{K}}{\pi+6}}+\frac{3}{\sqrt{\pi}}\left[\frac{6 \mathrm{~K}-\pi \mathrm{K}}{\sqrt{\frac{\pi \mathrm{K}}{\pi+6}}}\right]$

$=6 \sqrt{\frac{\mathrm{K}}{\pi+6}}+\frac{3}{\pi \sqrt{K}}[(6 \mathrm{~K}-\pi \mathrm{K}) \sqrt{\pi+6}]>0$

So, it is the minima.

Therefore, the required ratio is 1: 1 when the combined volume is minimum.