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


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?



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 $=\frac{4}{3} \pi r^{3}$


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.


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