The sum of the surface areas of a sphere and a cuboid with sides x/3, x and 2x is constant. Show that the sum of their volumes is minimum if x is equal to three times the radius of sphere.

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$$\begin{gathered} \mathrm{For} \mathrm{cuboid} \\ \mathrm{l}=\frac{\mathrm{x}}{3} \\ \mathrm{b}=\mathrm{x} \\ \mathrm{h}=2\mathrm{x} \\ \mathrm{Surface} \mathrm{Area}=2\left(\mathrm{lb}+\mathrm{bh}+\mathrm{hl}\right)=2\left(\frac{\mathrm{x}}{3}.\mathrm{x}+\mathrm{x}.2\mathrm{x}+2\mathrm{x}.\frac{\mathrm{x}}{3}\right)=2\left(\frac{{\mathrm{x}}^2}{3}+2{\mathrm{x}}^2+\frac{2{\mathrm{x}}^2}{3}\right) \\ =6{\mathrm{x}}^2 \\ \mathrm{Let} \mathrm{radius} \mathrm{of} \mathrm{sphere} \mathrm{be} \text{'}\mathrm{r}\text{'} \\ \mathrm{Surface} \mathrm{Area}=4{\mathrm{\pi r}}^2 \\ \mathrm{According} \mathrm{to} \mathrm{question}: \\ 6{\mathrm{x}}^2+4{\mathrm{\pi r}}^2=\mathrm{C} \left[\mathrm{Constant}\right] \\ \Rightarrow 4{\mathrm{\pi r}}^2=\mathrm{C}-6{\mathrm{x}}^2 ;\left(\mathrm{i}\right) \\ \Rightarrow \mathrm{r}=\frac{\sqrt{\mathrm{C}-6{\mathrm{x}}^2}}{2\sqrt{\mathrm{\pi }}} ;\left(\mathrm{ii}\right) \\ \mathrm{Sum} \mathrm{of} \mathrm{volume} \\ =\mathrm{S}=\frac{4}{3}{\mathrm{\pi r}}^3+\mathrm{l}.\mathrm{b}.\mathrm{h}=\frac{4}{3}{\mathrm{\pi r}}^3+\left(\frac{\mathrm{x}}{3}\right)\left(\mathrm{x}\right)\left(2\mathrm{x}\right)=\frac{1}{3}\left[4{\mathrm{\pi r}}^3+2{\mathrm{x}}^3\right] \\ \mathrm{Using} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}: \\ \mathrm{S}=\frac{1}{3}\left[4\mathrm{\pi }{\left(\frac{\sqrt{\mathrm{C}-6{\mathrm{x}}^2}}{2\sqrt{\mathrm{\pi }}}\right)}^3+2{\mathrm{x}}^3\right]=\frac{1}{3}\left[4\mathrm{\pi }{\frac{\left(\sqrt{\mathrm{C}-6{\mathrm{x}}^2}\right)}{8\mathrm{\pi }\sqrt{\mathrm{\pi }}}}^3+2{\mathrm{x}}^3\right] \\ \mathrm{S}=\frac{1}{3}\left[\frac{{\left(\mathrm{C}-6{\mathrm{x}}^2\right)}^{{\displaystyle \frac{3}{2}}}}{2\sqrt{\mathrm{\pi }}}+2{\mathrm{x}}^3\right] \\ \mathrm{Differentiating} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x}, \mathrm{we} \mathrm{get}: \\ 3\frac{\mathrm{dS}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[{\frac{\left(\mathrm{C}-6{\mathrm{x}}^2\right)}{2\sqrt{\mathrm{\pi }}}}^{\frac{3}{2}}\right]+\frac{\mathrm{d}}{\mathrm{dx}}\left(2{\mathrm{x}}^3\right) \\ \Rightarrow 3\frac{\mathrm{dS}}{\mathrm{dx}}=\frac{1}{2\sqrt{\mathrm{\pi }}}.\frac{3}{2}{\left(\mathrm{C}-6{\mathrm{x}}^2\right)}^{\frac{3}{2}-1}.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{C}-6{\mathrm{x}}^2\right)+6{\mathrm{x}}^2 \\ \Rightarrow 3\frac{\mathrm{dS}}{\mathrm{dx}}=\frac{1}{2\sqrt{\mathrm{\pi }}}.\frac{3}{2}{\left(\mathrm{C}-6{\mathrm{x}}^2\right)}^{\frac{1}{2}}\left(-12\mathrm{x}\right)+6{\mathrm{x}}^2 \\ \mathrm{For} \mathrm{maxima} \mathrm{or} \mathrm{minima} \frac{\mathrm{dS}}{\mathrm{dx}}=0 \\ \Rightarrow -\frac{1}{2\sqrt{\mathrm{\pi }}}.\frac{3}{2}{\left(\mathrm{C}-6{\mathrm{x}}^2\right)}^{\frac{1}{2}}\left(12\mathrm{x}\right)+6{\mathrm{x}}^2=0 \\ \Rightarrow \frac{1}{\sqrt{\mathrm{\pi }}}.3{\left(\mathrm{C}-6{\mathrm{x}}^2\right)}^{\frac{1}{2}}\left(12\mathrm{x}\right)=24{\mathrm{x}}^2 \\ \Rightarrow \frac{3}{\sqrt{\mathrm{\pi }}}{\left(\mathrm{C}-6{\mathrm{x}}^2\right)}^{\frac{1}{2}}=2\mathrm{x} \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides} \mathrm{we} \mathrm{get}: \\ \frac{9}{\mathrm{\pi }}\left(\mathrm{C}-6{\mathrm{x}}^2\right)=4{\mathrm{x}}^2 \\ \Rightarrow \mathrm{C}-6{\mathrm{x}}^2=\frac{4\mathrm{\pi }}{9}{\mathrm{x}}^2 \\ \left[\mathrm{Note}: \mathrm{You} \mathrm{have} \mathrm{to} \mathrm{check} \mathrm{that} \frac{{\mathrm{d}}^2\mathrm{S}}{{\mathrm{dS}}^2}<0 \mathrm{when} \mathrm{C}-6{\mathrm{x}}^2=\frac{4\mathrm{\pi }}{9}{\mathrm{x}}^2\right] \\ \mathrm{Put} \mathrm{this} \mathrm{value} \mathrm{in} \left(\mathrm{i}\right) \\ 4{\mathrm{\pi r}}^2=\frac{4\mathrm{\pi }}{9}{\mathrm{x}}^2 \\ \Rightarrow {\mathrm{r}}^2=\frac{{\mathrm{x}}^2}{9} \\ \Rightarrow \mathrm{r}=\frac{\mathrm{x}}{3} \\ \Rightarrow \mathbf{x}\mathbf{=}\mathbf{3}\mathbf{r}\mathbf{ }\left[\mathbf{Hence}\mathbf{ }\mathbf{proved}\right] \end{gathered}$$

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