Question

Show that the height of the cylinder of maximum volume that can be inscribed a sphere of radius R is $\frac{2R}{\sqrt{3}}.$

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{height}\mathrm{and}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{base}\mathrm{of}\mathrm{the}\mathrm{cylinder}\mathrm{be}h\mathrm{and}r,\mathrm{respectively}.\mathrm{Then}, \\ \frac{h^2}{4}+r^2=R^2 \\ \Rightarrow h=2\sqrt{R^2-r^2}...\left(1\right) \\ \mathrm{Volume}\mathrm{of}\mathrm{cylinder},V=\pi r^{2}h \\ \text{Squaring both sides, we get} \\ \Rightarrow V^2={\pi }^2{r}^4{h}^2 \\ \Rightarrow V^2=4{\pi }^2{r}^4\left(R^2-r^2\right)\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \mathrm{Now}, \\ Z=4{\pi }^2\left(r^4{R}^2-r^6\right) \\ \Rightarrow \frac{dZ}{dr}=4{\pi }^2\left(4r^3{R}^2-6r^5\right) \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}Z,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dZ}{dr}=0 \\ \Rightarrow 4{\pi }^2\left(4r^3{R}^2-6r^5\right)=0 \\ \Rightarrow 4r^3{R}^2=6r^5 \\ \Rightarrow 6r^2=4R^2 \\ \Rightarrow r^2=\frac{4R^2}{6} \\ \Rightarrow r=\frac{2R}{\sqrt{6}} \\ \text{Substituting the value of}\mathit{\text{r}}\text{in}\mathrm{eq}.\left(\text{1}\right)\text{, we get} \\ \Rightarrow h=2\sqrt{R^2-{\left(\frac{2R}{\sqrt{6}}\right)}^2} \\ \Rightarrow h=2\sqrt{\frac{6R^2-4R^2}{6}} \\ \Rightarrow h=2\sqrt{\frac{R^2}{3}} \\ \Rightarrow h=\frac{2R}{\sqrt{3}} \\ \mathrm{Now}, \\ \frac{d^{2}Z}{d{r}^2}=4{\pi }^2\left(12r^2{R}^2-30r^4\right) \\ \Rightarrow \frac{d^{2}Z}{d{r}^2}=4{\pi }^2\left(12{\left(\frac{2R}{\sqrt{6}}\right)}^2{R}^2-30{\left(\frac{2R}{\sqrt{6}}\right)}^4\right) \\ \Rightarrow \frac{d^{2}Z}{d{r}^2}=4{\pi }^2\left(8R^4-\frac{80R^4}{6}\right) \\ \Rightarrow \frac{d^{2}Z}{d{r}^2}=4{\pi }^2\left(\frac{48R^4-80R^4}{6}\right) \\ \Rightarrow \frac{d^{2}Z}{d{r}^2}=4{\pi }^2\left(-\frac{16R^4}{3}\right)<0 \\ \text{So, volume of the cylinder is maximum when}\mathit{\text{h}}=\frac{2R}{\sqrt{3}}. \\ \text{Hence proved.} \end{gathered}$$