Question

Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius $5\sqrt{3cm}is500\mathrm{\pi }{\mathrm{cm}}^3.$

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{height},\mathrm{radius}\mathrm{of}\mathrm{base}\mathrm{and}\mathrm{volume}\mathrm{of}\mathrm{a}\mathrm{cylinder}\mathrm{be}h,r\mathrm{and}V,\mathrm{respectively}.\mathrm{Then}, \\ \frac{h^2}{4}+r^2=R^2 \\ \Rightarrow h^2=4\left(R^2-r^2\right) \\ \Rightarrow r^2=R^2-\frac{h^2}{4}...\left(1\right) \\ \mathrm{Now}, \\ V=\pi r^{2}h \\ \Rightarrow V=\pi \left(h{R}^2-\frac{h^3}{4}\right)\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow \frac{dV}{dh}=\pi \left(R^2-\frac{3h^2}{4}\right) \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}V,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dV}{dh}=0 \\ \Rightarrow \pi \left(R^2-\frac{3h^2}{4}\right)=0 \\ \Rightarrow R^2-\frac{3h^2}{4}=0 \\ \Rightarrow R^2=\frac{3h^2}{4} \\ \Rightarrow h=\frac{2R}{\sqrt{3}} \\ \frac{d^{2}V}{d{h}^2}=\frac{-3\pi h}{2} \\ \frac{d^{2}V}{d{h}^2}=\frac{-3\pi }{2}\times \frac{2R}{\sqrt{3}} \\ \Rightarrow \frac{d^{2}V}{d{h}^2}=\frac{-3\pi R}{\sqrt{3}}<0 \\ \mathrm{So},\mathrm{the}\mathrm{volume}\mathrm{is}\mathrm{maximum}\mathrm{when}\mathrm{h}=\frac{2\mathrm{R}}{\sqrt{3}}. \\ \mathrm{Maximum}\mathrm{volume}=\pi h\left(R^2-\frac{h^2}{4}\right) \\ =\pi \times \frac{2R}{\sqrt{3}}\left(R^2-\frac{4R^2}{12}\right) \\ =\frac{2\pi R}{\sqrt{3}}\frac{8R^2}{12} \\ =\frac{4\pi R^3}{3\sqrt{3}} \\ =\frac{4\pi {\left(5\sqrt{3}\right)}^3}{3\sqrt{3}} \\ =500\pi {\mathrm{cm}}^3 \end{gathered}$$