Question

Show that the cone of the greatest volume which can be inscribed in a given spher has an altitude equal to $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ of the diameter of the sphere.

Answer 1

$$\begin{gathered} \mathrm{Let}h,r\mathrm{and}R\mathrm{be}\mathrm{the}\mathrm{height},\mathrm{radius}\mathrm{of}\mathrm{base}\mathrm{of}\mathrm{the}\mathrm{cone}\mathrm{and}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{sphere},\mathrm{respectively}.\mathrm{Then}, \\ h=R+\sqrt{R^2-r^2} \\ \Rightarrow {\left(h-R\right)}^2=R^2-r^2 \\ \Rightarrow h^2+R^2-2hr=R^2-r^2 \\ \Rightarrow r^2=2hR-h^2...\left(1\right) \\ \text{Volume of cone}=\frac{1}{3}\pi r^{2}h \\ \Rightarrow V=\frac{1}{3}\pi h\left(2hR-h^2\right)\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow V=\frac{1}{3}\pi \left(2h^{2}R-h^3\right) \\ \Rightarrow \frac{dV}{dh}=\frac{\pi }{3}\left(4hR-3h^2\right) \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}V,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dV}{dh}=0 \\ \Rightarrow \frac{\pi }{3}\left(4hR-3h^2\right)=0 \\ \Rightarrow 4hR=3h^2 \\ \Rightarrow h=\frac{4R}{3} \\ \text{Substituting the value of}\mathit{\text{y}}\text{in}\mathrm{eq}.\left(\text{1}\right)\text{, we get} \\ {x}^2=4\left(r^2-{\left(\frac{r}{\sqrt{2}}\right)}^2\right) \\ \Rightarrow x^2=4\left(r^2-\frac{r^2}{2}\right) \\ \Rightarrow x^2=4\left(\frac{r^2}{2}\right) \\ \Rightarrow x^2=2r^2 \\ \Rightarrow x=r\sqrt{2} \\ \mathrm{Now}, \\ \frac{d^{2}V}{d{h}^2}=\frac{\pi }{3}\left(4R-6h\right) \\ \Rightarrow \frac{d^{2}V}{d{h}^2}=\frac{\pi }{3}\left(4R-6\times \frac{4R}{3}\right) \\ \Rightarrow \frac{d^{2}V}{d{h}^2}=\frac{-4\pi R}{3}<0 \\ \mathrm{So},\mathrm{the}\mathrm{volume}\mathrm{is}\mathrm{maximum}\mathrm{when}h=\frac{4R}{3}. \\ \Rightarrow h=\frac{2}{3}\left(\mathrm{Diameter}\mathrm{of}\mathrm{sphere}\right) \\ \text{Hence proved.} \end{gathered}$$