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Question Number 21840 by Joel577 last updated on 05/Oct/17

$$\mathrm{A}\mathrm{cone}\mathrm{is}\mathrm{placed}\mathrm{inside}\mathrm{a}\mathrm{sphere}.$$$$\mathrm{If}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cone}\mathrm{is}\mathrm{maximum},$$$$\mathrm{find}\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{radius}\mathrm{from}\mathrm{the}\mathrm{cone}\mathrm{and}\mathrm{sphere}$$

Answered by mrW1 last updated on 05/Oct/17

$$\mathrm{R}=\mathrm{radius}\mathrm{of}\mathrm{sphere}$$$$\mathrm{r}=\mathrm{radius}\mathrm{of}\mathrm{cone}$$$$\mathrm{V}=\mathrm{volume}\mathrm{of}\mathrm{cone}$$$$\mathrm{V}=\frac{1}{3}\pi {\mathrm{r}}^2\times \mathrm{h}=\frac{1}{3}\times \pi {\mathrm{r}}^2\times (\mathrm{R}+\sqrt{{\mathrm{R}}^2-{\mathrm{r}}^2})$$$$\mathrm{V}=\frac{\pi }{3}{\mathrm{r}}^2(\mathrm{R}+\sqrt{{\mathrm{R}}^2-{\mathrm{r}}^2})$$$$\frac{\mathrm{dV}}{\mathrm{dr}}=\frac{2\pi \mathrm{r}}{3}(\mathrm{R}+\sqrt{{\mathrm{R}}^2-{\mathrm{r}}^2}-\frac{{\mathrm{r}}^2}{2\sqrt{{\mathrm{R}}^2-{\mathrm{r}}^2}})=0$$$$\mathrm{R}+\sqrt{{\mathrm{R}}^2-{\mathrm{r}}^2}-\frac{{\mathrm{r}}^2}{2\sqrt{{\mathrm{R}}^2-{\mathrm{r}}^2}}=0$$$$1+\sqrt{1-{\left(\frac{\mathrm{r}}{\mathrm{R}}\right)}^2}-\frac{{\left(\frac{\mathrm{r}}{\mathrm{R}}\right)}^2}{2\sqrt{1-{\left(\frac{\mathrm{r}}{\mathrm{R}}\right)}^2}}=0$$$$\mathrm{with}\mathrm{x}=\frac{\mathrm{r}}{\mathrm{R}}$$$$1+\sqrt{1-{\mathrm{x}}^2}-\frac{{\mathrm{x}}^2}{2\sqrt{1-{\mathrm{x}}^2}}=0$$$$\Rightarrow \mathrm{x}\approx 0.943$$

Commented by Joel577 last updated on 05/Oct/17

$$\mathrm{I}\mathrm{dont}\mathrm{understand}\mathrm{why}V=4\pi r^2\sqrt{R^2-r^2}$$$$\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{is}V=\frac{1}{3}\pi r^{2}h$$

Commented by mrW1 last updated on 05/Oct/17

$$\mathrm{you}\mathrm{are}\mathrm{right}\mathrm{sir}.$$$$\mathrm{I}\mathrm{have}\mathrm{corrected}.$$

Commented by Joel577 last updated on 05/Oct/17

$$thankyouverymuch$$

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