A given right circular cone has volume p and the largest right circular cylinder that can be inscribed in the cone has a volume q . Find the ratio of p and q.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Let} \mathrm{r} \mathrm{and} \mathrm{h} \mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{and} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{right} \mathrm{circlular} \mathrm{cone} \\ \mathrm{and} \mathrm{R} \mathrm{and} \mathrm{H} \mathrm{be} \mathrm{the} \mathrm{radius} \mathrm{and} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{greatest} \mathrm{right} \mathrm{circlular} \mathrm{cylinder} \\ \mathrm{that} \mathrm{can} \mathrm{be} \mathrm{inscribed} \mathrm{in} \mathrm{the} \mathrm{cone}. \end{gathered}$$



$$\begin{gathered} \mathrm{Then} \angle \mathrm{GAO}=\mathrm{\alpha }, \mathrm{OG}=\mathrm{r}, \mathrm{OA}=\mathrm{h}, \mathrm{OE}=\mathrm{R}, \mathrm{CE}=\mathrm{H} \\ \mathrm{We} \mathrm{have} \\ \mathrm{r}=\mathrm{h} \mathrm{tan\alpha } \\ \mathrm{Since} △\mathrm{AOG}~△\mathrm{CEG}, \mathrm{we} \mathrm{get} \\ \frac{\mathrm{AO}}{\mathrm{OG}}=\frac{\mathrm{CE}}{\mathrm{EG}} \\ \Rightarrow \frac{\mathrm{h}}{\mathrm{r}}=\frac{\mathrm{H}}{\mathrm{r}-\mathrm{R}} \\ \Rightarrow \mathrm{H}=\frac{\mathrm{h}}{\mathrm{r}}\left(\mathrm{r}-\mathrm{R}\right) \\ =\frac{\mathrm{h}}{\mathrm{h} \mathrm{tan\alpha }}\left(\mathrm{h} \mathrm{tan\alpha }-\mathrm{R}\right) \\ =\frac{1}{\mathrm{tan\alpha }}\left(\mathrm{h} \mathrm{tan\alpha }-\mathrm{R}\right) \\ \mathrm{So} \mathrm{the} \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylinder} \mathrm{is}, \\ \mathrm{V}={\mathrm{\pi R}}^2\mathrm{H} \\ =\frac{{\mathrm{\pi R}}^2}{\mathrm{tan\alpha }}\left(\mathrm{h} \mathrm{tan\alpha }-\mathrm{R}\right) \\ ={\mathrm{\pi R}}^2\mathrm{h}-\frac{{\mathrm{\pi R}}^3}{\mathrm{tan\alpha }} \\ \mathrm{The} \mathrm{derivative} \mathrm{of} \mathrm{this} \mathrm{function} \mathrm{is}, \\ \frac{\mathrm{dV}}{\mathrm{dR}}=2\mathrm{\pi Rh}-\frac{3{\mathrm{\pi R}}^2}{\mathrm{tan\alpha }} \\ \mathrm{Now} \frac{\mathrm{dV}}{\mathrm{dR}}=0 \mathrm{implies} \mathrm{that}, \\ 2\mathrm{\pi Rh}-\frac{3{\mathrm{\pi R}}^2}{\mathrm{tan\alpha }}=0 \\ \Rightarrow 2\mathrm{\pi Rh}=\frac{3{\mathrm{\pi R}}^2}{\mathrm{tan\alpha }} \\ \Rightarrow 2\mathrm{h} \mathrm{tan\alpha }=3\mathrm{R} \\ \Rightarrow \mathrm{R}=\frac{2\mathrm{h}}{3}\mathrm{tan\alpha } \\ \mathrm{The} \mathrm{second} \mathrm{derivative} \mathrm{is}, \\ \frac{{\mathrm{d}}^2\mathrm{V}}{{\mathrm{dR}}^2}=2\mathrm{\pi h}-\frac{6\mathrm{\pi R}}{\mathrm{tan\alpha }} \\ \mathrm{At} \mathrm{R}=\frac{2\mathrm{h}}{3}\mathrm{tan\alpha }, \mathrm{we} \mathrm{have} \\ \frac{{\mathrm{d}}^2\mathrm{V}}{{\mathrm{dR}}^2}=2\mathrm{\pi h}-\frac{6\mathrm{\pi }}{\mathrm{tan\alpha }}\left(\frac{2\mathrm{h}}{3}\mathrm{tan\alpha }\right) \\ =2\mathrm{\pi h}-4\mathrm{\pi h} \\ =-2\mathrm{\pi h}<0 \\ \mathrm{So} \mathrm{by} \mathrm{second} \mathrm{derivative} \mathrm{test} \mathrm{the} \mathrm{volume} \mathrm{of} \mathrm{cylinder} \mathrm{is} \mathrm{maximum} \mathrm{when} \\ \mathrm{R}=\frac{2\mathrm{h}}{3}\mathrm{tan\alpha } \\ \mathrm{Now} \mathrm{when} \mathrm{R}=\frac{2\mathrm{h}}{3}\mathrm{tan\alpha }, \mathrm{we} \mathrm{get} \\ \mathrm{H}=\frac{1}{\mathrm{tan\alpha }}\left(\mathrm{h} \mathrm{tan\alpha }-\frac{2\mathrm{h}}{3}\mathrm{tan\alpha }\right) \\ =\frac{1}{\mathrm{tan\alpha }}\left(\frac{\mathrm{h}}{3}\mathrm{tan\alpha }\right) \\ =\frac{\mathrm{h}}{3} \\ \mathrm{So} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cylinder} \mathrm{is} \mathrm{one}-\mathrm{third} \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cone} \\ \mathrm{The} \mathrm{maximum} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cylinder} \mathrm{is}, \\ \mathrm{q}={\mathrm{\pi R}}^2\mathrm{H} \\ =\mathrm{\pi }{\left(\frac{2\mathrm{h}}{3}\mathrm{tan\alpha }\right)}^2\left(\frac{\mathrm{h}}{3}\right) \\ =\frac{4}{27}\mathrm{\pi } {\mathrm{h}}^3\mathrm{tan\alpha } \\ \mathrm{The} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cone} \mathrm{is}, \\ \mathrm{p}=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ =\frac{1}{3}\mathrm{\pi }{\left(\mathrm{h} \mathrm{tan\alpha }\right)}^2\left(\mathrm{h}\right) \\ =\frac{1}{3}\mathrm{\pi } {\mathrm{h}}^3\mathrm{tan\alpha } \\ \mathrm{So} \mathrm{the} \mathrm{required} \mathrm{ratio} \mathrm{is}, \\ \frac{\mathrm{p}}{\mathrm{q}}=\frac{\frac{1}{3}\mathrm{\pi } {\mathrm{h}}^3\mathrm{tan\alpha }}{\frac{4}{27}\mathrm{\pi } {\mathrm{h}}^3\mathrm{tan\alpha }}=\frac{1}{3}\times \frac{27}{4}=\frac{9}{4}=\mathbf{9}\mathbf{:}\mathbf{4} \end{gathered}$$

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