Question

Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is ${\cot }^{-1}\left(\sqrt{2}\right)$. [CBSE 2014]

Answer 1

Let:
Radius of the base = r,
Height = h,
Slant height = l,
Volume = V,
Curved surface area = C

$$\begin{gathered} \mathrm{As},\mathrm{Volume},V=\frac{1}{3}\mathrm{\pi }{r}^{2}h \\ \Rightarrow h=\frac{3V}{\mathrm{\pi }{r}^2} \\ \mathrm{Also},\mathrm{the}\mathrm{slant}\mathrm{height},l=\sqrt{h^2+r^2} \\ =\sqrt{{\left(\frac{3V}{\mathrm{\pi }{r}^2}\right)}^2+r^2} \\ =\sqrt{\frac{9V^2}{{\mathrm{\pi }}^2{r}^4}+r^2} \\ =\sqrt{\frac{9V^2+{\mathrm{\pi }}^2{r}^6}{{\mathrm{\pi }}^2{r}^4}} \\ \Rightarrow l=\frac{\sqrt{9V^2+{\mathrm{\pi }}^2{r}^6}}{\mathrm{\pi }{r}^2} \\ \mathrm{Now}, \\ \mathrm{CSA},C=\mathrm{\pi }rl \\ \Rightarrow C\left(r\right)=\mathrm{\pi }r\frac{\sqrt{9V^2+{\mathrm{\pi }}^2{r}^6}}{\mathrm{\pi }{r}^2} \\ \Rightarrow C\left(r\right)=\frac{\sqrt{9V^2+{\mathrm{\pi }}^2{r}^6}}{r} \\ \Rightarrow C\text{'}\left(r\right)=\frac{r\times {\displaystyle \frac{6{\mathrm{\pi }}^2{r}^5}{2\sqrt{9V^2+{\mathrm{\pi }}^2{r}^6}}}-\sqrt{9V^2+{\mathrm{\pi }}^2{r}^6}}{r^2} \\ =\frac{{\displaystyle \left[\frac{3{\mathrm{\pi }}^2{r}^6\mathit{-}\left(\mathit{9}{V}^{\mathit{2}}\mathit{+}{\pi }^{\mathit{2}}{r}^{\mathit{6}}\right)}{\sqrt{9V^2+{\mathrm{\pi }}^2{r}^6}}\right]}}{r^2} \\ =\frac{3{\mathrm{\pi }}^2{r}^6\mathit{-}9V^2-{\mathrm{\pi }}^2{r}^6}{r^2\sqrt{9V^2+{\mathrm{\pi }}^2{r}^6}} \\ =\frac{2{\mathrm{\pi }}^2{r}^6\mathit{-}9V^2}{r^2\sqrt{9V^2+{\mathrm{\pi }}^2{r}^6}} \\ \mathrm{For}\mathrm{maxima}\mathrm{or}\mathrm{minima},C\text{'}\left(r\right)=0 \\ \Rightarrow \frac{2{\mathrm{\pi }}^2{r}^6\mathit{-}9V^2}{r^2\sqrt{9V^2+{\mathrm{\pi }}^2{r}^6}}=0 \\ \Rightarrow 2{\mathrm{\pi }}^2{r}^6\mathit{-}9V^2=0 \\ \Rightarrow 2{\mathrm{\pi }}^2{r}^6=9V^2 \\ \Rightarrow V^2=\frac{2{\mathrm{\pi }}^2{r}^6}{9} \\ \Rightarrow V=\sqrt{\frac{2{\mathrm{\pi }}^2{r}^6}{9}} \\ \Rightarrow V=\frac{\mathrm{\pi }{r}^3\sqrt{2}}{3}\mathrm{or}r={\left(\frac{3V}{\mathrm{\pi }\sqrt{2}}\right)}^{\frac{1}{3}} \\ \mathrm{So},h=\frac{3}{\mathrm{\pi }{r}^2}\times \frac{\mathrm{\pi }{r}^3\sqrt{2}}{3} \\ \Rightarrow h=r\sqrt{2} \\ \Rightarrow \frac{h}{r}=\sqrt{2} \\ \Rightarrow \cot \theta =\sqrt{2} \\ \therefore \theta ={\cot }^{-1}\left(\sqrt{2}\right) \\ \mathrm{Also}, \\ \mathrm{Since},\mathrm{for}r<{\left(\frac{3V}{\mathrm{\pi }\sqrt{2}}\right)}^{\frac{1}{3}},C\text{'}\left(r\right)<0\mathrm{and}\mathrm{for}r>{\left(\frac{3V}{\mathrm{\pi }\sqrt{2}}\right)}^{\frac{1}{3}},C\text{'}\left(r\right)>0 \\ \mathrm{So},\mathrm{the}\mathrm{curved}\mathrm{surface}\mathrm{for}r={\left(\frac{3V}{\mathrm{\pi }\sqrt{2}}\right)}^{\frac{1}{3}}\mathrm{or}V=\frac{\mathrm{\pi }{r}^3\sqrt{2}}{3}\mathrm{is}\mathrm{the}\mathrm{least}. \end{gathered}$$