If a right circularcone having maximum volume, is inscribed in a sphere of radius $3 \mathrm{~cm}$, then the curved surface area (in $\mathrm{cm}^2$ ) of this cone is

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Aicharya · Accepted Answer

Sphere of radius $r=3 \mathrm{~cm}$ Let $b, h$ be base radius and height of cone respectively. So, volume of cone $=\frac{1}{2} \pi b^2 h$

In right angled $\triangle A B C$ by Pythagoras theorem $ \begin{aligned} &(h-r)^2+b^2=r^2 \\ \Rightarrow & b^2=r^2-(h-r)^2=r^2-\left(h^2-2 h r+\mathrm{r}^2\right)=2 h r-h^2 \\ \therefore & \text { Volume }(v) \frac{1}{3} \pi h\left[2 h r-h^2\right]=\frac{1}{3}\left[2 h r^2-h^3\right] \\ & \frac{d v}{d h}=\frac{1}{3}\left[4 h r-3 h^2\right]=0 \Rightarrow h(4 r-3 h)=0 \\ & \frac{d^2 v}{d h^2}=\frac{1}{3}[4 r-6 h] \end{aligned} $ At $h=\frac{4 r}{3}, \frac{d^2 v}{d h^2}=\frac{1}{3}\left[4 r-\frac{4 r}{3} \times 6\right]=\frac{1}{3}[4 r-8 r < ] 0$ $\Rightarrow$ maximum volume ocurs at $h$ $ =\frac{4 r}{3}=\frac{4}{3} \times 3=4 \mathrm{~cm} $ As from (1), $ (h-r)^2+b^2=r^2 $ $ \Rightarrow b^2=2 h r-h^2=2 \cdot \frac{4 r}{3} r-\frac{16 r^2}{9} $ $ \begin{aligned} &=\frac{8 r^2}{3}-\frac{16 r^2}{9} \\ &=\frac{(24-16) r^2}{9}=\frac{8 r^2}{9} \\ &\Rightarrow b=\frac{2 \sqrt{2}}{3} r=2 \sqrt{2} \mathrm{~cm} \end{aligned} $ Therefore curved surface area $ \begin{aligned} &=\pi b l=\pi b \sqrt{h^2+r^2} \\ &=\pi 2 \sqrt{2} \sqrt{4^2+8}=8 \sqrt{3} \pi \mathrm{cm}^2 \end{aligned} $

If a right circularcone having maximum volume, is inscribed in a sphere of radius $3 \mathrm{~cm}$, then the…

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