Q. If a right circular cone, having maximum volume, is inscribed in a sphere of radius $3 \,cm$, then the curved surface area (in $cm^2$) of this cone is :
Solution:
$V=\frac{1}{3}\pi r^{2}h$
where r is radius and h is height of coin
$\Rightarrow V=\frac{1}{3}\,\pi\left(3\,sin 2\theta\right)^{2}\left(3+3\,cos2\theta\right)$
$=72\pi\,sin^{2}\theta\,cos^{4}\theta$
$\frac{dv}{d\theta}=72\pi\left[2sin\theta\,cos^{5}\theta-4sin^{3}\theta\,cos^{2}\theta\right]=0 \Rightarrow 0\,tan^{2}\theta=\frac{1}{2}$
$V_{mx} if \,tan=\frac{1}{\sqrt{2}}$
Hence curved surface area S $=\pi r\ell$
$=\pi r \sqrt{\left(3+3\,cos\,20\right)^{2}+\left(3\,sin\,2\theta\right)^{2}}$
$=\pi\left(3sin2\theta\right)\sqrt{36\,sin^{2}\,\theta}=18\pi\left(2sin\theta\,cos^{2}\theta\right)=36\pi.$
$ \frac{1}{\sqrt{3}}. \frac{2}{3}=\frac{24\pi}{3}=8\sqrt{3}\pi$
