Question

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 :

• A. $6\sqrt{2}\pi$
• B. $6\sqrt{3}\pi$
• C. $8\sqrt{2}\pi$
• D. $8\sqrt{3}\pi$

Answer 1

The correct option is D $8\sqrt{3}\pi$


From the figure,
$r^2+(h-3{)}^2=3^2\Rightarrow r^2+h^2-6h=0\Rightarrow r^2=6h-h^2$
Volume of the cone
$V=\frac{1}{3}\pi r^{2}h=\frac{\pi }{3}h(6h-h^2)=\frac{\pi }{3}(6h^2-h^3)$
Finding the maximum volume,
$\frac{dV}{dh}=0\Rightarrow (12h-3h^2)=0\Rightarrow h=0\text{or}h=4$
So, $h=4\Rightarrow r=\sqrt{6h-h^2}=2\sqrt{2}$
The curved surface area
$=\pi rl=\pi r\sqrt{r^2+h^2}=\pi \times 2\sqrt{2}\times \sqrt{8+16}=8\sqrt{3}\pi$