Question

A sector with a central angle $\alpha$ is cut out of a circle to make a cone. Then the value of the angle which will yield the greatest possible volume is

• A. $2\pi \sqrt{\frac{2}{3}}$
• B. $2\pi \sqrt{\frac{2}{5}}$
• C. $2\pi \sqrt{\frac{3}{5}}$
• D. $2\pi \sqrt{\frac{5}{7}}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{2}{3}}$

Let the radius of circle be r and r' be the radius of cone
Length of arc $=r\alpha$
For maximum volume , length of arc must be maximum.
$\Rightarrow 2\pi r^{\prime }=r\alpha$
$\Rightarrow r^{\prime }=\frac{r\alpha }{2\pi }$
Now , $h^2+(r^{\prime }{)}^2=r^2$
$\Rightarrow h=\sqrt{r^2-{r^{\prime }}^2}$
$\Rightarrow h=\sqrt{r^2-\frac{r^2{\alpha }^2}{4{\pi }^2}}$
Volume of cone $V=\frac{1}{3}\pi {r^{\prime }}^{2}h$
$V=\frac{r^2{\alpha }^2}{12\pi }\sqrt{r^2-\frac{r^2{\alpha }^2}{4{\pi }^2}}$
$=\frac{r^3{\alpha }^2}{12\pi }\sqrt{1-\frac{{\alpha }^2}{4{\pi }^2}}$
Let $f\left(\alpha \right)=V^2=k{\alpha }^4\left(1-\frac{{\alpha }^2}{4{\pi }^2}\right)$
$\Rightarrow f\left(\alpha \right)=k{\alpha }^4-\frac{{k\alpha }^6}{4{\pi }^2}$
$f^{\prime }\left(\alpha \right)=4{\alpha }^3-\frac{{6\alpha }^5}{4{\pi }^2}$
For maxima or minima,
$f^{\prime }\left(\alpha \right)=0$
$\Rightarrow \alpha =2\pi \sqrt{\frac{2}{3}}$