Question

The height of the maximum cone that can be obtained by revolving a right angled triangle of hypotenuse $1$ unit about a side is

• A. $\frac{1}{3}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}$

Let r and h be the radius and height of the cone respectively.
Then $r^2+h^2=1$
Volume of cone $V=\frac{1}{3}\pi r^{2}h$
$V=f\left(h\right)=\frac{1}{3}\pi (h-h^3)$
$f^{\prime }\left(h\right)=\frac{1}{3}\pi (1-3h^2)$
For maxima or minima,
$f^{\prime }\left(h\right)=0$
$\Rightarrow h=\pm \frac{1}{\sqrt{3}}$
$f^{\prime \prime }\left(h\right)<0$ at $h=\frac{1}{\sqrt{3}}$
Hence, volume is maximum at $h=\frac{1}{\sqrt{3}}$