Question

Find the volume of the greatest right circular cone obtained by rotating a right angled triangle of hypotenuse of $1$ foot about a side.

• A. $\frac{\sqrt{2}\pi }{9\sqrt{\left(3\right)}}$ cu. ft.
• B. $\frac{2\pi }{9\sqrt{\left(3\right)}}$ cu. ft.
• C. $\frac{2\pi }{9}$ cu. ft.
• D. $\frac{\sqrt{2}\pi }{9}$ cu. ft.

Answer 1

The correct option is B $\frac{2\pi }{9\sqrt{\left(3\right)}}$ cu. ft.

Let $ABC$ be a right triangle with $AC=1foot$.
When the triangle is rotated along one of its legs, a cone is generated.
So, let $x$ be the height of the cone and slant height will be 1 foot.
Now, $r^2=x^2+1$
$\Rightarrow r=\sqrt{1-x^2}$
So, volume of cone $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (1-x^2)x$
Now,$\frac{dV}{dx}=\frac{1}{3}\pi \left[x\right(-2x)+1-x^2]=\frac{1}{3}\pi (1-3x^2)$
For maximum or minimum,
$\frac{dV}{dx}=0$
$\Rightarrow 1-3x^2=0$
$\Rightarrow x=\pm \frac{1}{\sqrt{3}}$
Now, $\frac{d^{2}V}{d{x}^2}=\frac{1}{3}\pi (-6x)==-2x\pi$
Clearly $\frac{d^{2}V}{d{x}^2}<0$ for $x=\frac{1}{\sqrt{3}}$
Hence, V is maximum at $x=\frac{1}{\sqrt{3}}$
Maximum volume $=\frac{1}{3}\pi (1-\frac{1}{3})\frac{1}{\sqrt{3}}$
$=\frac{2\pi }{9\sqrt{3}}$ cu. ft