Question

Find the cylinder of greatest volume which can be inscribed in a cone.

• A. $\frac{4}{27}\pi b^{3}ta{n}^2\alpha .$
• B. $\frac{4}{9}\pi b^{3}ta{n}^2\alpha .$
• C. $\frac{1}{27}\pi b^{3}ta{n}^2\alpha .$
• D. $\frac{4}{3}\pi b^{3}ta{n}^2\alpha .$

Answer 1

The correct option is A $\frac{4}{27}\pi b^{3}ta{n}^2\alpha .$

Let $b$ be the height of the cone and $\alpha$ be its semi-vertical angle.

$LD=x=$radius of the inscribed cylinder and $LM=h$ be its height $LM=OM-OL=b-xcot$ $\alpha$.

$V=$volume of cylinder $=\pi r^{2}h$ $\pi x^2(b-xcot\alpha )$

$V=\pi (b{x}^2-x^{3}cot\alpha ).$

$dv/dx=\pi (2bx-3x^{2}cot\alpha )$

$\therefore x=0orx=\frac{2}{3}btan\alpha .$

Clearly $x=0$ is inadmissible and hence we consider the value $x=\frac{2}{3}btan\alpha$

$\frac{d^{2}V}{d{x}^2}=\pi (2b-6xcot\alpha )$ $=\pi (2b-4b)=-ive$ and hence max. when $x=\frac{2}{3}btan\alpha$

$h=b-xcot\alpha =b-\frac{2}{3}b=\frac{1}{3}b$

$\therefore$ Max, volume is $\pi r^{2}h=\pi x^{2}h=\pi .\frac{4}{9}{b}^2\alpha \frac{1}{3}b$ or $V=\frac{4}{27}\pi b^{3}ta{n}^2\alpha .$

Ans: A