Question

The volume of the greatest cylinder which can be inscribed in a cone of height $h$ and semi-vertical angle $Î\pm$ is (in cubic units)

• A. $\frac{4}{27}Ï€h^{3}ta{n}^2Î\pm$
• B. $\frac{4}{27}Ï€h^{2}ta{n}^2Î\pm$
• C. $Ï€h^{3}ta{n}^2Î\pm$
• D. $\frac{5}{36}Ï€htanÎ\pm$

Answer 1

Answer: (A)

Let $VAB$ be a given cone of height $h$, semi-vertical angle $Î\pm$ and let $x$ be the radius of the base of the cylinder $A^{′}{B}^{′}DC$ which is inscribed in the cone $VAB$. Then,
$O{O}^{′}=$ Height of the cylinder
$⇒O{O}^{′}−VO−V{O}^{′}=h−xcotÎ\pm …\left(i\right)$
Let $V$ be the volume of the cylinder. Then,
$V=Ï€x^2\left(h−xcotÎ\pm \right)…\left(ii\right)$
$⇒\frac{dV}{dx}=2Ï€xh−3Ï€x^{2}cotÎ\pm$
For maximum or minimum $V$, we must have
$\frac{dV}{dx}=0$
$⇒2Ï€xh−3Ï€x^{2}cotÎ\pm =0$



$⇒x=\frac{2h}{3}tanÎ\pm \left[âˆ\mu xî€=0\right]$
Now, $\frac{d^{2}V}{d{x}^2}=2Ï€h−6Ï€xcotÎ\pm$
When $x=\frac{2h}{3}tanÎ\pm$, we have
$\frac{d^{2}V}{d{x}^2}=π\left[2h−4h\right]=2πh<0$.
Hence, $V$ is maximum when $x=\frac{2h}{3}tanÎ\pm$.
The maximum volume of the cylinder is given by
$V=Ï€{\left(\frac{2h}{3}tanÎ\pm \right)}^2\left(h−\frac{2h}{3}\right)$ [From $\left(ii\right)$]
$⇒V=\frac{4}{27}Ï€h^{3}ta{n}^2Î\pm$.