Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

Find the volume of the largest cylinder that can be inscribed in a sphere of radius r.

Can you answer this question?

1 Answer

0 votes
  • Volume =$\pi R^2h$
Step 1:
Let $h$ be the height and $R$ be the radius of the base of the inscribed cylinder.
Let $V$ be the volume of the cylinder
$V=\pi R^2h$
From the figure :
By applying pythagoras theorem we get,
Where $r$ is the radius of the sphere
$\therefore R^2=r^2-\big(\large\frac{h^2}{4}\big)$
Substituting this in volume $V$ we get,
$\Rightarrow V=\pi r^2h-\large\frac{\pi}{4}$$h^3$
Step 2:
Now differentiating w.r.t $h$ we get,
$\large\frac{dV}{dh}$$=\pi r^2-\large\frac{3\pi h^2}{4}$
Again differentiating w.r.t $h$ we get,
$\large\frac{d^2V}{dh^2}$$=-\large\frac{3\pi h}{2}$
Since $\large\frac{d^2V}{dh^2} $$< 0$,V is maximium.
Step 3:
For maximum values of $V$,we have
$\Rightarrow \pi r^2-\large\frac{3\pi h^2}{4}$$=0$
$\Rightarrow \pi r^2=\large\frac{3\pi h^2}{4}$
$\Rightarrow r^2=\large\frac{3h^2}{4}$
Step 4:
Take square root on both sides
$r=\large\frac{\sqrt 3h}{2}$
$\therefore h=\large\frac{2}{\sqrt 3}$$r$
Hence $V$ is maximum when $h=\large\frac{2}{\sqrt 3}$$r$
Put $h=\large\frac{2r}{\sqrt 3}$ in $R^2=r^2-\large\frac{h^2}{4}$
Step 5:
We obtain $R=\sqrt{\large\frac{2}{3}}r$
The maximum volume of the cylinder is given by
$V=\pi R^2h$
$\;\;=\pi\big(\large\frac{2}{3}$$r^2\big)\big(\large\frac{2r}{\sqrt 3}\big)$
$\;\;=\large\frac{4\pi r^3}{3\sqrt 3}$
answered Sep 23, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App