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

Toolbox:
• 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,
$r^2=\big(\large\frac{h}{2}\big)^2$$+R^2 Where r is the radius of the sphere \therefore R^2=r^2-\big(\large\frac{h^2}{4}\big) Substituting this in volume V we get, V=\pi\big(r^2-\large\frac{h^2}{4}\big)$$h$
$\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
$\large\frac{dV}{dh}$$=0 \Rightarrow \pi r^2-\large\frac{3\pi h^2}{4}$$=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)$
$\;\;=\large\frac{4\pi r^3}{3\sqrt 3}$