The height of a right circular cylinder of maximum

Solution:

Let radius of base and height of cylinder be $r$ and $h$ respectively.

$\therefore r^{2}+\frac{h^{2}}{4}=9$ ...........(1)

Now, volume of cylinder, $V=\pi r^{2} h$

Substitute the value of $\mathrm{r}^{2}$ from equation (i),

$V=\pi h\left(9-\frac{h^{2}}{4}\right) \Rightarrow V=9 \pi h-\frac{\pi}{4} h^{3}$

Differentiating w.r.t. $h$,

$\frac{d V}{d h}=9 \pi-\frac{3}{4} \pi h^{2}$

For maxima/minima,

$\frac{d V}{d h}=0 \Rightarrow h=\sqrt{12}$

and $\frac{d^{2} V}{d h^{2}}=-\frac{3}{2} \pi h$

$\therefore\left(\frac{d^{2} V}{d h^{2}}\right)_{h=\sqrt{12}}<0$

$\Rightarrow$ Volume is maximum when $h=2 \sqrt{3}$