The maximum volume (in cubic units) of the cylinder which can be inscribed in a sphere of diameter 6 units is

Question

Aicharya · Accepted Answer

If $h$ is the height and $V$ be the volume, required cylinder then from the given figure. In $\triangle O A M$, $\begin{aligned} & r^2+\left(\frac{h}{2}\right)^2=(3)^2 \ & \Rightarrow \quad r^2+\frac{h^2}{4}=9 \ & \Rightarrow \quad r^2=9-\frac{h^2}{4} \ & \text {Now, volume }=\pi r^2 h=\pi\left(9-\frac{h^2}{4}\right) h \ & V=\pi\left(9 h-\frac{h^3}{4}\right) \ & \Rightarrow \quad \frac{d V}{d h}=\left(9-\frac{3 h^2}{4}\right) \pi \end{aligned}$ For maximum or minimum, $\frac{d V}{d h}=0$ $\begin{array}{rlrl} \Rightarrow & 9-\frac{3 h^2}{4} & =0 \Rightarrow \frac{3 h^2}{4}=9 \ \Rightarrow & h^2 & =12 \Rightarrow h= \pm 2 \sqrt{3} \ & \therefore & r^2 & =9-3 \Rightarrow r^2=6 \end{array}$ The maximum value, $\left[\because \frac{d^2 V}{d h^2}=-\frac{3 h}{2} < 0\right]$ $=\pi(6)(2 \sqrt{3})=12 \sqrt{3} \pi \text { sq units }$

The maximum volume (in cubic units) of the cylinder which can be inscribed in a sphere of diameter 6 units is

Solution