The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius =√3 is:

Given, radius of sphere $=\sqrt{3}$=3​ Now, In $\triangle \mathrm{OAB}$△OAB, by Pythagoras theorem $$(\mathrm{OA})^2=(\mathrm{OB})^2+(\mathrm{AB})^2$$(OA)2=(OB)2+(AB)2

$$\begin{aligned} &(\sqrt{3})^2=\left(\frac{h}{2}\right)^2+r^2 \\ &3=\frac{h^2}{4}+r^2 \Rightarrow r^2=3-\frac{h^2}{4} \end{aligned}$$​(3​)2=(2h​)2+r23=4h2​+r2⇒r2=3−4h2​​ Now, volume of cylinder $=\pi r^2 h$=πr2h $$\begin{aligned} &V=\pi\left(3-\frac{h^2}{4}\right) h \\ &V=3 \pi h-\frac{\pi h^3}{4} \end{aligned}$$​V=π(3−4h2​)hV=3πh−4πh3​​ Now, for largest possible right circular cylinder the volume must be maximum $\therefore$∴ For maximum volume, $\frac{d V}{d h}=0$dhdV​=0 Now, Differentiating eq. (2) w.r.t. $h$h $V^1=\frac{d V}{d h}=3 \pi-\frac{3}{4} \pi h^2$V1=dhdV​=3π−43​πh2 or $3 \pi-\frac{3}{4} \pi h^2=0 \Rightarrow 3 \pi=\frac{3}{4} \pi h^2$3π−43​πh2=0⇒3π=43​πh2 $\Rightarrow h^2=4 \Rightarrow h=2$⇒h2=4⇒h=2 Now, volume $(V)$(V) of the cylinder $$=\pi\left(3-\frac{h^2}{4}\right) h=\pi(6-2)=4 \pi$$=π(3−4h2​)h=π(6−2)=4π