Mathematics - Application of Derivative Question with Solution | TestHub

Given, radius of sphere $=\sqrt{3}$ Now, In $△\mathrm{OAB}$, by Pythagoras theorem $$(\mathrm{OA}{)}^2=(\mathrm{OB}{)}^2+(\mathrm{AB}{)}^2$$
$$\begin{array}{rl}& (\sqrt{3}{)}^2={\left(\frac{h}{2}\right)}^2+r^2\\ & 3=\frac{h^2}{4}+r^2\Rightarrow r^2=3-\frac{h^2}{4}\end{array}$$ Now, volume of cylinder $=\pi r^{2}h$ $$\begin{array}{rl}& V=\pi \left(3-\frac{h^2}{4}\right)h\\ & V=3\pi h-\frac{\pi h^3}{4}\end{array}$$ Now, for largest possible right circular cylinder the volume must be maximum $\therefore$ For maximum volume, $\frac{dV}{dh}=0$ Now, Differentiating eq. (2) w.r.t. $h$ $V^1=\frac{dV}{dh}=3\pi -\frac{3}{4}\pi h^2$ or $3\pi -\frac{3}{4}\pi h^2=0\Rightarrow 3\pi =\frac{3}{4}\pi h^2$ $\Rightarrow h^2=4\Rightarrow h=2$ Now, volume $(V)$ of the cylinder $$=\pi \left(3-\frac{h^2}{4}\right)h=\pi (6-2)=4\pi$$