What fraction of the volume of a sphere is taken up by the largest cylinder that can be fit inside the sphere?
Let radius of sphere $=R$
Let radius of cylinder $=r$
Let height of cylinder $=2h$
From the figure, $R^2 = h^2 + r^2$
$\therefore$ $r^2 = R^2 - h^2$ $\;\; \cdots$ (1)
Volume of sphere $=V_s = \dfrac{4}{3}\pi R^3$ $\;\; \cdots$ (2)
Volume of cylinder $= V_c = 2\pi r^2 h$ $\;\; \cdots$ (3)
In view of equation (1), equation (3) becomes
$V_c = 2 \pi h \left(R^2 - h^2\right)$
i.e. $V_c = 2\pi R^2 h - 2 \pi h^3$ $\;\; \cdots$ (4)
For cylinder with maximum volume, $\dfrac{dV_c}{dh} = 0$
Now, from equation (4), $\dfrac{dV_c}{dh} = 2 \pi R^2 - 6\pi h^2$ $\;\; \cdots$ (5)
$\therefore$ $\dfrac{dV_c}{dh} = 0$ $\implies$ $2\pi R^2 - 6\pi h^2 = 0$
i.e. $3h^2 = R^2$ $\implies$ $h = \dfrac{R}{\sqrt{3}}$ $\;\; \cdots$ (6)
Now, from equation (5), $\dfrac{d^2 V_c}{dh^2} = -12 \pi h$
$\therefore$ $\dfrac{d^2 V_c}{dh^2} \bigg |_{h=\frac{R}{\sqrt{3}}} = \dfrac{-12\pi R}{\sqrt{3}} < 0$
$\implies$ $h = \dfrac{R}{\sqrt{3}}$ gives a cylinder with maximum volume.
Now, from equation (1), when $h = \dfrac{R}{\sqrt{3}}$,
$r = \sqrt{R^2 - \dfrac{R^2}{3}} = \dfrac{R\sqrt{2}}{\sqrt{3}}$ $\;\; \cdots$ (7)
$\therefore$ From equation (2), volume of cylinder $=V_c = 2 \pi \times \dfrac{2R^2}{3} \times \dfrac{R}{\sqrt{3}}$
i.e. $V_c = \dfrac{4\pi R^3}{3\sqrt{3}}$ $\;\; \cdots$ (8)
$\therefore$ From equations (2) and (8),
$\dfrac{V_c}{V_s} = \dfrac{4\pi R^3 / 3\sqrt{3}}{4\pi R^3 / 3} = \dfrac{1}{\sqrt{3}}$
$\therefore$ Fraction of the volume of a sphere is taken up by the largest cylinder that can be fit inside the sphere $= \dfrac{1}{\sqrt{3}} \times 100 = 57.5 \%$