Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $8/27$ of the volume of the sphere.

Let $PAB$ be a cone of greatest volume inscribed in the sphere.

Let $OC = x$. Then in $\Delta OAC,AC = \sqrt {{R^2} - {x^2}}$

and $PC = PO + OC = R + x =$height of the cone.

Let $V$ be the volume of the cone. Then,
$V = \cfrac{1}{3}\pi {(AC)^2}(PC)$
$\Rightarrow V = \cfrac{1}{3}\pi ({R^2} - {x^2})(R + x)$

$= \cfrac{1}{3}\pi ({R^3} + x{R^2} - R{x^2} - {x^3})$ …(i)

Differentiating (i) w.r.t. $x$, we get $\cfrac{{dV}}{{dx}} = \cfrac{1}{3}\pi \left[ {({R^2} - {x^2}) - 2x(R + x)} \right]$
$\Rightarrow \cfrac{{dV}}{{dx}} = \cfrac{1}{3}\pi ({R^2} - 2Rx - 3{x^2})$

To find the maximum or minimum of $V$ we must have $\cfrac{{dV}}{{dx}} = 0$
$\Rightarrow {R^2} - 2Rx - 3{x^2} = 0 \Rightarrow (R - 3x)(R + x) = 0$

$\Rightarrow R - 3x = 0 \Rightarrow x = \cfrac{R}{3}$
$\cfrac{{{d^2}V}}{{d{x^2}}} = \cfrac{1}{3}\pi ( - 2R - 6x)$ and

${\left( {\cfrac{{{d^2}V}}{{d{x^2}}}} \right)_{x = R/3}} = - \cfrac{4}{3}R\pi < 0$

Thus, $V$ is maximum when $x = \cfrac{R}{3}$.

Putting the value of $x = \cfrac{R}{3}$ in $V = \cfrac{1}{3}\pi ({R^2} - {x^2})(R + x)$ , we get

Maximum volume of the cone $= \cfrac{1}{3}\pi \left( {{R^2} - \cfrac{{{R^2}}}{9}} \right)\left( {R + \cfrac{R}{3}} \right)$
$\Rightarrow$ Maximum volume of the cone $= \cfrac{{32\pi {R^3}}}{{81}}$

$\Rightarrow$ Maximum volume of the cone

$= \cfrac{8}{{27}}\left( {\cfrac{4}{3}\pi {R^3}} \right)$

$= \cfrac{8}{{27}}$ (Volume of the sphere)
Hence proved.