Differential Calculus: largest cone inscribed in a sphere

$V = \frac{1}{3}\pi r^2 h$

$r^2 + (h - R)^2 = R^2$

$r^2 = R^2 - (h - R)^2$
 

 

$r^2 = R^2 - (h^2 - 2hR + R^2)$

$r^2 = 2hR - h^2$

$V = \frac{1}{3}\pi (2hR - h^2)h$

$V = \frac{1}{3}\pi (2h^2R - h^3)$

$\dfrac{dV}{dh} = \frac{1}{3}\pi (4hR - 3h^2) = 0$

$4R - 3h = 0$

$h = \frac{4}{3}R$   ←   height of cone
 

$r^2 = R^2 - (\frac{4}{3}R - R)^2$

$r^2 = R^2 – \frac{1}{9}R^2$

$r^2 = \frac{8}{9}R^2$

$r = \frac{2}{3}\sqrt{2}R$   ←   radius of cone