Solid Geometry – Problem 22: Find and that maximize the cone’s volume, and give the maximum volume

A right circular cone is inscribed in a sphere of radius $R$. The cone’s axis passes through the sphere center, and its vertex is at the “north pole” of the sphere. Let $h$ be the cone’s height and $r$ its base radius.

(1) Express $r$ in terms of $h$ and $R$.

(2) Find $h$ and $r$ that maximize the cone’s volume, and give the maximum volume.

Set up a coordinate system with the sphere center $O$ at the origin and the axis as the $z$-axis. Then the sphere is $$x^2+y^2+z^2=R^2.$$ Let the cone’s vertex be $V(0,0,R)$. If the cone has height $h$, then its base plane is $$z=R-h.$$ The base circle lies on the sphere, so at $z=R-h$ we have $$r^2=x^2+y^2=R^2-(R-h)^2=2Rh-h^2.$$ This proves (1).

(2) The cone volume is $$V(h)=\frac13\pi r^2h=\frac13\pi(2Rh-h^2)h=\frac13\pi(2Rh^2-h^3),\qquad 0<h<2R.$$ Differentiate: $$V'(h)=\frac13\pi(4Rh-3h^2)=\frac13\pi h(4R-3h).$$ So the maximum occurs at $h=\frac{4R}{3}$. Substitute into $r^2=2Rh-h^2$: $$r^2=2R\cdot\frac{4R}{3}-\left(\frac{4R}{3}\right)^2=\frac{8R^2}{9},\quad r=\frac{2\sqrt2}{3}R.$$ The maximum volume is $$V_{\max}=\frac13\pi\cdot\frac{8R^2}{9}\cdot\frac{4R}{3}=\frac{32}{81}\pi R^3.$$

$r^2=2Rh-h^2$. The maximum volume occurs at $h=\dfrac{4R}{3}$, $r=\dfrac{2\sqrt2}{3}R$, and $V_{\max}=\dfrac{32}{81}\pi R^3$.