jmc

The ideal cone must have its vertex on the surface of the sphere or else a larger cone will be constructible. Likewise the circumference of the base must be tangent to the sphere.



Let $d$ denote the distance from the center of the sphere to the center of the base of the cone.



Since the sphere has radius 1, we can use the Pythagorean Theorem to find other values.



If $r$ is the radius of the base of the cone, then $$r^2+d^2=1^2,$$and the height of the cone is $$h=1+d.$$Therefore, the volume of the cone is $$V=\frac{\pi }{3}{r}^{2}h=\frac{\pi }{3}(1-d^2)(1+d)=\frac{\pi }{3}(1-d)(1+d{)}^2.$$Thus, we want to maximize $(1-d)(1+d{)}^2$.

We need a constraint between the three factors of this expression, and this expression is a product. Let's try to apply the AM-GM inequality by noting that $$(1-d)+\frac{1+d}{2}+\frac{1+d}{2}=2.$$Then $$\begin{array}{rl}{\left(\frac{2}{3}\right)}^3& ={\left[\frac{(1-d)+\frac{1+d}{2}+\frac{1+d}{2}}{3}\right]}^3\\ & \ge (1-d)\cdot \frac{1+d}{2}\cdot \frac{1+d}{2},\end{array}$$so $$(1-d)(1+d)(1+d)\le 4{\left(\frac{2}{3}\right)}^3=\frac{32}{27}.$$and $$V=\frac{\pi }{3}(1-d)(1+d{)}^2\le \frac{\pi }{3}\cdot \frac{32}{27}=\frac{32\pi }{81}.$$The volume is maximized when the AM-GM inequality is an equality. This occurs when $$1-d=\frac{1+d}{2}=\frac{1+d}{2}$$so $d=\frac{1}{3}.$ In this case $h=\frac{4}{3}$ and $$r=\sqrt{1-d^2}=\sqrt{\frac{8}{9}}.$$Indeed, in this case $$V=\frac{\pi }{3}{r}^{2}h=\frac{\pi }{3}\cdot \frac{8}{9}\cdot \frac{4}{3}=\boxed{\frac{32\pi }{81}}.$$