Inscribed cone

Let the radius of the base of the cone be $a$ , and the height of the cone be $h$ . Since the radius of the sphere is 6, we can see that the sides $h-6$ , $r$ , and 6 make a right triangle, with 6 as it's hypotenuse. By the pythagorean theorem, we have $(h-6)^2+r^2=36$ , and so $r^2=-h^2+12h$ . The volume of the cone is $\frac{1}{3}$ $πr^2h$ . Substituting $r^{2}$, we have Volume= $\frac{1}{3}$ $π(-h^2+12h)h$ = $\frac{1}{3}$ $π(-h^3+12h^2)$ . To find the maximum of this, we need to find the maximum of $-h^3+12h^2$ . It is easy to see that for $0<h<12$ the cubic function has a maximum point at a local maxima and then decreases. We use differentiation to find the local maxima. $\frac{dV}{dh}=\frac{1}{3}π(-3h^2+24h)=π(-h^2+8h)=πh(8-h)$ . Hence since $0<h<12$ when $h=8$ we have the local maxima, or the maximum area. Thus, the height is 8 units.