Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4r}{3}$.

(A) Let the sphere have a fixed radius $r$. Let the cone have base radius $R$ and height $h$. The center of the sphere is $B$. Let $A$ be the vertex of the cone and $CD$ be the radius of the base of the cone. In $\triangle BCD$,by the Pythagorean theorem,$BC^2 + CD^2 = BD^2$. Since $BD = r$ and $CD = R$,we have $BC = \sqrt{r^2 - R^2}$. The height of the cone is $h = AB + BC = r + \sqrt{r^2 - R^2}$.
The volume $V$ of the cone is $V = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi R^2 (r + \sqrt{r^2 - R^2}) = \frac{1}{3} \pi R^2 r + \frac{1}{3} \pi R^2 \sqrt{r^2 - R^2}$.
To maximize $V$,we differentiate with respect to $R$:
$\frac{dV}{dR} = \frac{1}{3} \pi [2Rr + 2R\sqrt{r^2 - R^2} + R^2 \cdot \frac{-2R}{2\sqrt{r^2 - R^2}}] = \frac{1}{3} \pi [2Rr + \frac{2R(r^2 - R^2) - R^3}{\sqrt{r^2 - R^2}}] = \frac{1}{3} \pi [2Rr + \frac{2Rr^2 - 3R^3}{\sqrt{r^2 - R^2}}]$.
Setting $\frac{dV}{dR} = 0$,we get $2Rr\sqrt{r^2 - R^2} = 3R^3 - 2Rr^2$. Dividing by $R$ (since $R \neq 0$):
$2r\sqrt{r^2 - R^2} = 3R^2 - 2r^2$.
Squaring both sides: $4r^2(r^2 - R^2) = (3R^2 - 2r^2)^2 = 9R^4 - 12R^2r^2 + 4r^4$.
$4r^4 - 4r^2R^2 = 9R^4 - 12R^2r^2 + 4r^4$.
$9R^4 - 8R^2r^2 = 0 \Rightarrow R^2(9R^2 - 8r^2) = 0$.
Since $R \neq 0$,$R^2 = \frac{8r^2}{9}$.
Substituting $R^2$ into the height formula:
$h = r + \sqrt{r^2 - \frac{8r^2}{9}} = r + \sqrt{\frac{r^2}{9}} = r + \frac{r}{3} = \frac{4r}{3}$.
Thus,the altitude of the cone of maximum volume is $\frac{4r}{3}$.