Answer
$r=4\ ft$, $h=4\ ft$
Work Step by Step
Step 1. Using the figure given in the exercise, we let $\theta$ be the angle of the large cone, and we have $tan\frac{\theta}{2}=\frac{r}{12-h}=\frac{6}{12}$ (this relation can also be obtained using similar triangles). Thus we have $h=12-2r$
Step 2. The volume of the small cone is given by $V=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi r^2 (12-2r) =\frac{2}{3}\pi (6r^2-r^3) $
Step 3. Letting $V'=0$, we have $\frac{2}{3}\pi (12r-3r^2)=0$, which gives $r=4\ ft$ (discard $r=0$); thus $h=12-8=4\ ft$
Step 4. Check $V''(4)=\frac{2}{3}\pi (12-6r)=-8\pi \lt0$; we confirm the above dimensions give a maximum volume.
