Problem:

What is the volume of the largest sphere that can be placed inside a cone with a slant height of 10?

What I've done so far:

- Recognize that finding the largest radius will suufice; once I find it, I can find the volume easily.
- Reduce the problem to a right triangular cross-section with an inscribed semicircle. the semicircle has a diameter that lies on the height of the triangle. A radius has been drawn to the right angle vertex of the triangle, and to the hypotenuse of ten (at a right angle).

If I call the far angle across from the semicircle $\displaystyle \theta$, then:

$\displaystyle 10 sin(\theta) = r + \frac{r}{cos(\theta)}$

Or, $\displaystyle 5 sin(2\theta) = rsin(\theta) + r$

Past this, I am stumped. An Implicit derivative doesn't seem to get me anywhere, but it is impossible to fully separate my two variables. Any thoughts?