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?

let $\displaystyle \theta$ be the base angle of the cone.

Note that a line from the center of the circle to the vertex of $\displaystyle \theta$ bisects $\displaystyle \theta$.

$\displaystyle R = r \tan\left(\frac{\theta}{2}\right)$

$\displaystyle r = 10\cos{\theta}$


$\displaystyle R = 10\cos{\theta} \cdot \tan\left(\frac{\theta}{2}\right)$

find $\displaystyle \frac{dR}{d\theta}$ and maximize ...

I get the value $\displaystyle \theta \approx 52^\circ$