Hello, Neverquit!

This is an intricate (and badly worded) problem . . .

Find the sector angle (where cone volume is maximized)

when a sector is removed and the remainder is joined to form a cone. Code:

* * *
* *
* *
* *
* O *
* o *
* * θ * R *
* *
* * * *
A o o B
* *
* * *
Rθ

The sector angle is $\displaystyle \theta \,=\,\angle AOB.$

The radius of the circle is $\displaystyle R.$

The arc length of $\displaystyle AB$ is: .$\displaystyle R\theta.$

This is the circumference of the circular base of our cone.

. . $\displaystyle 2\pi r \:=\:R\theta \quad\Rightarrow\quad r \:=\:\frac{R\theta}{2\pi}$ .[1]

A cross-section of our cone looks like this:

Code:

*
/|\
/ | \
/ h| \ R
/ | \
/ | \
* - - + - - *
r

We have: .$\displaystyle h^2 \;=\;R^2 - r^2 \;=\;R^2 - \left(\frac{R\theta}{2\pi}\right)^2 \;=\;\frac{R^2(4\pi^2 - \theta^2)}{4\pi^2}$

. . Hence: .$\displaystyle h \;=\;\frac{R}{2\pi}\sqrt{4\pi^2 - \theta^2}$ .[2]

The volume of a cone is: .$\displaystyle V \;=\;\frac{\pi}{3}r^2h$

Substitute [1] and [2]: .$\displaystyle V \;=\;\frac{\pi}{3}\left(\frac{R\theta}{2\pi}\right )^2\cdot\frac{R}{2\pi}\sqrt{4\pi^2 - \theta^2}$

. . Therefore: .$\displaystyle V \;=\;\frac{R^3}{24\pi^2}\theta^2\sqrt{4\pi^2-\theta^2}$

And that is the function we must maximize.