A sector with central angle $\theta$ is cut from a circle of radius 12 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of $\theta$ so that the volume of the cone is a maximum.
See attached figure
thanks

2. Originally Posted by Mr. Engineer
A sector with central angle $\theta$ is cut from a circle of radius 12 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of $\theta$ so that the volume of the cone is a maximum.
See attached figure
thanks
There's stuff in this link that should solve-enable you: Math Forum - Ask Dr. Math

3. ## Tell if im in the right direction

$Vcone=\frac{1}{3}\pi r^2 h$
$\frac{length of sector}{2\pi}=\frac{\theta}{360}$

$length of sector=\frac{2\pi\theta}{360}$this would be the circumference of the cone

then from the example in the previous post, the height would be
$cos\frac{\theta}{2}=\frac{h}{12}$
$h=12cos\frac{\theta}{2}$

am i right so far?

4. Originally Posted by Mr. Engineer
$Vcone=\frac{1}{3}\pi r^2 h$
$\frac{length of sector}{2\pi {\color{red}\cdot r}}=\frac{\theta}{360}$

$length of sector=\frac{2\pi\theta}{360} {\color{red}\cdot 12}$ this would be the circumference of the cone

then from the example in the previous post, the height would be
$cos\frac{\theta}{2}=\frac{h}{12}$
$h=12cos\frac{\theta}{2}$

am i right so far?
You are dealing with two different radi:
- The radius of the sector which is the slanted line from the tip of the cone to it's edge
- the radius R of the base circle of the cone.

You know the length l of the arc of the sector: $l = \frac{2\pi\theta}{360} \cdot 12= \frac{\pi \cdot \theta}{15}$

which has the same length as the circumference of the base circle:

$\frac{\pi \cdot \theta}{15} = 2\pi \cdot R~\implies~R=\frac{\theta}{30}$

The slanted line s, the radius of the base cicle R and the height of the cone form a right triangle. Use Pythagorean theorem:

$h^2+R^2=s^2~\implies~h=\sqrt{144-\left(\frac{\theta}{30} \right)^2}$

Plug in the terms of R and h into the equation of the volume. You'll get a function with respect of $\theta$ :

$V(\theta)=\frac13 \cdot \pi \cdot \left(\frac{\theta}{30}\right)^2 \cdot \sqrt{144-\left(\frac{\theta}{30} \right)^2}$

To calculate the maximum volume you have to derive this function and solve for $\theta$ the equation:

$V'(\theta) = 0$

EDIT: Removed my typo in the last equation.