# Math Help - Optimization problem, stuck

1. ## Optimization problem, stuck

A conical drinking cup is made from a circular piece of paper of radius R by cutting out a sector and joining the edges CA and CB. Find the maximum capacity of such a cup

link of picture of the cup is found in the attachement

2. Hello, aaasssaaa!

A conical drinking cup is made from a circular piece of paper of radius $R$
by cutting out a sector and joining the edges $CA$ and $CB.$
Find the maximum capacity of such a cup.
Code:
            ..*.*.*..
.*:::::::::::*.
A*:::::::::::::::*B
*  *:::::::::::*  *
R *:::::::* R
*       *:::*       *
*         *         *
*         C         *

*                 *
*               *
*           *
* * *

The side view of the cup looks like this:
Code:
               r
*-----+-----*
\    :    /
\  h:   /
\  :  / R
\ : /
\:/
*

where $r$ is the radius of the cone and $h$ is its height.

We see that: . $r^2 + h^2\;=\;R^2\quad\Rightarrow\quad r^2 \;=\;R^2 - h^2$ [1]

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

Substitute [1] into [2]: . $V \;=\;\frac{\pi}{3}\left(R^2-h^2\right)h\quad\Rightarrow\quad V \;=\;\frac{\pi}{3}\left(Rh - h^3\right)$

Maximize $V:\;\;V' \;=\;\frac{\pi}{3}\left(R - 3h^2\right)\;=\;0\quad\Rightarrow\quad h^2 \:=\:\frac{R^2}{3}$

. . and we get: . $\boxed{h \:= \:\frac{\sqrt{3}R}{3}}$

Substitute into [1]: . $r^2\;=\;R^2 - \left(\frac{R}{\sqrt{3}}\right)^2\quad\Rightarrow\ quad\boxed{ r^2\;=\;\frac{2R^2}{3}}$

Substitute into [2]: . $V \;=\;\frac{\pi}{3}r^2 h\;=\;\frac{\pi}{3}\left(\frac{2R^2}{3}\right)\lef t(\frac{\sqrt{3}R}{3}\right)$

Therefore, the maximum volume is: . $V \;=\;\boxed{\frac{2\pi\sqrt{3}}{27}R^3}$

3. Hey what program did you use to right your numbers, roots and pie in

4. Originally Posted by aaasssaaa
Hey what program did you use to right your numbers, roots and pie in
The mathematical typesetting here is done using a LaTeX see this
(to see the code generating the equations quote the message and
you will see the typesetting codes embedded in the message).

RonL