Hello, aaasssaaa!
A conical drinking cup is made from a circular piece of paper of radius $\displaystyle R$
by cutting out a sector and joining the edges $\displaystyle CA$ and $\displaystyle CB.$
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 $\displaystyle r$ is the radius of the cone and $\displaystyle h$ is its height.
We see that: .$\displaystyle r^2 + h^2\;=\;R^2\quad\Rightarrow\quad r^2 \;=\;R^2 - h^2$ [1]
The volume of a cone is: .$\displaystyle V \;=\;\frac{\pi}{3}r^2h$ [2]
Substitute [1] into [2]: .$\displaystyle V \;=\;\frac{\pi}{3}\left(R^2-h^2\right)h\quad\Rightarrow\quad V \;=\;\frac{\pi}{3}\left(Rh - h^3\right)$
Maximize $\displaystyle V:\;\;V' \;=\;\frac{\pi}{3}\left(R - 3h^2\right)\;=\;0\quad\Rightarrow\quad h^2 \:=\:\frac{R^2}{3}$
. . and we get: .$\displaystyle \boxed{h \:= \:\frac{\sqrt{3}R}{3}}$
Substitute into [1]: .$\displaystyle r^2\;=\;R^2 - \left(\frac{R}{\sqrt{3}}\right)^2\quad\Rightarrow\ quad\boxed{ r^2\;=\;\frac{2R^2}{3}}$
Substitute into [2]: .$\displaystyle V \;=\;\frac{\pi}{3}r^2 h\;=\;\frac{\pi}{3}\left(\frac{2R^2}{3}\right)\lef t(\frac{\sqrt{3}R}{3}\right)$
Therefore, the maximum volume is: .$\displaystyle V \;=\;\boxed{\frac{2\pi\sqrt{3}}{27}R^3} $