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### Conical Drinking Cup

```
Date: 06/20/97 at 21:41:33
From: Brett & Leslie Gammill
Subject: Maximum capacity of a conical drinking cup

Dear Dr. Math,

Here's the problem: 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.

I know that I will probably need to know how to find the volume of a
cone and perhaps I will also have to use a derivative, but I am lost
from here!

I hope you can help me! Thanks a million!
```

```
Date: 06/21/97 at 14:30:52
From: Doctor Anthony
Subject: Re: Maximum capacity of a conical drinking cup

If you cut a sector from the circle with angle x at the center, the
length of the arc of the sector is xr.

The cone formed from this sector therefore has a base circumference
equal to xr, and if R is the radius of the cone, we get:

2pi*R = xr
R = xr/(2pi)

The slant height of the cone is r, so the perpendicular height is:

h = sqrt[r^2 - R^2]

The volume = (1/3)pi*R^2*h

Then dV/dx = (1/3)pi[2R*dR/dx*h + R^2*dh/dx] = 0 (for max or min)

Divide out by (1/3)pi*R and we get:

2dR/dx*h + Rdh/dx = 0

dR/dx = r/(2pi)    dh/dx = (1/2)[r^2-R^2]^(-1/2)*(-2Rr/(2pi)]

-Rr
=  ---------------
2pi*sqrt(r^2-R^2)

So   2hr            R^2*r
-------  -  ----------------    =  0
2pi       2pi*sqrt(r^2-R^2)

2h*sqrt(r^2-R^2) =  R^2

Since h = sqrt(r^2-R^2) we get:

2(r^2-R^2) = R^2
2r^2 = 3R^2
R^2 = (2/3)r^2

So volume = (1/3)pi*R^2*h    h = sqrt(r^2-(2/3)r^2) = sqrt[(1/3)r^2]

=  r/sqrt(3)

= (1/3)pi*(2/3)r^2*r/sqrt(3)

2pi*r^3
=  -----------
9sqrt(3)

This is the maximum volume of the cone.  You can find the angle x for
cutting out the sector from (R/r)(2pi) =  sqrt(2/3)(2pi)

= .81649(2pi)

=  294 degrees (approximately)

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 06/21/97 at 16:19:26
From: Brett & Leslie Gammill
Subject: Re: Maximum capacity of a conical drinking cup

Dear Dr. Math,

Thank you for your prompt response. However, I am afraid that maybe I
misled you a little in my wording of the problem. The drinking cup is
being made from the remainder of the circle after the sector is
removed. The drinking cup is not being made from the sector.

I understood most of your response to my original question except the
part about "the cone formed from this sector therefore has a base
circumference equal to xr, and if R..."  Does your solution reflect
the maximum capacity of a cup made from the sector that was cut away?

Thanks again.
```

```
Date: 06/21/97 at 18:31:03
From: Doctor Anthony
Subject: Re: Fw: maximum capacity of a conical drinking cup

When you cut away a sector, the remaining portion is also a sector and
it is a matter of preference whether you think of using the piece that
is cut away or the piece that is left.  The important thing is the
angle of the sector you actually use for making the cone, and you will
see that this angle x is 294 degrees for the maximum volume.  So we
have used most of the available material to make the cone (294/360 =
.82 of the whole).

The length of arc is xr (x in radians) and this arc is now curved
round to join up and make the circumference of the base of the cone.
So if R is the radius of the base of the cone, then 2pi*R = xr.  I
hope you will agree that the arc of the sector of the original circle
becomes the circumference of the base of the cone.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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