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### Ice Cream Cone Problem

```
Date: 02/27/2000 at 19:57:39
From: Tripp Ratcliff
Subject: Maximum volume

I'm not sure if you've heard of the ice cream cone problem, but it
goes like this: You are to place a sphere of ice cream into a cone of
height 1. What radius of the sphere will give the most volume of ice
cream inside the cone (as opposed to above the cone) for a cone with a
base angle of 30 degrees? I can not figure out how to solve it. Any
suggestions?
```

```
Date: 03/15/2000 at 16:56:34
From: Doctor Fwg
Subject: Re: Maximum volume

Dear Tripp,

Here is one possible solution to this problem. Some of the details
have been deliberately left out so you will have to go through this
pretty carefully to be sure I haven't made any mistakes and to also be
sure you understand this solution. You may write back if you still
have any questions.

RESTATEMENT OF THE CONE PROBLEM: Find the size of a perfect sphere of
ice cream that will result in the most volume of ice cream within a
perfect cone. For this particular case, let the cone height be H and
the cone base angle be 30 degrees.

First, one may assume that the sphere that satisfies these conditions
must lie only partially inside and partially outside of the cone. In
the extreme, a very tiny sphere near the bottom of the cone will
certainly occupy less volume then a slightly larger sphere. So the
sphere volume within the cone will increase with the sphere's radius
until, at some point, the radius becomes so large that most of the
sphere volume will lie outside of the cone. However, as long as the
center of the sphere lies at, below, or only slightly above an
imaginary plane covering the top of the cone, there is a simple
relation between the sphere's radius (R), the cone height (H), and the
perpendicular distance between the sphere's center and the plane
covering the top of the cone (A). In referring to the following
figure:

it may be seen that this relation is:

R = (H - A)[sin(Theta)],

where: A is positive if measured between the imaginary plane and the
inside of the cone, and negative if measured from the imaginary plane
and outside of the cone, and Theta equals (30/2 = 15 degrees). This
expression for R is only valid for values of A between +H and ca.
-0.0718H. Do you see why?

Since, under these conditions, it is possible to find an expression
for the volume of the sphere that lies inside the cone (as a function
of the sphere's radius and the perpendicular distance between the
sphere's center and the plane surface covering the top of the cone),
it is also possible to set the first derivative of this expression
equal to zero to see if a maximum can be found. Note: H and Theta are
both constants.

One next needs to find the sphere volume within the cone, as a
function of R and A. One way to obtain this expression is to find the
sphere volume outside of the cone, then subtract that result from the
total sphere volume. If one allows the sphere volume lying outside of
the cone to be represented by Vout, it can be shown that:

Vout = Pi[(2/3)R^3 - AR^2 + A^3/3].

This expression is a result of integrating the expression:

dVout = Pi[y^2]dx = Pi[R^2 - x^2]dx,

between the x limits of A and R.

Subtracting Vout from the total sphere volume of (4/3)Pi R^3, to
produce an expression for the sphere volume within the cone (Vin),
produces the following:

Vin = Pi[(2/3)R^3 + AR^2 - A^3/3].

Taking the derivative of Vin, with respect to A, produces:

(dVin/dA) = Pi[2R^2(dR/dA) + R^2 + 2AR(dR/dA) - A^2].

Rearranging slightly produces:

(dVin/dA) = Pi[2R(R + A)(dR/dA) +  (R + A)(R - A)]
= Pi(R + A)[2R(dR/dA) + (R - A)].

Setting the last expression above for (dVin/dA) equal to zero and
solving for A produces:

Pi(R + A)[2R(dR/dA) +  (R - A)] = 0.

But, Pi(R + A) cannot equal zero (as long as A is not is too
negative), so:

[2R(dR/dA) +  (R - A)] = 0.

If solving the expression above for A produces a reasonable value for
A (i.e., a value for A between +H and ca. -0.718H, then a maximum
sphere volume within the cone (for that value of A) will have been
found.

Note that (dR/dA) = [-Sin(Theta)], from the first equation written
above.  Using that first expression for R and (dR/dA) = [-Sin(Theta)]
in the equation above should produce the following result for A:

A = [H Sin(Theta)][1.0 - 2 Sin(Theta)]/[1.0 + Sin(Theta) -
2 Sin^2(Theta)].

For H = 1.0 and Theta = 15 degrees, A is approximately = +0.111, or

- Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/
```

```Date: 01/11/2010 at 16:56:34
From: Doctor Jerry
Subject: Re: Maximum volume

As a matter of personal opinion, while I see the value of the variable A,
it must be remembered that we are to find the radius R that maximizes the
volume.  I would prefer an argument that (1) uses R as a variable, and
(2) takes as reasonably obvious that the domain of R should be [R0,R1].
The value R0 satisfies the equation

R0
sin(theta) = ------
H - R0

This corresponds to the position of the sphere when it first emerges from
the cone.  The value R1 satisfies the equation

B
cos(theta) = --
R1

where B is the radius of the base of the cone.  This corresponds to the
position of the sphere when it is tangent to the top edge of the cone
(thinking of the cone as in standard position, tip down).

Using the diagram already supplied, take the vertex of the cone as the
origin.  Using the 'disk method', the volume of that part of the sphere
that is inside the cone is

H
/
V® = |  pi(R^2 - (x - R csc(theta))^2) dx
/
R(csc(theta)-1)

= (1/3) pi (2R + R csc(theta) - H)(H - R csc(theta) + R)^2

Taking H = 1 and theta = pi/6 and then solving the equation dV/dR = 0, we
find exactly one root between

~
R0 = 0.205605

and

~
R1 = 0.277401

namely,

~
R = 0.230093

One's confidence in this domain is increased by noting that dV/dR is
positive at R0 and negative at R1.

- Dr. Jerry
```
Associated Topics:
College Calculus
College Higher-Dimensional Geometry

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