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 
about 11% of H.

- 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