Ice Cream Cone ProblemDate: 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 |
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