CANS WITH A CONE BRIEF ABSTRACT The goal is to produce a cone of maximum volume by cutting a sector from a fixed circle and connecting the two radial edges. GENERAL INFORMATION FileName: CANCONE Full title: Cans With a Cone - Forming a Cone of Maximum Volume from a Flat Disk Last Revision Date: 31 May 1996. Developers: Robert Feller, Richmond High School, 380 Hub Etchison Blvd., Richmond IN 47374 USA and Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Contact: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Aaron D. Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Aaron.Klebanoff@Rose-Hulman.Edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM A machine that cuts pie shaped slices (sectors) from circular pieces of tin is set by entering the size of the angle to be cut in radians [0, 2 Pi] . The resultant piece of tin is then welded together along the edges of the cuts to form a cone. The machine is now set to the appropriate angle to form cones of maximum volume from circular pieces 4 cm. in radius. 1) What is the current setting of the machine in radians and what is the volume of the resulting cone, i.e. what is the maximum volume cone we can obtain from this process? 2) A cylindrical tin can that presently holds exactly 12 fl. ounces (354 ml) has a height of 12 cm. with a flat top and base. If a cone of the above type (i.e. cut from circle of radius 4 cm) replaces the flat top, how much more volume will be added and what will be the size of the angle of the cut needed to construct it? KEYWORDS Optimization, volume, circle, cone. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Students need to know area and perimeter formulae for circle and volume formulae of a cone and a cylinder. Further, students need to know how to eliminate variables from a function of two variables and they also need to know how to optimize a function of one variable. Time allotment - time management Each of these problems can be accomplished in half a class period. Expectations Students will play with the shapes, perhaps in paper cut-outs or in their head and see the volumes forming with a reasonable expectation as to what might be optimal. Future payoffs Students will gain practice in setting up optimization problems, in this case geometric optimization and thus should be better suited to tackle such problems. Extensions Put some cost functions associated with cutting, scrap, and volume obtained. References and Sources Source: Problem from Exploring Optimization, by Millie J. Johnson, Western Washington University In, Focus on Calculus, Fall 1992, Issue No. 2 published by John Wiley & Sons, Inc. POSSIBLE SOLUTION(S) 1. We define the following variables: R = radius of the resultant cone; r = radius of the original circular piece of metal; h = height of the cone; t = angle of the sector cut from the circle. We know the resulting perimeter when we cut the sector out of the cone must be the same as that of the perimeter of the base of the resulting cone. eq1 = 2 Pi r - t r == 2 Pi R; And hence we can write the radius, R, of the resulting cone in terms of the original circles radius, r, and the angle of the cut. sol = Solve[eq1,R] 2 Pi r - r t {{R -> ------------}} 2 Pi Using the fact that the volume of a cone is 1/3 the area of its base times the height of the cone and the right triangle geometry of a triangle with sides h and R and hypotenuse r we can write the volume V as a function of t Again, here r is the constant radius of the original circle and R is the radius of the cone. This right triangle we obtain by taking one half of the sliced triangle obtained by cutting the resulting cone along its axis. Thus the length of its height is h, the length of its base radius is R, and the length of its lateral edge is r (the radius of the original circle.) We seek to determine the t which maximizes V[t]. And we can do this by replacing R in a computation of the volume of the cone by an expression (from the solution of the perimeter equation for R above) involving r and t alone. V[t_] = 1/3 Pi R^2 Sqrt[r^2 - R^2]/.sol[[1]]; We seek to maximize V[t] with respect to t, i.e. find t the angle of the cut so that V[t], the resulting volume is maximum. Thus we take the derivative with respect to t and set it equal to zero. Vp[t_] = V'[t]; And we set this derivative equal to 0 to determine the extrema. sol1 = Solve[Vp[t]==0,t]//N {{t -> 6.28319}, {t -> 1.15299}, {t -> 11.4134}} We determine the volume at the various extrema - even though we note that the solution t = 11.4134 is out of range (for we only consider cutting for t in the interval [0, 2 Pi]). Solution 1 is t = 2 Pi, which says cut all the circle out and obtain a (minimal!) 0 volume. V1 = V[t]/.sol1[[1]] 2 ((-6.28319 r + 2 Pi r) 2 2 (-6.28319 r + 2 Pi r) Sqrt[r - ----------------------]) / (12 Pi) 2 4 Pi V2 = V[t]/.sol1[[2]]//N 2 2 0.403067 r Sqrt[r ] V3 = V[t]/.sol1[[3]]//N 2 2 0.403067 r Sqrt[r ] We consider the case of r = 4. Our volume formula is now defined for r = 4. Vr[t_] = V[t]/.{r->4}; We plot volume as a function of the angle t. Plot[Vr[t],{t,0, 2 Pi}] -Graphics- And in order to maximize this volume we find where the derivative is 0. Vrp[t_] = Vr'[t]; solt = Solve[Vrp[t]==0,t]//N {{t -> 6.28319}, {t -> 1.15299}, {t -> 11.4134}} Again t = 11.4134 is out of range and t = 6.28319 gives the 0 volume. Therefore we select solution 2 - 1.15299 radians or 66.0612 degrees (a rather flat cone) and determine the maximum volume to be 25.7963 cm^3. Vr3 = Vr[t]/.solt[[2]]//N 25.7963 t 180/Pi/.solt[[2]]//N 66.0612 2. We need to find the dimensions of the given cylinder without the new conical top so that we can find the radius of our cone top. solR = Solve[ Pi R^2 12 == 354,R] 59 59 {{R -> -Sqrt[----]}, {R -> Sqrt[----]}} 2 Pi 2 Pi Thus the radius R of our tin can must be 3.06433 cm. R = R/.solR[[2]]//N 3.06433 Therefore we want to construct a cone top that will have a base radius of 3.06433 cm. Recall the relationship between the outer perimeter of the cut out circle and the perimeter of the base of the cone from equation 1 above. We reeneter this formula with our R in it. eq1 = 2 Pi r - t r == 2 Pi R; We solve for t, the angle of the cut, for this specific radius, R, of the base of the cone. sol3 = Solve[eq1,t] -19.2538 + 6.28319 r {{t -> --------------------}} r And substitute the radius of the tin can r = 4 cm into this expression to find that our angle for the cut must be 1.46974 radians or 84.21 degrees. a1=t/.sol3[[1]]/.{r->4}//N 1.46974 a1 180/Pi//N 84.21 Then find the additional volume from the cone atop the can is 25.2812 cm^3 coneVol = V[t]/.{t->a1,r->4}//N 25.2812 And the total volume of the can cone combination is 628.467 cm^3. TotalVol = coneVol + Pi 4^2 12//N 628.467 ISSUES IN SOLUTION Problem 1 asks students to optimize the volume of a cone manufactured from cutting a sector from a fixed circle of radius r = 4 cm and welding the cut lines together to form a cone. Problem 2 asks the students to use the results/formulae of problem 1 to determine the optimal cone which can be cut from this same 4 cm radius circle if the base of the cone is also prescribed. The latter is prescribed by requiring the cone to sit atop a cylinder of known height and known volume and hence attainable radius.