CONEMESH BRIEF ABSTRACT From a fixed length of wire make a "Pac-Man" shaped perimeter (circle with sector cut out) so that if we fill the interior with paper mesh, cut out the paper, and fold to make a cone by connecting the two radial edges, we have a maximim volume cone. GENERAL INFORMATION FileName: CONEMESH Full title: Optimal Volume Paper Cone from Wire Paper Mold Last Revision Date: 2 June 1996. Developer: 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 Given a 100 cm length of thin wire, make a closed boundary consisting of a circle of radius r with a sector of angle theta cut out of it. We place this closed, planar, wire form in a paper-making, mesh mold. When the two edges of the resulting paper sector from the inner area bounded by the wire are attached there will be formed a cone. Design the closed boundary so that the paper cone thus formed will have maximum volume. KEYWORDS Perimeter, area, volume, functions of two and one variable, eliminating variables, optimization of function of one variable, derivative. TEACHER NOTES ISSUES RELATED TO THE PROBLEM This problem is an extension on the older problem of cutting out a sector from a fixed radius circle such that when the two edges of the sector are attached a cone of maximum volume will be formed. In this problem apparently there are two variables - the radius of the region and the angle of the sector. However, the student should be able to eliminate one of these variables by using the total length of the wire as a constraint and thus create a function of one variable to optimize. Prerequisites Familiarity with volume and perimeter formulae. Use of the derivative of a function of one variable to optimize. Elimination of a variable in one expression using a constraint equation. Time allotment - time management We have given this problem in class and students working in groups have finished in in 20 minutes. Expectations We expect students to have some difficulty in bringing the problem to a one variable function to optimize. Future payoffs Students will improve their ability to set up and complete optimization problems. Extensions Perhaps make two equal cones with cutting out two equal wedges. References and Sources POSSIBLE SOLUTION(S) This appears to be a function of two variables but if one writes out all the information one finds there is but one variable, we used theta - the angle of the cut sector. We offer a plot of a typical bounded region. The following code will generate the figure above. c = ParametricPlot[{Cos[t],Sin[t]},{t,Pi/6,2 Pi - Pi/4}, Axes->False, AspectRatio->Automatic] l1 = Graphics[Line[{{0,0}, {Cos[Pi/6], Sin[Pi/6]}}]] l2 = Graphics[Line[{{0,0}, {Cos[2 Pi - Pi/4], Sin[2 Pi - Pi/4]}}]] let = Show[Graphics[{Text["Angle theta",{.5,0}], Text["Radius r",{.4,.45}]}]] Show[l1,l2,c,let,AspectRatio->Automatic] We define our variables: R = radius of the base of proposed cone; r = radius of the circle we intend to make of the wire; theta = angle of the sector we will cut out of the circle of wire; length = length of the wire we have to work with (in our case this is 100 cm); ConeVol = volume of cone of base radius R and lateral surface length r; ConeBase = perimeter of the base of the cone; CVol[theta] = the volume of the cone in terms of the one variable theta. We presume a sector of angle theta and a circle of radius r. We specify that the length of the wire has to be the sum of the two radii (the edges of the sector) and the resulting reduced circumference of the remaining sector. length = (2 Pi - theta) r + 2 r == L; We give the formula for the volume of a cone of base radius R and lateral surface length r. ConeVol = 1/3 Pi R^3 Sqrt[ r^2 - R^2]; We eliminate the variable r using the length constraint, at least write r in terms of L and theta only. solr = Solve[length,r] L {{r -> ----------------}} 2 + 2 Pi - theta We set the perimeter of the base of the proposed cone (2 Pi R) equal to the available perimeter of the wire formed sector (2 Pi - theta)r. ConeBase = 2 Pi R == (2 Pi - theta) r/.solr[[1]] L (2 Pi - theta) 2 Pi R == ---------------- 2 + 2 Pi - theta We eliminate the variable R or at least write R in terms of L and theta only. solR = Solve[ConeBase,R] -(L (-2 Pi + theta)) {{R -> -----------------------}} 2 Pi (2 + 2 Pi - theta) We now substitute into the ConeVol formula for r and R our expressions which involve theta only to obtain the volume of the cone in terms of theta alone. CVol[theta_] = ConeVol/.solR[[1]]/.solr[[1]] 3 3 -(L (-2 Pi + theta) 2 L Sqrt[------------------- - 2 (2 + 2 Pi - theta) 2 2 L (-2 Pi + theta) -------------------------]) / 2 2 4 Pi (2 + 2 Pi - theta) 2 3 (24 Pi (2 + 2 Pi - theta) ) We specify the length of wire we have, 100 cm. L = 100; We plot the volume of the cone formula as a function of theta to see is there is a clear maximum which emerges. Plot[CVol[theta],{theta,0, 2 Pi}] -Graphics- To determine the maximum value of this volume function CVol[theta] we take the derivative of the function with respect to theta and set it equal to zero. derCVol[theta_] = CVol'[theta]; We solve for when the derivative of CVol[theta] = 0. solMaxtheta = FindRoot[derCVol[theta] == 0,{theta,2.5}] {theta -> 2.7759} We now determine the parameters (radius and theta) of our formed shape of wire. radiusOfCut = r/.solr[[1]]/.solMaxtheta[[1]]//N 18.1578 thetaOfCut = theta/.solMaxtheta[[1]] 2.7759 And finally we compute the maximum volume for a cone we can make out of paper from the inside of the wire mesh. MaxVol = CVol[thetaOfCut]//N 16427.7 We sketch our solution region. cAnswer = ParametricPlot[{Cos[t],Sin[t]}, {t,0,2Pi - thetaOfCut}, Axes->False, AspectRatio->Automatic] -Graphics- l1Answer = Graphics[Line[{{0,0}, {Cos[0], Sin[0]}}]] -Graphics- l2Answer = Graphics[Line[{{0,0}, {Cos[2 Pi - thetaOfCut], Sin[2 Pi - thetaOfCut]}}]] -Graphics- Show[cAnswer,l1Answer,l2Answer,Axes->True] -Graphics- ISSUES IN SOLUTION Basically this is a problem in setting up a function of one variable out of a complex situation.