BOXMAKE BRIEF ABSTRACT We seek to determine first the maximum volume box cut from a fixed planar region and then the minimum cost box cut from comparable region. GENERAL INFORMATION FileName: BOXMAKE Full title: Determine the process for manufacturing the most profitable enclosed box given associated costs and revenue parameters. Last Revision Date: 18 November 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. Contacts: 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 We consider the manufacturing process in which we need to make an enclosed, rectangular box from a flat (8.5" by 11" rectangle) sheet of metal. The box will have a rectangular bottom Teacher's Note: This description gives the variable formulation and may be omitted in favor of a diagram to be labeled. The basic approach will be to make a cross shaped flat piece consisting of a y" (vertical) by z" (horizontal) rectangle at the center with x"(horizontal) by y" (vertical) rectangles on both sides and x" by z" rectangles on top and bottom of the center rectangle. Finally a y" by z" rectangle will be centrally located at the bottom of the bottom x" by y" rectangle and will serve as a cover for the enclosed box. (1) Find the dimensions of the box which will make the volume of the enclosed, rectangular box maximum. (2) Economic considerations in manufacturing process. We have an incentive in forming the box to make a large contained volume but we also have costs associated with this manufacturing process. We can charge $0.10/in^3 for the box. The costs of materials is $0.015/square inch when new and any scrap left over from the process will be generating $0.004/square inch. We note that on the box we need to file and glue the eight vertical edges at a cost of $0.01/linear inch, file all exposed edges (the three edges of the open top and the open flap itself) at a cost of $0.02/linear inch, and insert hinges in the open flap edge. The latter is a fixed cost of $0.40 no matter what geometry we use. Now, find the dimensions of the enclosed box which will make profit maximum. Recall Profit = Revenue - Cost. (3) Compare the outcomes of (1) and (2). KEYWORDS Optimization - volume and revenue, geometry. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Formation of function from description, optimization of a function of one variable. Time allotment - time management This problem is a somewhat standard optimization problem and with a diagram the students can formulate the geometric and economic optimization function in 10 minutes, the latter being a bit more difficult as they have to assign the various costs to specific pieces. Expectations We would expect the student to formulate an objective function - in the first problem one of volume and in the second problem one of profit = revenue - cost. Further we would expect the student to be able to optimize such objective functions and compare the results. Future payoffs This problem demands careful reading, extraction of a drawing from a description if no drawing is provided, and a step-by-step accounting for all costs involved. Such care in accounting for terms in the objective function will be useful throughout future modeling activities. Extensions This IS an extension of the standard maximum volume box out of a rectangle - only this time we require a top and are concerned with economics. One could require double thickness of the bottom, using a piece from the unused material and costing for the glueing of the two pieces. One could discuss marginal increases in say total cost, for marginal increases in any component's cost. References and Sources We appreciate the suggestions and corrections offered by Paul H. Bouknecht who teaches AP Calculus at Eau Gallie High School in Melbourne FL. POSSIBLE SOLUTION(S) (1) We label the edges of the box. We write down the constraints of the sheet ht = 2 x + z == 8.5 2 x + z == 8.5 len = 2 x + 2 y == 11 2 x + 2 y == 11 And we determine the sides y and z as functions of x. yxlen = y/.Solve[len,y][[1]] 11 - 2 x -------- 2 zxht = z/.Solve[ht,z][[1]] 8.5 - 2. x Thus we have the volume as x y z, but since y and z are functions of x we have the volume as a function of x alone, vol(x). vol[x_] = x yxlen zxht (11 - 2 x) (8.5 - 2. x) x ------------------------- 2 We plot vol(x) to see if there is a maximum volume. It appears to be around x = 1.5 in. Plot[vol[x],{x,0, 8.5/2}] -Graphics- We determine the dimensions x = 1.59 in, y = yxlen = 3.91 in, and z = zxht = 5.33 in. Also we determine the maximum volume vol = 33.07 in^3 when vol'(x) = 0. maxvol = {x, yxlen,zxht,vol[x]}/. Solve[vol'[x]==0,x][[1]] {1.58542, 3.91458, 5.32916, 33.0741} (2) We consider the problem of maximizing profit. We recall the data. We can charge $0.10/in^3 for the box. The costs of materials is $0.015/square inch when new and any scrap left over from the process will be generating $0.004/square inch. We note that we need to file and glue the eight vertical edges (x) at a cost of $0.01/linear inch, file all exposed edges (the three edges (z + 2 x) of the open top and the three edges of the open flap itself (z + 2x)) at a cost of $0.02/linear inch, and insert small hinges in the open flap edge which is connected to the box top. The latter is a fixed cost of $0.40 no matter what geometry we use. We represent the cost using the x, y, and z variables - solved for x. BoxCost[x_] = .01 8 x + .02 (2 z + 4 y) + .015 8.5 11 + .4/.{y->yxlen,z->zxht} 1.8025 + 0.02 (2 (11 - 2 x) + 2 (8.5 - 2. x)) + 0.08 x And finally we compute the revenue for our manufacturing process. BoxRevenue[x_] = Expand[.10 vol[x] + y .004 (8.5 11 - (4 x^2 + 2 x y))- BoxCost[x]/.{y->yxlen,z->zxht}] 2 3 -0.5255 + 4.139 x - 1.95 x + 0.208 x We plot the revenue as a function of variable x. We note the possibilities of negative revenue in our feasible region 0 <= x <= 8.5) - we never got negative volume in this region. Plot[BoxRevenue[x],{x,0, 8.5/2}] -Graphics- We determine the dimensions x = 1.35 in, y = yxlen = 4.14 in, and z = zxht = 5.78 in. Also we determine the maximum volume vol = 32.60 in^3 when vol'(x) = 0. Our maximum revenue is $2.02 per box. max = {x, yxlen, zxht,vol[x], BoxRevenue[x]}/.Solve[BoxRevenue'[x]==0,x][[1]] {1.35508, 4.14492, 5.78984, 32.5198, 2.02006} In comparison with our simpler volume only Problem (1) we note that for maximum profit x decreases by .23 in, y increases by.2 in, and z increases by .4 in while the volume decreases by 0.56 in^3. We repeat the answers from the Problem (1) below. max = {x, xlen,xht,vol[x]}/.Solve[vol'[x]==0,x][[1]] {1.58542, 3.91458, 5.32916, 33.0741} We note that the volume for our most economical shape has decreased and is not the maximum volume possible. ISSUES IN SOLUTION We got mixed up in our own problem formulation as was pointed out to us by Paul H. Bouknecht of Melbourne FL. Thus a good diagram and careful accounting of costs is essential.