Maximum Volume of a Cut Off Box
One of probably most regular problems in a beginning calculus class is this: given a rectangular piece of carton. If by cuts parallel to the sides of the rectangle equal squares are removed from each corner, and the remaining shape is folded into a box, how big the volume of the box can be made? A rare calculus source would miss an opportunity to discuss such a practical application of differential calculus. A temptation is also great to use modern technology, say, graphic calculators or dynamic geometry software, like Geometer's Sketchpad, to solve the problem graphically in an attempt to reduce the proficiency required of students to understand the problem and approach its solution. Indeed, one such attempt intended for the middle school has been made at the On-Math, the Online Journal of School Mathematics. The article, however, as the whole magazine, is only available by subscription to the NCTM members. Which supplied me with a motivation to create a publicly available online discussion.
Let a and b denote the horizontal and vertical dimensions of the rectangle. At the outset, these may be reasonably designated as its length and width. However, technology permits modifications: there are three points that control the dimensions of the rectangle and the squares. In particular, it is possible to make the horizontal dimension of the rectangle smaller than its vertical dimension, which would beg for a reverse appellation: width (horizontal) and length (vertical).
|What if applet does not run?|
In any event, assume that x is the common side of the squares. The potential base of the box under construction has dimensions: horizontal
|v(x)||= x(a - 2x)(b - 2x)|
|= 4x³ - 2(a + b)x² + abx.|
This is a third degree polynomial with three real roots
What we are looking for is the point of local maximum that lies between 0 and the smallest of the two roots: a/2, b/2. We are thus interested in the values of x that satisfy
0 ≤ x ≤ min(a/2, b/2).
To find the local maximum we differentiate:
v'(x) = 12x² - 4(a + b)x + ab.
and then equate the derivative to 0. This leads to a quadratic equation
12x² - 4(a + b)x + ab = 0.
We expect the equation to have two roots: one corresponding to the local maximum and the other to the local minimum of v(x). Of interest to us is the smallest of the two. We can apply the quadratic formula to find both. The smallest of the two is obtained by selecting the minus sign.
It's important to understand that technology here provides only an approximate solution to the problem. For example, by dragging the uppermost point, which changes x, we may make the dot on the graph that corresponds the volume of the box for a given x, reach the local maximum point. For, say,
Finally, there are differences in implementation between the above applet and that by Ortiz and Popovich. On one hand, the latter shows only the modifiable rectangle and a lot of textual (in part, computed) information, but no graph. I believe that a volume graph adds an essential dimension to the problem presentation, which might be especially important in middle school. The second difference is the other applet displays, in my view, an enormous amount of information. In particular, the areas of the flaps provide no additional insight into the problem and constitute a distraction. The numeric value of the area of the base, while mildly relevant to the problem, is rather a side effect of computations and, again, does not help students gain an insight.
Thirdly, I am completely at odds with the introductory paragraph of their article:
This problem can be solved using many representations: construction-paper models and beans to approximate the volume of different boxes; value tables for different box sizes; graphical representations of the volume as the box size changes; formulas to calculate the volume of the boxes by substituting different values in the formula; or derivatives of the formulas to find the maximum values. In this article, we demonstrate the use of applets as another way to present or explore possible solutions for this and other variations of this problem.
None of the above, applets including, provides a solution to the problem. Claims to the contrary are misleading. Some mathematical problems do not admit exact solutions and, if necessary and if possible, could be solved approximately. Exact solutions to other problems can't often be obtained as exact quantities and may be presented by their approximations. Whenever mathematics models a practical problem, having an exact solution may not be even important. It is still important to know the difference between solving a problem and finding an approximate solution.
- E. Ortiz, A. Popovich, Solving a Volume Problem Using the Geometer's Sketchpad, NCTM, Online Journal of School Mathematics, Winter 2004-05, Volume 3, Number 2
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