Optimization: The Box Problem

by

Laura Kimbel

In this exploration, we will investigate various methods of maximizing the volume of a box.

Let's begin by looking at a box:

The rectangle above has dimensions of 5 X 8. In order to construct a box, we have cut squares out of the corners of our rectangle that are x X x in size. We want to see what value of x will produce the largest volume when constructed into a box.

Let us first find a way to express the volume of our box. Since V = lwh (length * width * height), we must first determine how to express these dimensions in terms of x. Beginning with the length, l, of our rectangle, we can see that there are 2 lengths of x cut out of that side. Therefore we can express l as l = 8 - 2x. Similarly, our width is w = 5 - 2x. When our box is constructed, we can see that the height of our box is the length of the cutout so, h = x.

By substitution we now have an equation for volume in terms of x:

V(x) = (8 - 2x)(5 - 2x)x.

Using a table, we can investigate how different values of x affect the volume of our box.

It is important to consider the domain of x. Since one of our sides can be a maximum of 5, and 2 values of x are taken out of each side, x must be less than 2.5 (since at x = 2.5, our width is 0). Similarly, x must be greater than 0 in order to actually form a box. This is called an "implied domain" of our function V(x). Ordinarily, a cubic function would have a domain that spans all of the reals, however most of the numbers in that domain produce invalid volumes. We can also plot these points in a graph to explore another interpretation of what a "maximum" is.

From the table, the largest volume is 18 cubic units. (This can also be observed from our graph, although it may not be as easy to read the exact values.) This is obtained when x =1. Is this precise? From the table, we can see that the values increase until x = 1 and then 1.1 is the first place it decreases. Could there be a value between .9 and 1.1 that gives us a greater volume than 18?

Another way to solve an optimization problem is by using calculus.

By finding the derivative of V(x) and setting it equal to zero, we can determine the "critical points," also known as the maximums and minimums of our function.

when

.

However, from the graph below, we can see that x = 1 is a maximum, and x = 10/3 is a minimum. We also know that x = 1 is our maximum since out of the 2 solutions, it is the only one that is in the domain of our problem.

Therefore, at x = 1, we do have a maximum of 18 cubic units.

The dimensions of the box with a volume of 18 are 1 X 3 X 6.

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