# THE BOX PROBLEM

Our goal is to find the maximum volume of a box given a rectangular piece of cardboard that measures 15 inches by 25 inches. It is imperative that we are able to understand what we are looking for, so we shall begin by drawing a picture of the situation.

In order to fold the sides to make the box, it is necessary to cut out squares from each corner.

In the picture above, the length of each side of the square measures x. Originally, the length of the rectangular piece of cardboard was 25 inches; however, if we cut a section measuring x from one corner and a second section measuring x from the other corner, we have essentially removed 2x from the 25 inches. It follows that the "new" length of the rectangular region is 25 - 2x. Similarly, the width of the rectangular region is now 15 - 2x.

At this point, we want to formulate a way to calculate the maximum volume of the box. One familiar formula that is used quite often is V = lwh. We know that the length l = 25 - 2x and the width w = 15 - 2x. However, the height is not obvious! Or is it? The height of the box will be determined once the box has been folded. Since the height is simply the measure from the base to the lip of the box, the height is given by x. It follows that

Of course, we must know a value of x in order to calculate the volume. Suppose that x = 2. This means that the square region we are removing has a side measuring 2 inches. Using the formula, we see that

Unfortunately, this formula will not generate the maximum area without going through a great deal of trial-and-error. However, we may use a spreadsheet, such as EXCEL, to examine different values of x and the volume that a particular value of x will yield. Although we wish to create an extensive list, we should also keep in mind that upon examining these values, x may have to be restricted. Let's begin by booking at values of x in the interval [1,30].

 x 25-2x 15-2x Volume 1 23 13 299 2 21 11 462 3 19 9 513 4 17 7 476 5 15 5 375 6 13 3 234 7 11 1 77 8 9 -1 -72 9 7 -3 -189 10 5 -5 -250 11 3 -7 -231 12 1 -9 -108 13 -1 -11 143 14 -3 -13 546 15 -5 -15 1125 16 -7 -17 1904 17 -9 -19 2907 18 -11 -21 4158 19 -13 -23 5681 20 -15 -25 7500 21 -17 -27 9639 22 -19 -29 12122 23 -21 -31 14973 24 -23 -33 18216 25 -25 -35 21875 26 -27 -37 25974 27 -29 -39 30537 28 -31 -41 35588 29 -33 -43 41151 30 -35 -45 47250

It is important to note that if we use values of x greater than or equal to 7.5, then the width (given as 15 - 2x) will either be 0 or will be negative. Obviously, if the width is 0 this reduces our figure to a two-dimensional object; if the width is negative, this situation is impossible. Hence, x must be less than 7.5. In addition, x must be greater than 0 because we do not want to choose a height of 0 or a negative height.

Now that it has been stipulated that x is in the interval (0, 7.5), we may now examine the chart for the maximum volume. Notice that if x = 2, then the volume = 462 cubic inches; if x = 3, then the volume = 513 cubic inches. However, if x = 4, then the volume = 476 cubic inches. So, it appears as though the maximum volume is produced when x = 3. Unfortunately, this chart only reveals integral values of x. It is certainly possible that the maximum volume could be produced for some value of x between 3 and 4. Let's explore some of those values.

 x 25 - 2x 15 - 2x Volume 3 19 9 513 3.01 18.98 8.98 513.025604 3.02 18.96 8.96 513.042432 3.03 18.94 8.94 513.050508 3.04 18.92 8.92 513.049856 3.05 18.9 8.9 513.0405 3.06 18.88 8.88 513.022464 3.07 18.86 8.86 512.995772 3.08 18.84 8.84 512.960448 3.09 18.82 8.82 512.916516 3.1 18.8 8.8 512.864 3.11 18.78 8.78 512.802924 3.12 18.76 8.76000000000001 512.733312 3.13 18.74 8.74000000000001 512.655188 3.14 18.72 8.72000000000001 512.568576 3.15 18.7 8.70000000000001 512.4735 3.16 18.68 8.68000000000001 512.369984 3.17 18.66 8.66000000000001 512.258052 3.18 18.64 8.64000000000001 512.137728 3.19 18.62 8.62000000000001 512.009036 3.2 18.6 8.60000000000001 511.872

Notice that the chart reveals that if x = 3.03, then the maximum volume is approximately 513.05 cubic inches.

We also want to know the size(s) of the box that would produce a volume of 400 cubic inches. If we look at the original spreadsheet above, we can easily see that if x = 1, then the volume is 299 cubic inches; if x = 2, then the volume is 462 cubic inches. It follows that the value of x that would produce such a box is in the interval (1,2). Let's explore these values further.

 1.52 21.96 11.96 399.215 1.53 21.94 11.94 400.804

This small portion of the chart reveals that x must be between 1.52 and 1.53. So, let's explore a bit further.

 1.52492 21.9502 11.9502 399.999 1.52493 21.9501 11.9501 400 1.52494 21.9501 11.9501 400.002

Notice that if x = 1.5249301, then the volume is approximately 400 cubic inches.

We should also note that if x = 4, then the volume of the box is 476 cubic inches and if x = 5, the volume is 375 cubic inches. Hence, a second value of x will yield a volume of 400 cubic inches as well and that value is located in the interval from (4,5). Let's examine those values.

 4.791 15.418 5.418 400.215 4.792 15.416 5.416 400.099 4.793 15.414 5.414 399.983

Again, this is only a small portion of the spreadsheet, however it seems as though if x is approximately 4.79 inches, then the volume of the box would be 400 cubic inches.

This box problem may also be explored from a graphical standpoint. We stated earlier that the function

represents the volume of the lidless box for given values of x. Now, let's take this same function and graph it.

Although the graph of the volume function would typically as a cubic function, we only want to view the portion of the graph where x is the interval from (0,7.5)--as we stated earlier. Hence, this portion of the graph appears to be an upside down parabola. The x-axis represents the length of a side of the square in inches, and the y-axis represents the volume of the rectangular box. Moreover, the vertex will provide us with the needed information; i.e., the x-coordinate will tell us the required length of one side of the box in order to yield the maximum volume (y-coordinate of the vertex). Let's try to look at a close-up picture of the vertex of this parabola.

Just "eyeballing" this picture it appears as though the vertex is given by the point (3.0,513). It follows that we must cut-out a square with a side measuring approximately three inches or so in order to obtain a box whose volume is approximately 513 cubic inches.

Let's also try to find the size(s) of the square that would produce a box volume of 400 cubic inches. Indeed, we are looking for points on the graph where the y-coordinate is 400. We may visually locate these points by drawing the line y =400 and looking for points of intersection with the volume function. Again, we will place the same restrictions on x as before.

Notice that the horizontal line y = 400 intersects the volume function in two points, so there are two values of x that will produce a box of the desired volume. At this point, it is necessary to focus on those two points of intersection.

It appears as though x = 1.5 or x = 4.7 would yield a box whose volume is 400 cubic inches. Obviously, these are approximate values.

A third approach would be to actually construct and animate the box using Geometer's Sketchpad. It is also necessary to take into account the meaures of each dimesion of the box.

In order to be as accurate as possible, it is necessary to multiply by the appropriate scale factor for the drawing. Indeed, the results should be the same: 1) the maximum volume of the box is approximately 513 cubic inches and 2) x should be approximatley 1.5 or approximately 4.7 in order to construct a box whose volume is 400 cubic inches.

Exam Part 2