## Part 1.2

Given a rectangular sheet of cardboard 15 in. by 25 in. If a small square of the same size is cut from each corner and each side folded up along the cuts to form a lidless box:

• What is the maximum value of the box?
• What size(s) of the square would produce a box of volume equal to 400 cubic inches?
• Prepare demonstrations/solutions to this problem using: GSP, Algebra Xpresser and Excel.

This investigation is an example of a problem that can be found in many high school mathematics curriculum. It is an excellent situational problem that can integrate the use of each of the aforementioned software programs. Each software supports the findings of the investigation in their own way.

I chose to begin this investigation through The Geometer's Sketchpad. This software allowed me to visualize the figure and determine the dimensions needed to solve the problem.

This approach needed one adjustment involving the measurements of the rectangular sheet of cardboard. Obviously, it is impossible to create a figure with the given dimensions. To work with accurate measurements the figure is similar to the original rectangle. The dimensions are based on the ratio of the original figure.

The ratio of the dimensions are

Therefore, I created a rectangle with a length that measures 3 inches and a width that measures 5 inches.

The final measurements need to be multiplied by the same factor I used in order to determine the ratio. More specifically, I created a rectangle whose length, width, and height were changed by a magnitude of 5 inches. My final volume must be mulitplied by 5 x 5 x 5 to solve the problem.

GSP allows for a visual discussion of the problem. The length of the side of the square is represented by the length x (shown above). Through the animation, students should be able to estimate the maximum volume of the box. They can also estimate the size of the square that should be cut out of the box to ensure the maximum value is met. The above image shows a static picture that has approximated a volume that is close to the maximum volume.

The next exploration of the volume of the rectangular box dealt with finding the size of the square that would produce a box whose volume equals 400 cubic inches. Again, the animation in this GSP image allows for such an investigation. One of the beneficial features of the animation is the continuous manipulation of the formula for the values of x. The animation shows that the volume of the rectangular box reaches 400 cubic inches for two values of x. (Click on the link to GSP to investigate furthur).

In order to connect the geometrical perspective to algebra, I would advance to using Algebra Xpresser as the next software program. This will also allow for a closer examination of the values of x that give the box a maximum volume and a volume of 400 cubic inches.

The Geometer's Sketchpad is a natural lead-in for Algebra Xpressor. It is easy for students to determine the dimensions of the rectangular box.

The general formula for volume is as follows

Volume = (length)(width)(height)

The volume of the rectangular box will be

Volume = (15 - 2x)(25 - 2x)(x)

Multiplying through will produce the equation

The following shows a graph of this equation

The graph clearly shows the values at which the volume has hit a maximum value and when the volume has hit its minimum value of zero (in other words - the box has no volume).

The relationship between the graph of the parabola and the volume of the rectangular box can be seen with the support of the GSP work previously done or some knowledge about maximum and minumum values of parabolas. Because this is the graph of a function we can extrapolate information from it.

It is easy to compare the maximum volume we approximated using GSP to the point at which this parabola reaches its maximum value. Thus, the maximum volume of the rectangular box will be when the parabola reaches its maximum value. Along the same accord, when the parabola crosses the x-axis, this tells us that the volume of the box has reached its minimum. More specifically the volume is zero and it is simply a rectangular sheet.

Algebra Xpresser can also help to determine the size of the square that would produce a box of volume 400 cubic inches. Overlay the graph of y = 400 on the existing parabola.

The line y = 400 will cross the parabola at the values of x (x represents the length of a side of the square) that produce the volume of 400 cubic inches. From the graph, it can be interpreted that the volume of the box reaches 400 cubic inches when x is approximately 1.6 inches and also approximately 4.75 inches.

Support of this approach can be discussed when the volume of the box is zero. The line y = 0, or the x-axis, crosses the parabola at two values of x. One of these values is when x = 0 ,that is, when there is no height. This in turn creates only two dimensions...a rectangular sheet.

Again Algebra Xpresser is able to show a lot through the graph, however, the values seem once again to be estimates. Using a spreadsheet will be able to show values that are more accurate.

The next part of the investigation was completed using Microsoft Excel. The approach is a bit different than the above explorations. While The Geometer's Sketchpad and Algebra Xpresser allow explorations using variables, a spreadsheet needs to have data that is inputed into functions.

The first column in the spreadsheet should contain the range of the function that is being explored. The beginning cell in this spreadsheet is the minimum value that could be inputed into the function (volume of the rectangular box). The beginning value of x that was inputed was 0. This means that there was no height and therefore no volume. Once again this is when we have a rectangular sheet. For the initial investigation, the values were increased by 1. The following columns determined the length of the box, the height of the box, and the volume of the box respective to the original value.

Values for the volume of a rectangular box 15 in by 25 in.
 height length width volume 0 15 25 0 1 13 23 299 2 11 21 462 3 9 19 513 4 7 17 476 5 5 15 375 6 3 13 234 7 1 11 77 8 -1 9 -72

The spreadsheet at this point, continues to produce estimations of the maximum volume for the rectangular box. It is clear that when x is approximately 3 we reach a maximum volume. For accuracy, we can continue inputing values in the spreadsheet that are between 2 and 3 and between 3 and 4.

 height length width volume 2.5 10 20 500 2.6 9.8 19.8 504.504 2.7 9.6 19.6 508.032 2.8 9.4 19.4 510.608 2.9 9.2 19.2 512.256 2.95 9.1 19.1 512.7395 3 9 19 513 3.05 8.9 18.9 513.0405 3.06 8.88 18.88 513.022464 3.07 8.86 18.86 512.995772 3.08 8.84 18.84 512.960448 3.09 8.82 18.82 512.916516 3.1 8.8 18.8 512.864 3.2 8.6 18.6 511.872 3.3 8.4 18.4 510.048 3.4 8.2 18.2 507.416 3.5 8 18 504 3.6 7.8 17.8 499.824 3.7 7.6 17.6 494.912 3.8 7.4 17.4 489.288 3.9 7.2 17.2 482.976

The spreadsheet is able to calculate new data with ease. The values approximate the maximum value to be approximately 513.04 cubic inches when the length of a side of the square is 3.05 inches. This evaluation can be even more accurate with furthur inputed data.

The spreadsheet also illustrates where the volume of the rectangular box becomes 400 cubic inches (approximately). To determine these values more accurately, the range should become more acute. The following shows another spreadsheet that examines the values between 1 and 2 and between 4 and 5.

 height length width volume 1 13 23 299 1.05 12.9 22.9 310.1805 1.1 12.8 22.8 321.024 1.15 12.7 22.7 331.5335 1.2 12.6 22.6 341.712 1.25 12.5 22.5 351.5625 1.3 12.4 22.4 361.088 1.35 12.3 22.3 370.2915 1.4 12.2 22.2 379.176 1.45 12.1 22.1 387.7445 1.5 12 22 396 1.55 11.9 21.9 403.9455 1.6 11.8 21.8 411.584 1.65 11.7 21.7 418.9185 1.7 11.6 21.6 425.952 1.75 11.5 21.5 432.6875 1.8 11.4 21.4 439.128 1.85 11.3 21.3 445.2765 1.9 11.2 21.2 451.136 1.95 11.1 21.1 456.7095 2 11 21 462 4 7 17 476 4.05 6.9 16.9 472.2705 4.1 6.8 16.8 468.384 4.15 6.7 16.7 464.3435 4.2 6.6 16.6 460.152 4.25 6.5 16.5 455.8125 4.3 6.4 16.4 451.328 4.35 6.3 16.3 446.7015 4.4 6.2 16.2 441.936 4.45 6.1 16.1 437.0345 4.5 6 16 432 4.55 5.9 15.9 426.8355 4.6 5.8 15.8 421.544 4.65 5.7 15.7 416.1285 4.7 5.6 15.6 410.592 4.75 5.5 15.5 404.9375 4.8 5.4 15.4 399.168 4.85 5.3 15.3 393.2865 4.9 5.2 15.2 387.296 4.95 5.1 15.1 381.1995 5 5.0 15 375

Furthur investigation using the spreadsheet shows that the volume of the box is 400 cubic inches when the side of the square cut out is equal to either 1.524929 or 4.79284954. These of course are still just approximations.

A spreadsheet can help to input exacting values for the functions. The benefit of the spreadsheet is its easy manipulation of the values and the affect they have on the functions. Students are able to quickly conjecture and receive feedback while investigating this problem.

I believe this exploration to be a substantial investigation if the above software programs are able to be used to support each other. The investigation brings together a geometric, algebraic and data/function approach to one problem. The three software programs help to connect various forms of mathematics in one exploration.