Assignment 12: Volume of a Box

by

Melissa Silverman

Suppose we begin with a 5x8 sheet of cardboard. We want to make an open box from this sheet, with the largest volume possible. How can we construct the box with the largest volume? To begin, let's draw a picture of what happens when we create a three dimensional figure from a flat piece of cardboard.

By cutting a square piece from each corner, we get a box whose base is smaller than the original box (the green rectangle). The width of the base is 5 units minus two of the squares' lengths. Similarly, the length of the base is 8 units, minus two of the squares' lengths. The depth of the box is the same as one side of the square. Thus the dimensions of the squares cut from each corner determine the depth of the box.

How big should the squares be in each corner to achieve the maximum volume of the box? We can use a spreadsheet to calculate the volume of the box given the dimensions of the square to be cut from each corner. The length of the base is 8-2*(Size of Square). The width of the base is 5-2*(Size of Square). The area of the base is the product of the new length and width. The volume is the product of the area of the base and the dimension of the square.

Looking at the chart, we can see that cutting a 1 unit square from each corner will create a box with the largest volume. Cutting a square that is larger or smaller than that results in a smaller. The squares cut from each corner cannot be greater than 2.5 units. At that point, the width of the base would be 0 units, and a box would not be formed. If the squares cut from each corner are small, the area of the base is large, but the height of the box is small, resulting in a smaller volume.

What if we began with an equilateral triangle, with side lengths of 10 units? We could construct a triangular box by cutting a rhombus from each corner. We will leave tabs on each of the three side pieces (see the red dashed lines in the picture below). How could we cut the triangle so that we get a triangular box with the largest volume?

We can use a spreadsheet to calculate the volume of the box given the dimensions of the rhombus. CLICK HERE to open the spreadsheet on your own computer.

We begin with the height of the rhombus. Then we can use the properties of a 30-60-90 triangle to figure out the length of the tab. The length of the new base is one rhombus length and one tab length shorter on each side. Subtracting two rhombus lengths and two base lengths from the triangle's base of 10 units yields the length of the new base. To figure out the height of the new triangle, use the Pythagorean theorem. Subtract the square of half the base length from the square of the base length. The square root of that value is the base height. The area of the box base is the product of one half, the base height, and the base length. The volume of the box is the product of the area of the base and the height of the rhombus.

We can see from the chart that the rhombus that will give us the largest volume is a rhombus whose height is 1 unit, just like the rectangular box that was also maximized with a square of 1 unit cut from each corner.

This problem is often seen in calculus curriculum as a min/max problem using derivatives. Using a simple diagram and a spreadsheet, middle school students could develop the same ideas using their math background. This problem is important in middle school curriculum for several reasons.

First of all, it is a great way to use both area and volume in one problem. Having students combine the concepts of 2 dimensional area and 3 dimensional volume is a great exercise. The students must practice visualizing both area and volume. Making a 2 dimensional pattern for a three dimensional figure is often in the middle school curriculum. Secondly, this problem does involve some simple algebra, with the dimension of the square as one of the variables. This is a great application of algebra in that it connects geometry and algebra. Lastly, this problem develops the ideas they will need in higher level mathematics, such as algebra and calculus.

I also like the potential for exploration built into this problem. Students may ask about many things. What would happen if we started with a square sheet instead of a retangular one? What if we tried the same procedure with a triangular piece? Or any polygon? Why do we have to cut a square out of the corner? Asking students to use both computer technology and actual construction with paper to test their ideas could lead to some really great mathematical discovery.