EMT 668
Final Project #3

Maximum volumes of a 15 x 25 x ? lidless box
Brian Seitz

PROBLEM:: I am 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 along the cuts to form a lidless box, what size squares can be cut from each corner to produce a box with a volume of 400 cubic in? Also, what size square would produce the maximum volume?

SOLUTION: First, I solved the problem using Excel and a spreadsheet. The formula I derived is
x*(15-2x)*(25-2x). x is the length of a side of the square cut out of the original cardboard. 15 - 2x is the side of the cardboard minus the side of square on both sides of the piece of cardboard. 25 - 2x is the third side of the box and is equal to the 25 inch side of cardboard minus the two times the side of the square. The following is a spreadsheet for the value of x (the side of the square cut out) and the volume of the lidless box for the given x value.

One can see that the maximum volume of the box occurs when the value of x=3. At this value for x, the box has a volume of 513 cubic inches. The box dimensions would be 3 in. x 9 in. x 19 in.

To find the values of x when the volume equals 400 is rather simple now. By examining the spreadsheet, the volume 400 is found at approximately 1.5 and approximately 4.8. To narrow the value of x, I can adjust x on the spreadsheet. Now using, .01 as my increment for the x value, the volume is found precisely.

The volume of 400 is obviously when x = 1.53 and when x = 4.79. This would make the dimensions of the box 1.53 in. x 11.94 in. x 21.94 in. or 4.79 in. x 5.42 in. x 15.42. in. ONe may wonder why there are two answers to when the volume equals 400, and by looking at a graph of the function explains this very clearly.

With the formula known, I can use Algebra Xpresser to plot the graph of the equation. The following graph is the graph for x*(15-2x)*(25-2x)

This graph reinforces what the spreadsheet told us, that the maximum volume occurs at x =3 and the volume equals 400 at x=1.53 and x=4.79. The graph is a parabola with a maximum value of x =3. The parabola crosses y=400 at two distinct locations, one on the increasing side of the parabola and one on the decreasing side.

Geometers Sketchpad (GSP) can also be used a tool in solving this particular problem. Constructing a rectangle of any given size, then cutting out equal squares from each corner gives us the ability to see the volume of our lidless box. Click below, then click animate. Observe how the volume of the lidles box increases, peaks, then decreases as the segment AB is stretched. The maximum volume and any desired volume (i.e 400 cubic inches) can be found in tlis manner. The given rectangle of 15 in x 25 in. could not be constructed because it is too large for GSP, but one can plainly see how GSP can be used to find the volume of a lidless box.

Click here to play with lidless box using GSP

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