Investigate the volume of an open box constructed by one piece of rectangular card that has all four corners having had squares cut out of them.

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Introduction

The aim of this coursework is to investigate the volume of an open box constructed by one piece of rectangular card that has all four corners having had squares cut out of them.  Firstly I will be studying the volume whilst changing the side of one length of the cut out square and the size of the original rectangle card.  After I have investigated this relationship I will try to find out the formula for finding the cut size to get the largest volume for any specified original card size.

Square card size

I am going to begin by investigating a square card because this will give me a basic formula which I can elaborate on.  I will start with a round number of 20cm for the length.  This means that the maximum cut out square length I can cut out will be 9cm else I will have no box left.

        The formula for the volume of any box is as follows:

                

        Below, there is a diagram to explain where all these figures come from.

        I have also included in this diagram the labels c, x and y, these show the cut out size and the original length and width of the card, I will now need to show the values of the width, length and height in terms of c, x and y.

        

Therefore I can replace these sub formulae into the first formula.

However for the values of a square, y=x therefore:

 As I have already said, the value of x is 20cm and I will use a range of values for c, then I will plot a graph with c on the x-axis and the volume (V) on the y-axis.  At the peak, the value on the x-axis will be the cut size that gives the maximum volume.

x=12

For this value of x, the maximum cut size I could cut out would be 5cm, yet these results were clear and showed a very good peak at c=2.  As 12/2=6, I predict that to get the maximum value for V, the cut size needs to be a sixth of x. To verify this prediction, I will repeat the results for 2 more values of x.

x=18

The next value I am using for x will be 30, using my prediction, I predict the graph will peak at c=5

x=30

I have proven my prediction to be correct, therefore I can construct a new formula so I only have to input one figure, the length of the original square, and I can find the largest volume possible for the specified square length.

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Revised Formulae

To find the largest volume of an open top box I first need to state that:

This means the cut size needs to be one sixth of the original length.

In the original formula, I stated that, to find the volume of an open top box, where x is one length:

                

Then if we replace c with x/6:

Multiply out the brackets:

Multiply the brackets by x/6:

Divide by a common denominator:

Simplify:

Finally divide by a factor to simplify:

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