The Open Box Problem

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The Open Box Problem

An open box is to be made from a sheet of card. Identical squares are cut off the four corners of the card, which is then folded along the dotted lines to make the box.

The main aim is to find out the size of the square cut which makes the volume of the box as large as possible for any rectangular sheet of card.

I am going to begin by looking at which size of cut out square will give the largest volume for a square sheet of card.

I will now look at a square with a side of 6 cm.

The letter 'x' represents the length of the square cut out to form the box. The letter 'l' is used for the length of each side - 2x, this is going to be the length of the open box created. The letter 'h' is going to be the height of the open box created, and 'v' is the volume of the open box created.

I will use the formula Volume=Length×Width×Height or v=l²h to find out the volume of the box.

I will investigate the volume of boxes, where x is between 0.5 and 2.5, at intervals of 0.5. I cannot go higher than 2.5 to 3, because then the card would be cut in half, and would not form an open box.

I have used the information above to construct the spreadsheet below. The number in bold is the maximum volume. I intend to use spreadsheets to find the volumes given by different values of x, so I can find which value of x gives the maximum volume.

This shows that the maximum volume is obtained when x is between 1 and 1.5, I will now make another three spreadsheets, to find the maximum volume correct to three decimal places, as I believe that this first spreadsheet does not give an accurate result.

These show, that for a square of width 6cm, that the maximum volume is obtained when x=1.000.

I will now repeat the spreadsheets above for a square of width 12cm, to find the maximum volume for this size of card, and to attempt to find any patterns in maximum volumes.

These show that the maximum volume for a square of width 12cm is obtained when x=2.

Width of square Value of x to give maximum volume

6 1

2 2

I have noticed that so far, my results would suggest that the maximum volume if given when x is 1/6th of the width of the square sheet of card. To prove this, I will have to find the maximum volume for another square, and see whether that obeys with what I have predicted.

Therefore a square of width 10cm should give a maximum volume when:

X=10=1.667

6

This is the spreadsheet for the square of length 10cm, this will give me what value of x gives the maximum volume for this size sheet of card.

The maximum volume is in fact obtained when x=1.667, which is what I predicted. This proves that the maximum volume is in fact obtained when x=1/6th of the width of the square.

It is possible to confirm my results for the maximum area through using differentiation. Differentiation will tell us the turning points on a graph, where the gradient = 0, one of the two numbers we will get will be the maximum value (which I have already found through spreadsheets) and the other will be the minimum value, this maximum will be the same as the values I have already come to, proving that they are correct.
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I will now use differentiation for the square of width 6.

x(6 - 2x)² = V

x(6 - 2x)(6 - 2x) = V

x(4x² - 24x + 36) = V

4x³ - 24x² +36x = dv

dx

at maximum or minimum dv = 0

dx

2x² - 48x + 36 = 0

simplify by dividing by 12

x² - 4x + 3 = 0

(x - 1)(x - 3) = 0

x is therefore either 1 or 3

Since through my spreadsheets I have ...

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