# The Open Box Problem

Extracts from this document...

Introduction

An open box is to be made from a sheet of card. Identical squares are cut off the four corners of the card, as shown below. The card is then folded along the dotted lines to make the box. My main aim is to determine the size of the square cut which makes the volume of the box as large as possible for any given square sheet of card. I am going to begin by investigating a square with a side length of 24 cm. Using this side length, the maximum whole number I can cut off each corner is 11cm, as otherwise I would not be able to make an open cube. I am going to begin by looking into whole numbers being cut out of the box corners. The formula needed to get the volume of a box is: Volume = Length x Width x Height If I am to use a square of side length 24cm, then I can calculate the side lengths minus the cut out squares using the following equation. Volume = Length - (2 x Cut Out) x Width - (2 x Cut Out) x Height (Cut Out) Using a square, both the length & the width are equal. I am using a length and width of 24cm. I am going to call the cut out "x." ...read more.

Middle

18 D=10 8 18x18x8 2592 L = 19 D=10 9 19x19x9 3249 L = 20 D=20 0 20x20x0 0 L = 21 D=20 1 21x21x1 882 L = 22 D=20 2 22x22x2 968 As you can see, my formula did not work completely. I did not find any patterns between the number and so decided to restart, using an "easy" length. I am going to continue with 20 as L instead of 24. If I were using a cut out of length 1cm, the equation for this would be as follows: Volume = [20 - (2 x 1) x 20] - (2 x 1) x 1 So we can work out through this method that the volume of a box with corners of 1cm� cut out would be: (20 - 2) x (20 - 2) x 1 18 x 18 x 1 = 324cm� I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box. Below are the results I got through this spreadsheet. x=1 18x18x1 324 x=2 16x16x2 512 x=3 14x14x3 588 x=4 12x12x4 576 x=5 10x10x5 500 x=6 8x8x6 384 x=7 6x6x7 252 x=8 4x4x8 128 x=9 2x2x9 36 As you can see by the table above, the largest volume is achieved when an area of 3cm� is cut off each corner of the box. ...read more.

Conclusion

20 cm is the length and width of the square, therefore 20 = 2H + L Rearranging this it becomes L = 20 - 2H The formula for the volume is Width * Length * Height Or (20 - 2H) * (20 - 2H) * H The maximum area it found when the cut out is 1/6, meaning that: H = 20/6 If I substitute this into the equation for maximum volume then it will become V = (L - 2L/6)(L - 2L/6) L/6 I multiply the 2 in each bracket with L/6, giving you V = (L - 2L/6) (L - 2L/6) L/6 To multiply brackets out I'll separate them, and write the 2nd bracket out twice. V = L (L - 2L/6) - 2L/6 (L - 2L/6) L/6 Then complete multiplication leaving �V = (L� - 2L�/6 - 2L�/6 - 4L�/36) L/6 Then I multiply the whole thing by L/6, which gives me V = L�/6 - 2L�/36 - 2L�/36 + 4L�/216 I put the whole equation over a common denominator to help me with simplification, meaning the whole thing becomes over 216. This means that it becomes: V = 4L� + 36L� - 12L� - 12L� 216 This can be simplified so you end up with: 40L� - 24L� 216 Which then cancels out to give you 16L� 216 You then divide the top and bottom by 8, which is a factor of both, and you will end up with: 2L� 27 ...read more.

This student written piece of work is one of many that can be found in our GCSE Open Box Problem section.

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