# The Open Box Problem

Extracts from this document...

Introduction

The Open Box Problem

Maths Coursework

Jawhari Thomas 11.5

An open box is a box missing 1 of its six surfaces it can be made using either a squared or rectangular sheet of card. Identical squares are cut from the four corners of the card this creates the height of the box it is then folded as shown below. The card is folded along the dotted lines to form the box.

Stage 1 Stage 2 Stage 3

The aim of this exercise is to find the formulae that will enable someone determine the size of the squares cut from the corners of the sheet of card to give the greatest the volume of the box.

I am going to begin the investigation using squares, as this will most probably be easiest. I won’t build the boxes I am going to use simple mathematics to work out the volumes. Firstly I am going to use an example of how I will carry out the experiment using a square 20cm in length. Using this sized length will allow me to only cut off each corner up to 9.9cm as otherwise I will cause me to run out of card. I am going to begin by looking at cutting the squares off as whole numbers. To find the volume of any box we must use the formula:

V = L * W * H

When:

V is Volume

## L is Length

W is Width and

H is Height

Middle

704

30x30

12

24

6

6

432

30x30

13

26

4

4

208

30x30

14

28

2

2

56

30x30

15

30

0

0

0

30x30 | 5.1 | 10.2 | 19.8 | 19.8 | 1999.404 | |

30x30 | 5.2 | 10.4 | 19.6 | 19.6 | 1997.632 | |

30x30 | 5.3 | 10.6 | 19.4 | 19.4 | 1994.708 | |

30x30 | 5.4 | 10.8 | 19.2 | 19.2 | 1990.656 | |

30x30 | 5.5 | 11 | 19 | 19 | 1985.5 | |

30x30 | 5.6 | 11.2 | 18.8 | 18.8 | 1979.264 | |

30x30 | 5.7 | 11.4 | 18.6 | 18.6 | 1971.972 | |

30x30 | 5.8 | 11.6 | 18.4 | 18.4 | 1963.648 | |

30x30 | 5.9 | 11.8 | 18.2 | 18.2 | 1954.316 |

8x8

8x8 | 1 | 2 | 6 | 6 | 36 | |

8x8 | 2 | 4 | 4 | 4 | 32 | Max |

8x8 | 3 | 6 | 2 | 2 | 12 | |

8x8 | 4 | 8 | 0 | 0 | 0 |

8x8 | 2.1 | 4.2 | 3.8 | 3.8 | 30.324 | |

8x8 | 2.2 | 4.4 | 3.6 | 3.6 | 28.512 | |

8x8 | 2.3 | 4.6 | 3.4 | 3.4 | 26.588 | |

8x8 | 2.4 | 4.8 | 3.2 | 3.2 | 24.576 | |

8x8 | 2.5 | 5 | 3 | 3 | 22.5 | |

8x8 | 2.6 | 5.2 | 2.8 | 2.8 | 20.384 | |

8x8 | 2.7 | 5.4 | 2.6 | 2.6 | 18.252 | |

8x8 | 2.8 | 5.6 | 2.4 | 2.4 | 16.128 | |

8x8 | 2.9 | 5.8 | 2.2 | 2.2 | 14.036 |

15x15

15x15 | 1 | 2 | 13 | 13 | 169 | |

15x15 | 2 | 4 | 11 | 11 | 242 | |

15x15 | 3 | 6 | 9 | 9 | 243 | Max |

15x15 | 4 | 8 | 7 | 7 | 196 | |

15x15 | 5 | 10 | 5 | 5 | 125 | |

15x15 | 6 | 12 | 3 | 3 | 54 | |

15x15 | 7 | 14 | 1 | 1 | 7 |

15x15 | 2.1 | 4.2 | 10.8 | 10.8 | 244.944 | |

15x15 | 2.2 | 4.4 | 10.6 | 10.6 | 247.192 | |

15x15 | 2.3 | 4.6 | 10.4 | 10.4 | 248.768 | |

15x15 | 2.4 | 4.8 | 10.2 | 10.2 | 249.696 | |

15x15 | 2.5 | 5 | 10 | 10 | 250 | Max |

15x15 | 2.6 | 5.2 | 9.8 | 9.8 | 249.704 | |

15x15 | 2.7 | 5.4 | 9.6 | 9.6 | 248.832 | |

15x15 | 2.8 | 5.6 | 9.4 | 9.4 | 247.408 | |

15x15 | 2.9 | 5.8 | 9.2 | 9.2 | 245.456 | |

15x15 | 3 | 6 | 9 | 9 | 243 | |

15x15 | 3.1 | 6.2 | 8.8 | 8.8 | 240.064 | |

15x15 | 3.2 | 6.4 | 8.6 | 8.6 | 236.672 | |

15x15 | 3.3 | 6.6 | 8.4 | 8.4 | 232.848 | |

15x15 | 3.4 | 6.8 | 8.2 | 8.2 | 228.616 | |

15x15 | 3.5 | 7 | 8 | 8 | 224 | |

15x15 | 3.6 | 7.2 | 7.8 | 7.8 | 219.024 | |

15x15 | 3.7 | 7.4 | 7.6 | 7.6 | 213.712 | |

15x15 | 3.8 | 7.6 | 7.4 | 7.4 | 208.088 | |

15x15 | 3.9 | 7.8 | 7.2 | 7.2 | 202.176 |

25x25

25x25 | 1 | 2 | 23 | 23 | 529 | |

25x25 | 2 | 4 | 21 | 21 | 882 | |

25x25 | 3 | 6 | 19 | 19 | 1083 | |

25x25 | 4 | 8 | 17 | 17 | 1156 | Max |

25x25 | 5 | 10 | 15 | 15 | 1125 | |

25x25 | 6 | 12 | 13 | 13 | 1014 | |

25x25 | 7 | 14 | 11 | 11 | 847 | |

25x25 | 8 | 16 | 9 | 9 | 648 | |

25x25 | 9 | 18 | 7 | 7 | 441 | |

25x25 | 10 | 20 | 5 | 5 | 250 | |

25x25 | 11 | 22 | 3 | 3 | 99 | |

25x25 | 12 | 24 | 1 | 1 | 12 |

25x25 | 4.1 | 8.2 | 16.8 | 16.8 | 1157.184 | ## Max |

25x25 | 4.2 | 8.4 | 16.6 | 16.6 | 1157.352 | |

25x25 | 4.3 | 8.6 | 16.4 | 16.4 | 1156.528 | |

25x25 | 4.4 | 8.8 | 16.2 | 16.2 | 1154.736 | |

25x25 | 4.5 | 9 | 16 | 16 | 1152 | |

25x25 | 4.6 | 9.2 | 15.8 | 15.8 | 1148.344 | |

25x25 | 4.7 | 9.4 | 15.6 | 15.6 | 1143.792 | |

25x25 | 4.8 | 9.6 | 15.4 | 15.4 | 1138.368 | |

25x25 | 4.9 | 9.8 | 15.2 | 15.2 | 1132.096 | |

25x25 | 5 | 10 | 15 | 15 | 1125 | |

25x25 | 5.1 | 10.2 | 14.8 | 14.8 | 1117.104 | |

25x25 | 5.2 | 10.4 | 14.6 | 14.6 | 1108.432 | |

25x25 | 5.3 | 10.6 | 14.4 | 14.4 | 1099.008 | |

25x25 | 5.4 | 10.8 | 14.2 | 14.2 | 1088.856 | |

25x25 | 5.5 | 11 | 14 | 14 | 1078 | |

25x25 | 5.6 | 11.2 | 13.8 | 13.8 | 1066.464 | |

25x25 | 5.7 | 11.4 | 13.6 | 13.6 | 1054.272 | |

25x25 | 5.8 | 11.6 | 13.4 | 13.4 | 1041.448 | |

25x25 | 5.9 | 11.8 | 13.2 | 13.2 | 1028.016 |

25x25 | 4.11 | 8.22 | 16.78 | 16.78 | 1157.246124 | |

25x25 | 4.12 | 8.24 | 16.76 | 16.76 | 1157.298112 | |

25x25 | 4.13 | 8.26 | 16.74 | 16.74 | 1157.339988 | |

25x25 | 4.14 | 8.28 | 16.72 | 16.72 | 1157.371776 | |

25x25 | 4.15 | 8.3 | 16.7 | 16.7 | 1157.3935 | |

25x25 | 4.16 | 8.32 | 16.68 | 16.68 | 1157.405184 | |

25x25 | 4.17 | 8.34 | 16.66 | 16.66 | 1157.406852 | Max |

25x25 | 4.18 | 8.36 | 16.64 | 16.64 | 1157.398528 | |

25x25 | 4.19 | 8.38 | 16.62 | 16.62 | 1157.380236 |

25x25 | 4.161 | 8.322 | 16.678 | 16.678 | 1157.405801 | |

25x25 | 4.162 | 8.324 | 16.676 | 16.676 | 1157.406318 | |

25x25 | 4.163 | 8.326 | 16.674 | 16.674 | 1157.406735 | |

25x25 | 4.164 | 8.328 | 16.672 | 16.672 | 1157.407052 | |

25x25 | 4.165 | 8.33 | 16.67 | 16.67 | 1157.407269 | |

25x25 | 4.166 | 8.332 | 16.668 | 16.668 | 1157.407385 | Max |

25x25 | 4.167 | 8.334 | 16.666 | 16.666 | 1157.407402 | |

25x25 | 4.168 | 8.336 | 16.664 | 16.664 | 1157.407319 | |

Conclusion

I worked out the proportion that needs to be cut off the box to give maximum volume this was 0.24445, which is very close to ¼.

Ð stands for delta.

Firstly we should consider a graph of y = x² as shown below.

The line through X and Y has almost the correct gradient.

It’s gradient is

Increase in y-coordinate from X to Y

Increase in x-coordinate from X to Y

You have to find an expression for , which represents

the gradient of the graph at the point X.

So

y = x²

y + Ðy = (x + Ðx)(x +Ðx)

Multiply out brackets.

y+ Ðy = x² + x Ðx + x Ðx + (Ðx) ²

Add like terms together.

y + Ðy = x² + 2xÐx + (Ðx)²

Now here the x² at the end is y in terms of x.

Ðy = x² + 2xÐx + (Ðx) ² - x²

Then you divide by Ðx, which gives you

Ðy 2xÐx – (Ðx) ²

Ðx = Ðx

Ðy

Ðx = 2x + Ðx

Now because delta (Ð) is so tiny that it is insignificant, we forget all about it, which leaves us with

Ðy/Ðx = 2x = 0

I can use calculus to help me complete my calculations to solve the problem, for proof through exhaustion.

I have drawn a graph showing the proportions that I have worked out. I can see that they tend towards ¼. This is the amount that should be cut off each corner to give the maximum volume possible.

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

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month