# Open Box Investigation

Extracts from this document...

Introduction

25th January

Math | Sara Ellis |

The Open Box Investigation |

Introduction:

The open box problem is regarding an open box made from a sheet of card. Identical sized squares are removed from the four corners of the card (shown in the diagram below). The aim of this investigation is to verify the size of the square removed that would make the largest volume for any given sheet of card, both square and rectangular. I will do this by finding a general formula which would work for finding the best size of square cut.

Square Boxes

In order to determine the size of square cut off from the card, I tried many different sizes on square boxes to find the size that would give the largest volume. To find the volume I have used the formula:

Volume = Length x Width x Height

To find the length and width I take two of the length of square cut from the total length/width of the box. The height is the same as the side length of square cut.

Below are my results:

10x10 Square

Side Length of Square (cm) | 1 | 2 | 3 | 4 |

Length of Box | 8 | 6 | 4 | 2 |

Width of Box | 8 | 6 | 4 | 2 |

Height | 1 | 2 | 3 | 4 |

Volume | 64 | 72 | 48 | 16 |

To find the more precise length of square which I should cut, I have used the upper and lower bounds of the best length of square in the table above (in this case, 2cm).

Side Length of Square | 1.5 | 1.6 | 1.7 | 1.8 | 1.9 | 2 | 2.1 |

Length of Box | 7 | 6.8 | 6.6 | 6.4 | 6.2 | 6 | 5.8 |

Width of Box | 7 | 6.8 | 6.6 | 6.4 | 6.2 | 6 | 5.8 |

Height | 1.5 | 1.6 | 1.7 | 1.8 | 1.9 | 2 | 2.1 |

Volume | 73.5 | 73.98 | 74 | 73.7 | 73 | 72 | 70.6 |

I have used these results to plot a graph, as shown below. This illustrates the pattern of the results more clearly.

15x15 Square

Side Length of Square | 1 | 2 | 3 | 4 |

Length of Box | 13 | 11 | 9 | 7 |

Width of Box | 13 | 11 | 9 | 7 |

Height | 1 | 2 | 3 | 4 |

Volume | 169 | 242 | 243 | 196 |

Middle

2.6

2.6

2.8

2.9

3

3.1

Volume of Box

250

249.7

248.8

247.4

245.5

243

240.1

20x20 Square

Side Length of Square | 1 | 2 | 3 | 4 |

Length of Box | 18 | 16 | 14 | 12 |

Width of Box | 18 | 16 | 14 | 12 |

Height | 1 | 2 | 3 | 4 |

Volume | 324 | 512 | 588 | 576 |

Side Length of Square | 2.9 | 3 | 3.1 | 3.2 | 3.3 | 3.4 | 3.5 |

Length of Box | 14.2 | 14 | 13.8 | 13.6 | 13.4 | 13.2 | 13 |

Width of Box | 14.2 | 14 | 13.8 | 13.6 | 13.4 | 13.2 | 13 |

Height of Box | 2.9 | 3 | 3.1 | 3.2 | 3.3 | 3.4 | 3.5 |

Volume of Box | 584.8 | 588 | 590.4 | 591.9 | 592.6 | 592.4 | 591.5 |

Rectangular Boxes

Like what I did with the square cards, I will try cutting different sizes of squares from different sizes of rectangle to find out the square that would create the largest volume.

From looking at the formula, Volume= Length x Width x Height, and inserting the variables I have, I realized I could make a quicker formula to find out the volume. Since finding the length means subtracting two of side length of the square cut from the length of the card, this can be simplified into the formula, Length= a-2x (a=length of card, x=side length of square cut). The same method is applied to the width, Width= b-2x (b=width of the card). Because height of the box is equal to the length of square cut, Height= x.

By substituting these formula’s into the general formula, V=L x W x H, we would be able to calculate the formula just by knowing the length of square cut and the length and width of the sheet of card used, without the trouble of subtracting length and width all the time. The formula would be:

Volume = x (a-2x)(b-2x).

Using this formula, I have calculated the results below.

10x12 Rectangle

Side Length of Square | 1 | 2 | 3 | 4 |

Volume | 80 | 96 | 72 | 32 |

Side Length of Square | 1.5 | 1.6 | 1.7 | 1.8 | 1.9 | 2 | 2.1 |

Volume | 94.5 | 95.7 | 96.5 | 96.8 | 96.6 | 96 | 95 |

10x14 Rectangle

Side Length of Square | 1 | 2 | 3 | 4 |

Volume | 96 | 120 | 64 | 48 |

Conclusion

Verification

I substituted the general formula with some results I calculated previously to prove this formula is accurate.

10x12

𝑥

Since it is impossible to two 5.5cm squares off of a 10cm wide card, the answer is 1.8. To make the largest volume for an open box with piece of card 10cm by 12cm, a square of around 1.8cm by 1.8cm is to be cut from each corner. This is verified above in the results table.

20x20

If four 10cm by 10cm squares is be removed from a 20cm by 20cm square card, there will be nothing left. Therefore, 3.3cm is the correct answer which is proven to be true in the results table above.

Conclusion

The general formula I obtained above could be used to find the size of the side length needed to be removed from the sheet of card in order to create an open box a volume as large as possible. I have made sure this formula was accurate trying different sizes of card and all matches my results. While there are many other ways to find the answer, x, I believe by using calculus and quadratics, this formula could give the most accurate and precise answer.

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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## Here's what a teacher thought of this essay

An excellent piece of work which demonstrates good understanding of volume of cuboids, drawing graphs, use of the formula for solving quadratic equations and substitution into algebraic expressions. Also a good taste of calculus for extension work. Five stars.

Marked by teacher Mick Macve 18/03/2012