# The Open Box Problem

Extracts from this document...

Introduction

³The Open Box Problem

In this investigation, I am going to be given a square sheet of card. I will then cut four squares from each corner of this card; each of these squares will be identical in size. With the remaining shape, I will then fold the sides to make a box. The task is to work out what size the cut will need to be to make the box with the biggest volume for any size sheet of card.

24cm sheet of card

I am going to start with a square sheet of card of length 24cm. I am going to cut a square off each corner. I am then going to work out the volume of each box using the formula

Volume = L x B x H

L = length

B = base

H = height

Results Table

Height | Length | Width | Volume |

1cm | 22cm | 22cm | 22 x 22 x 1 = 484cm³ |

2cm | 20cm | 20cm | 20 x 20 x 2 = 800cm³ |

3cm | 18cm | 18cm | 18 x 18 x 3 = 972cm³ |

4cm | 16cm | 16cm | 16 x 16 x 4 = 1,024cm³ |

5cm | 14cm | 14cm | 14 x 14 x 5 = 980cm³ |

6cm | 12cm | 12cm | 12 x 12 x 6 = 864cm³ |

7cm | 10cm | 10cm | 10 x 10 x 7 = 700cm³ |

8cm | 8cm | 8cm | 8 x 8 x 8 = 512cm³ |

9cm | 6cm | 6cm | 6 x 6 x 9 = 324cm³ |

10cm | 4cm | 4cm | 4 x 4 x 10 = 160cm³ |

11cm | 2cm | 2cm | 2 x 2 x 11 = 44cm³ |

I am now going to draw a graph. This will help me to work out whether the highest number is a decimal or a whole number.

Because this graph is not symmetrical in shape I cannot be certain that the top point of the graph represents the biggest possible value. The biggest value may actually be just before or just after the top point of the graph.

I am now going to test whether the highest point represents the biggest value by taking the numbers immediately above and below the supposed biggest value.

Middle

6 x 6 x 15 = 540cm³

16cm

4cm

4cm

4 x 4 x 16 = 256cm³

17cm

2cm

2cm

2 x 2 x 17 = 68cm³

I am now going to form my data into a graph. This will help me too see whether the highest number is a decimal or a whole number

Because I cannot be sure whether or not the above graph is symmetrical, I am going to test for this by taking the numbers immediately above and below my highest point. If both of these numbers produce a volume of less than 3,456cm³ then 6cm will be the required cut-out size to achieve the biggest value.

Height | Length | Width | Volume |

5.9cm | 24.2cm | 24.2cm | 24.2 x 24.2 x 5.9 = 3,455.276cm³ |

6.1cm | 23.8cm | 23.8cm | 23.8 x 23.8 x 6.1 = 3,455.284cm³ |

This test has proved that cut-outs of length 6cm will give the box with the biggest possible volume. I have proved this because when I took cut-outs of length 5.9cm and 6.1cm the volumes achieved were smaller, so therefore 6cm must be the optimum value.

Conclusion

This result has proved my hypothesis, in that to quickly work out what size the square cut-outs will need to be in order to make a box with the biggest volume you take the size of the sheet of square card you are using and divide it by 6. This can be simply put into the algebraic formula:

X = L

6

X represents that size of the square cut-outs while L represents the length of the square sheet of card.

36cm using algebra

I am now going to prove that this formula is accurate by using it to work out the required cut-out size for a piece of card of length 36cm. I know that the answer should come out at 6cm, so if it does then this formula must be correct and accurate.

V = L x B x H

= x(36 – 2x)(36 – 2x)

= x(1296 – 72x – 72x + 4x²)

= x(4x² - 144x +1296)

V = 4x³ - 144x² + 1296x

I am now going to find the gradient through the use of differentiation. At the biggest value the gradient should be 0.

Gradient = 12x² - 288x + 1296

= 12x² - 288x + 1296 = 0

12x² - 288x + 1296 = 0

12

x² - 24x + 108 = 0

(x – 18)(x – 6) = 0

Either x – 18 = 0 or x – 6 = 0

+18 +18 +6 +6

x = 18 x = 6

x cannot be equal to 18 because if both the cut-outs were 18cm then there would be no cart left so x must be equal to 6cm.

This result has shown that my formula is correct. However, to make sure I am also going to test the formula on square sheets of card 24cm and 30cm in length.

24cm using algebra

I am now going to test my formula using a sheet of card 24cm square.

V = L x B x H

= x(24 – 2x)(24 – 2x)

= x(576 – 48x – 48x +4x²)

= x(4x² - 96x + 576)

V = 4x³ - 96x² + 576x

I am now going to find the gradient using differentiation. The gradient should be 0 at the biggest value.

Gradient = 12x² - 192x + 576

= 12x² - 192x + 576 = 0

12x² - 192x + 576 = 0

12

x² - 16x + 48 = 0

(x – 4)(x – 12) = 0

Either x – 12 = 0 or x – 4 = 0

+12 +12 +4 +4

x = 12 x = 4

12 cannot be equal to x because two cut-outs of 12cm would mean there was no card left so therefore x must be equal to 4cm.

30cm using algebra

I am now going to test my formula using a sheet of card 30cm square.

V = L x B x H

= x(30 – 2x)(30 – 2x)

= x(900 – 60x – 60x + 4x²)

= x(4x² - 120x + 900)

V = 4x³ - 120x² +900x

I am now going to differentiate to find the gradient because the gradient will be 0 at the biggest volume.

Gradient = 12x² - 240x + 900

= 12x² - 240x + 900 = 0

12x² - 240x + 900 = 0

12

x² - 20x + 75 = 0

(x – 5)(x – 15) = 0

Either x – 5 = 0 or x – 15 = 0

+5 +5 +15 +15

x = 5 x = 15

x cannot be equal to 15cm because two cut-outs of 15cm would mean there was no card left so x must be equal to 5.

Calculus and Algebra

I am now going to prove, through the use of calculus and algebra, that in the formula x = L

6

That x stands for the size of each of the four cu-outs and that L stands for the length of one side of the card.

V = L x B x H

= x(L – 2x)(L – 2x)

= x(L² - 2Lx – 2Lx + 4x²)

= x(L² - 4Lx + 4x²)

V = 4x³ - 4Lx² + L²x

At the biggest volume the gradient is always equal to 0 so I am going to find the gradient using differentiation and put it equal to 0.

Gradient = 12x² - 8Lx + L²

= 12x² - 8Lx + L² = 0

I am now going to factorise this formula.

-12Lx – Lx = -13Lx

-3Lx – 4Lx = -7Lx

-6Lx – 2Lx = -8Lx

(6x – L)(2x – L) = 0

Either 6x – L = 0 or 2x – L = 0

+L +L +L +L

6x = L2x = L

6 6 2 2

x = L x = L

6 2

L

2 cannot be right because if you cut off half the card twice then there will be no card left so L must be the correct rule.

6

I have proved using algebra and calculus that the cut-out required to achieve the biggest volume is the length of one side of the card divided by 6.

Rectangle

Length 2 times Width

I am now going to investigate the size of the cut-out square which will make an open box of the biggest volume for a rectangular piece of card.

I am first going to use a rectangle where the length is twice the size of the width. This rectangle has a length of 20 cm and a width of 10cm.

V = L x B x H

= x(20 – 2x)(10 – 2x)

= x(200 – 40x – 20x +4x²)

= x(200 – 60x +4x²)

V = 4x³ - 60x² +200x

I am now going to differentiate to find the gradient which will be 0 at the biggest volume.

Gradient = 12x² - 120x +200

= 12x² - 120x + 200 = 0

12x² - 120x +200 = 0

4

3x² - 30x +50 = 0

-25x – 6x = -31x

-75x – 2x = -77x

-50x – 3x = -53x

-x – 150x = -151x

-30x – 5x = -35x

-15x – 10x = -25x

This doesn’t factorise normally so I will have to make use of the quadratic formula to factorise it instead.

x = -b b² - 4ac

2a

a = 3

b = -30

c = 50

4ac = 600

x = --30 900 – 600

6

x = 30 300

6

Either x = 7.89 or x = 2.11

You cannot cut 7.89cm out twice because there isn’t enough card so the answer must be 2.11cm.

30cm by 15cm

I am now going to use a rectangle which has a length of 30cm and a width of 15cm.

V = L x B x H

= x(30 - 2x)(15 – 2x)

= x(450 – 60x – 30x + 4x²)

= x(4x² - 90x + 450)

V = 4x³ - 90x² + 450x

I am now going to differentiate to find the gradient which will be 0 at the biggest volume.

Gradient = 12x² - 180x + 450

= 12x² - 180x + 450 = 0

12x² - 180x + 450 = 0

6

2x² - 30x +75 = 0

-50x – 3x = -53x

-6x – 25x = -31x

This formula will not factorise in the normal way so I will have to use the quadratic formula to factorise it instead.

x = -b b² - 4ac

2a

a = 2

b = -30

c = 75

4ac = 600

x = --30 900 – 600

4

x = 30 300

4

Either x = 11.83 or x = 3.17

You cannot cut out 11.83cm twice from this sheet of card because there would be no card left so the answer must be 3.17cm.

10cm by 5cm

I am now going to use a rectangle that has a length of 10cm and a width of 5cm.

V = L x B x H

= x(10 – 2x)(5 – 2x)

= x(50 – 2x – 10x + 4x²)

= x(4x² - 30x + 50)

V = 4x³ - 30x² + 50x

I am now going to differentiate to find the gradient which will be 0 at the biggest volume.

Gradient = 12x² - 60x +50

= 12x² - 60x +50 = 0

12x² - 60x + 50 = 0

2

6x² - 30x + 25 = 0

This formula will not factorise normally so I will have to use the quadratic formula to factorise it instead.

x = -b b² - 4ac

2a

a = 6

b = -30

c = 25

4ac = 600

x = --30 900 – 600

12

X = 30 300

12

Either x = 3.94 or x = 1.06

x cannot be equal to 3.94cm because two cut-outs of that size would mean that there was no card left so x must be equal to 1.06cm.

I am now going to tabulate my results to see if there is any sort of pattern beginning to emerge.

Width of Rectangle | Size of Cut-out |

5cm | 1.06cm |

10cm | 2.11cm |

15cm | 3.17cm |

Conclusion

x = w

4.752

I then considered a rectangular piece of card which had a length 3 times the size of the width. For this piece of card I came up with the formula:

x = w

4.431

After that I considered a rectangular piece of card which had a length 4 times bigger than the width. For this particular piece of card I came up with the formula:

x = w

4.303

Finally, I investigated a rectangular piece of card which had a length 5 times bigger than the width. For this piece of card I came up with the formula:

x = w

4.233

I then came up with the hypothesis that the denominator, that is the number on the bottom of the formula, will never be less than four no matter what the dimensions of the rectangle are. I decided to test this hypothesis by taking a rectangle which had a length 145 times bigger than the width. This investigation resulted in the formula:

x = w

4.007

This result proved beyond reasonable doubt that my hypothesis was correct and accurate.

I then decided to look at rectangles where the length was less than 2 times the size of the width to investigate what happened to the denominator.

The first rectangle I decided to investigate was one where the length was 1.5 times bigger than the width. The formula this investigation resulted in was:

x = w

5.097

I then investigated a rectangle where the length was 1.2 times bigger than the width. This resulted in the formula:

x = w

5.523

Finally, I investigated a rectangle where the length was 1.1 times bigger than the width. This investigation resulted in the formula:

x = w

5.695

This has led me to conclude that as the ratio of length to width gets smaller, the rectangle becomes closer and closer in its dimensions to a square. Therefore, the denominator will become closer and closer to 6, which is the denominator for a square.

This student written piece of work is one of many that can be found in our GCSE Comparing length of words in newspapers section.

## Found what you're looking for?

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