# Investigate the volume of an open box constructed by one piece of rectangular card that has all four corners having had squares cut out of them.

Extracts from this document...

Introduction

Introduction

The aim of this coursework is to investigate the volume of an open box constructed by one piece of rectangular card that has all four corners having had squares cut out of them. Firstly I will be studying the volume whilst changing the side of one length of the cut out square and the size of the original rectangle card. After I have investigated this relationship I will try to find out the formula for finding the cut size to get the largest volume for any specified original card size.

Square card size

I am going to begin by investigating a square card because this will give me a basic formula which I can elaborate on. I will start with a round number of 20cm for the length. This means that the maximum cut out square length I can cut out will be 9cm else I will have no box left.

The formula for the volume of any box is as follows:

Below, there is a diagram to explain where all these figures come from.

I have also included in this diagram the labels c, x and y, these show the cut out size and the original length and width of the card, I will now need to show the values of the width, length and height in terms of c, x and y.

Middle

47.2

5195.77600000000

30

6.33

17.34

47.34

5196.14254800000

30

6.34

17.32

47.32

5196.15241600000

30

6.35

17.3

47.3

5196.14150000000

30

6.339

17.322

47.322

5196.152364876

30

6.34

17.32

47.32

5196.152416000

30

6.341

17.318

47.318

5196.152259284

30

6.3396

17.3208

47.3208

5196.15242049254

30

6.3397

17.3206

47.3206

5196.15242248709

30

6.3398

17.3204

47.3204

5196.15242240317

The value of c for the highest volume is 6.3397 only up to 4 decimal places.

x | c | x/c |

20 | 4.2265 | 4.732048 |

30 | 6.3397 | 4.732085 |

I noticed whilst evaluating the results that if you divide x by c the values are very similar:

This is a very strong relationship; I can therefore predict that if I construct a similar table for another number, if you divide that number by 4.732, it will give its largest volume. I will use number 17: 17/4.732=3.59 therefore I predict that the largest volume will be made with a cut size of approx 3.59.

x=17

x | c | x-2c | 2x-2c | c(x-2c)(2x-2c) |

17 | 3 | 11 | 28 | 924 |

17 | 4 | 9 | 26 | 936 |

17 | 5 | 7 | 24 | 840 |

17 | 3.5 | 10 | 27 | 945 |

17 | 3.6 | 9.8 | 26.8 | 945.504 |

17 | 3.7 | 9.6 | 26.6 | 944.832 |

17 | 3.58 | 9.84 | 26.84 | 945.498048 |

17 | 3.59 | 9.82 | 26.82 | 945.506916 |

17 | 3.6 | 9.8 | 26.8 | 945.504000 |

17 | 3.592 | 9.816 | 26.816 | 945.507274752 |

17 | 3.593 | 9.814 | 26.814 | 945.507277428 |

17 | 3.594 | 9.812 | 26.812 | 945.507162336 |

17 | 3.5924 | 9.8152 | 26.8152 | 945.50728995610 |

17 | 3.5925 | 9.815 | 26.815 | 945.50729081250 |

17 | 3.5926 | 9.8148 | 26.8148 | 945.50729049110 |

The value of c for the highest volume is 3.5925 only up to 4 decimal places.

x | c | x/c |

20 | 4.2265 | 4.732048 |

30 | 6.3397 | 4.732085 |

17 | 3.5925 | 4.732081 |

Average = | 4.732071 |

My prediction proved to be correct and I have added a third row to the data and the average of the three results gives me a formula of the cut size for maximum volume:

This I can include in the formula for area:

Replace c with formula:

Multiply out the brackets:

Multiply by x/4.732071:

Put formula over a common denominator:

Simplify:

This then is the formula for the maximum volume of any open box constructed from a rectangle that’s length is twice as large as its width.

Ratio 1:3

The following formulae shows the values of the width, length and height in terms of c, x and y for the ratio of 1:3 and this time I know that the length is exactly three times the width.

Therefore I can replace these sub formulae into the first formula.

I have constructed a table in excel where I can input the data for the cut size and the original width of the card and it will calculate the volume, I will use it to find the largest volume through the cut size.

x=20

x | c | x-2c | 3x-2c | c(x-2c)(3x-2c) |

20 | 4 | 12 | 52 | 2496 |

20 | 5 | 10 | 50 | 2500 |

20 | 6 | 8 | 48 | 2304 |

20 | 4.4 | 11.2 | 51.2 | 2523.136 |

20 | 4.5 | 11 | 51 | 2524.500 |

20 | 4.6 | 10.8 | 50.8 | 2523.744 |

20 | 4.50 | 11 | 51 | 2524.500000 |

20 | 4.51 | 10.98 | 50.98 | 2524.519404 |

20 | 4.52 | 10.96 | 50.96 | 2524.517632 |

20 | 4.513 | 10.974 | 50.974 | 2524.521094788 |

20 | 4.514 | 10.972 | 50.972 | 2524.521234976 |

20 | 4.515 | 10.97 | 50.97 | 2524.521163500 |

20 | 4.5141 | 10.9718 | 50.9718 | 2524.5212373529 |

20 | 4.5142 | 10.9716 | 50.9716 | 2524.5212376132 |

20 | 4.5143 | 10.9714 | 50.9714 | 2524.5212357568 |

20 | 4.51415 | 10.9717 | 50.9717 | 2524.52123774759 |

20 | 4.51416 | 10.97168 | 50.97168 | 2524.52123776304 |

20 | 4.51417 | 10.97166 | 50.97166 | 2524.52123775731 |

To achieve even more accurate results I managed to go to fifth decimal place where the value for c was, yet even this is not totally accurate as there will be a much larger number of decimals in the c column to get the maximum number for V.

x | c | x/c |

20 | 4.51416 | 4.430503128 |

Conclusion

c

x-2c

2x-2c

c(x-2c)(2x-2c)

30

6

18

78

8424

30

7

16

76

8512

30

8

14

74

8288

30

6.7

16.6

76.6

8519.452

30

6.8

16.4

76.4

8520.128

30

6.9

16.2

76.2

8517.636

30

6.76

16.48

76.48

8520.2391

30

6.77

16.46

76.46

8520.2589

30

6.78

16.44

76.44

8520.2470

30

6.77

16.46

76.46

8520.2589320

30

6.771

16.458

76.458

8520.2591680

30

6.772

16.456

76.456

8520.2590866

30

6.7711

16.4578

76.4578

8520.2591741857

30

6.7712

16.4576

76.4576

8520.2591771525

30

6.7713

16.4574

76.4574

8520.2591769444

30

6.77123

16.45754

76.45754

8520.25917742344

30

6.77124

16.45752

76.45752

8520.25917745025

30

6.77125

16.4575

76.4575

8520.25917744531

30

6.771242

16.457516

76.457516

8520.25917745180

30

6.771243

16.457514

76.457514

8520.25917745210

30

6.771244

16.457512

76.457512

8520.25917745208

30

6.7712433

16.4575134

76.4575134

8520.25917745213

30

6.7712434

16.4575132

76.4575132

8520.25917745214

30

6.7712435

16.4575130

76.4575130

8520.25917745213

The value of c for the highest volume is 6.7712435 only up to 7 decimal places.

x | c | x/c |

20 | 4.51416 | 4.430503128 |

30 | 6.7712434 | 4.430500903 |

Proving my prediction correct, the value of x divided by c the values are very similar:

As before, this is a very strong relationship; I can therefore predict that if I construct a similar table for another number, if you divide that number by 4.4305009, it will give its largest volume. I will use number 17:

17/4.4305009=3.837036 therefore I predict that the largest volume will be made with a cut size of approx 3.837036.

x=17

x | c | x-2c | 2x-2c | c(x-2c)(2x-2c) |

17 | 3 | 11 | 45 | 1485 |

17 | 4 | 9 | 43 | 1548 |

17 | 5 | 7 | 41 | 1435 |

17 | 3.7 | 9.6 | 43.6 | 1548.672 |

17 | 3.8 | 9.4 | 43.4 | 1550.248 |

17 | 3.9 | 9.2 | 43.2 | 1550.016 |

17 | 3.83 | 9.34 | 43.34 | 1550.3671 |

17 | 3.84 | 9.32 | 43.32 | 1550.3708 |

17 | 3.85 | 9.3 | 43.3 | 1550.3565 |

17 | 3.836 | 9.328 | 43.328 | 1550.3715082 |

17 | 3.837 | 9.326 | 43.326 | 1550.3716050 |

17 | 3.838 | 9.324 | 43.324 | 1550.3715219 |

17 | 3.8369 | 9.3262 | 43.3262 | 1550.3716034296 |

17 | 3.837 | 9.326 | 43.326 | 1550.3716050120 |

17 | 3.8371 | 9.3258 | 43.3258 | 1550.3716047952 |

17 | 3.83703 | 9.32594 | 43.32594 | 1550.37160513588 |

17 | 3.83704 | 9.32592 | 43.32592 | 1550.37160514119 |

17 | 3.83705 | 9.3259 | 43.3259 | 1550.37160512851 |

17 | 3.837037 | 9.325926 | 43.325926 | 1550.37160514149 |

17 | 3.837038 | 9.325924 | 43.325924 | 1550.37160514157 |

17 | 3.837039 | 9.325922 | 43.325922 | 1550.37160514147 |

17 | 3.8370378 | 9.3259244 | 43.3259244 | 1550.37160514157000 |

17 | 3.8370379 | 9.3259242 | 43.3259242 | 1550.37160514157000 |

17 | 3.837038 | 9.3259240 | 43.3259240 | 1550.37160514157000 |

The value of c for the highest volume is 3.8370379 up to 7 decimal places.

x | c | x/c |

20 | 3.8370379 | 4.430500934 |

30 | 4.5141600 | 4.430503128 |

17 | 6.7712434 | 4.430500903 |

Average = | 4.430501655 |

Revised Formulae

My prediction proved to be correct and I have added a third row to the data and the average of the three results gives me a formula of the cut size for maximum volume:

This I can include in the formula for area:

Replace c with formula:

Multiply out the brackets:

Multiply by x/4.732071:

Put formula over a common denominator:

Simplify:

I now have the formulae for both the ratios 1:2 and 1:3, with examples for each.

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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