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  • Level: GCSE
  • Subject: Maths
  • Word count: 2481

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:

image00.png

        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.

image01.png

...read more.

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:

image12.png

This I can include in the formula for area:

image13.png

Replace c with formula:

image14.png

Multiply out the brackets:

image15.png

Multiply by x/4.732071:

image16.png

Put formula over a common denominator:

image17.png

Simplify:

image18.png

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.

image19.png

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

image20.png

        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

...read more.

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:

image21.png

This I can include in the formula for area:

image23.png

Replace c with formula:

image24.png

Multiply out the brackets:

image25.png

Multiply by x/4.732071:

image26.png

Put formula over a common denominator:

image27.png

Simplify:

image28.png

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

...read more.

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