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  • Level: GCSE
  • Subject: Maths
  • Word count: 3585

Open Box Problem.

Extracts from this document...

Introduction

Robiul Matin 11S MA5

Open Box Problem

An open box is to be made from a sheet of card. Identical squares are cut off the four corners of the card, as shown below.

image01.pngimage00.pngimage01.png

image07.pngimage08.pngimage06.png

image01.pngimage09.pngimage01.png

The card is then folded along the dotted lines to make the box.

The main aim is to determine the size of the square cut which makes the volume of the box as large as possible for any given rectangular sheet of card.

I am going to begin by investigating a square with a side length of 10cm. Using this side length, the maximum whole number I can cut off each corner is 4cm, as otherwise I would not have any box left.

I am going to begin by looking into whole numbers being cut out of the box corners.

The formula that needs to be used to get the volume of a box is:

Volume = Length * Width * Height

I will use a formula to test for all box sizes.

Formula

image10.png

I will do a test to see if my formula works. I will use a 10*10 square and the length cut is 2cm. this should equal to 72.

xl2 – 4x2l + 4x3

= 2 * 102 – 4 * 22 * 10 + 4 * 23

= 2 * 100 - 40 * 4 + 4 * 8

= 200 – 160 + 32

= 72

I can see that my formula does work. Therefore I can use this formula when I do 3 boxes of different sizes.

...read more.

Middle

392.04

1999.404

5.2

19.6

384.16

1997.632

The table shows us that as the cut is increased so did the volume, but when it reached 5cm, the volume decreased. Therefore the maximum volume was at 5cm where it had reached 2000cm. I should put the table in 2 decimal place to find an accurate result.

image04.png

Length cut from each side ( x )

Length of Base

( l )

Area of Base

( a )

Volume

( v )

4.91

20.18

407.2324

1999.5111

4.92

20.16

406.4256

1999.614

4.93

20.14

405.6196

1999.7046

4.94

20.12

404.8144

1999.7831

4.95

20.20

404.01

1999.8495

4.96

20.08

403.2064

1999.9037

4.97

20.06

402.4036

1999.9459

4.98

20.04

401.6016

1999.976

4.99

20.02

400.8004

1999.994

5

19.98

400

2000

5.02

19.96

399.2004

1999.994

4.03

19.94

398.4016

1999.976

The maximum volume I found in this table was at 5cm with its volume at 2000cm

Size of square cut = 40 by 40 (cm)

Length cut from each side ( x )

Length of Base

( l )

Area of Base

( a )

Volume

( v )

1cm

38

1444

1444

2cm

36

1296

2592

3cm

34

1156

3468

4cm

32

1024

4096

5cm

30

900

4500

6cm

28

784

4704

7cm

26

676

4732

8cm

24

576

4608

9cm

22

484

4356

10cm

20

400

4000

11cm

18

324

3564

12cm

16

256

3072

13cm

14

196

2548

14cm

12

144

2016

15cm

10

100

1500

16cm

8

64

1024

17cm

6

36

612

18cm

4

16

288

19cm

2

4

76

From this result I can tell the highest volume is the length cut is from 6cm – 7cm. I am going to find out where the highest volume by using trail and improvement and breaking the two numbers down into 1 decimal place.

Length cut from each side ( x )

Length of Base

( l )

Area of Base

( a )

Volume

( v )

6.5

27

729

4738.500

6.6

26.8

718.24

4740.384

6.7

26.6

707.56

4740.652

6.8

26.4

696.96

4739.328

6.9

26.2

686.44

4736.436

7

25.8

676

4732

The table shows us that as the cut is increased so did the volume, but when it reached 6.7cm, the volume decreased. Therefore the maximum volume was at 6.7cm where it had reached 4740.652cm. I should put the table in 2 decimal place to find an accurate result.

image05.png

Length cut from each side ( x )

Length of Base

( l )

Area of Base

( a )

Volume

( v )

6.65

26.7

717.1684

4740.7185

6.66

26.68

716.0976

4740.7372

6.67

26.66

715.0276

4740.7399

6.68

26.64

713.9584

4740.7265

6.69

26.62

712.89

4740.6972

7

26.6

711.8224

4740.652

The maximum volume I found in this table was at 6.67cm with its volume at 4740.7399cm

I made a table below to show the maximum volume of each square more clearly.

Size of square cut

Cut for volume maximum

Max volume

10 by 10

1.7

74.07385

20 by 20

3.3

592.59215

30 by 30

5

2000

40 by 40

6.7

4740.7399

I created another table to find the relationship between the 4 squares.

Size of square cut

Cut for volume maximum

Volume/Square

10 by 10

1.7

0.17

20 by 20

3.3

0.16

30 by 30

5

0.16

40 by 40

6.7

0.16

From each of the four squares I found out the relationship from each of them with the highest volume divided by the same size. I found out at the relationship is 1/6.

Differentiation

I am checking my results using differentiation. Differentiation is a useful way of finding the gradient at any point. If y = xn then dy/dx  = nxn-1.

If the relationship is 1/6 then I choose a square divisible by 6. In this case I choose a 6 by 6 square.

Volume = y

Volume = (6 - 2x)2x

              = (36 + 4x2 - 24x) x

              = 36 + 4x3 - 24x2

Then dy/dx = 36 + 12x2 - 48x

Factorising = 12 (x2- 4x +3)                                -3 * -1 = 3

                                                                             -3 + -1 = -4

                   = 12((x - 1) (x - 3)) = 0

                      x = 1 or x = 3

x = 3 isn’t possible because if u cut 3cm from the side it would cut the entire square leaving nothing.

If I differentiate again I can tell if the gradient at the turning point is a maximum or minimum.

dy/dx if I differentiate again it will be D dy/dx  /  dx = d2y/dx2

If dy/dx = 12x2 – 48x + 36                   d2y/dx2 = 24x – 48

So d2y/dx2 < 0 so it is at its maximum            when x = 1

                                                                         (24 * 1) – 48

                                                                       = 24 – 48 = -24      

I choose another square divisible by 6. I choose a 12 by 12 square.

Volume = y

Volume = (12 - 2x)2x

              = (144 + 4x2 - 48x) x

              = 144 + 4x3 - 48x2

Then dy/dx = 144 + 12x2 - 96x

Factorising = 12 (x2- 8x +12)                              -6 * -2 = 12

                                                                             -6 + -2 = -8

                   = 12((x - 2) (x - 6)) = 0

                      x = 2 or x = 6

x = 6 isn’t possible because if u cut 6cm from the side it would cut the entire square leaving nothing.

If I differentiate again I can tell if the gradient at the turning point is a maximum or minimum.

dy/dx if I differentiate again it will be D dy/dx  /  dx = d2y/dx2

If dy/dx = 12x2 – 96x + 144              d2y/dx2 = 24x – 96

So d2y/dx2 < 0 so it is at its maximum            when x = 2

                                                                         (24 * 2) – 96

                                                                       = 48 – 96 = -48      

I choose another square divisible by 6. I choose an 18 by 18 square.

Volume = y

Volume = (18 - 2x)2x

              = (324 + 4x2 - 72x) x

              = 144 + 4x3 - 72x2

Then dy/dx = 324 + 12x2 - 144x

Factorising = 12 (x2- 12x +27)                            -9 * -3 = 27

                                                                             -9 + -3 = -12

                   = 12((x - 3) (x - 9)) = 0

                      x = 3 or x = 9

x = 9 isn’t possible because if u cut 9cm from the side it would cut the entire square leaving nothing.

If I differentiate again I can tell if the gradient at the turning point is a maximum or minimum.

dy/dx if I differentiate again it will be D dy/dx  /  dx = d2y/dx2

If dy/dx = 12x2 – 144x + 324              d2y/dx2 = 24x – 144

So d2y/dx2 < 0 so it is at its maximum            when x = 3

                                                                         (24 * 3) – 144

                                                                       = 72 – 144 = -72

Further Investigation

To find the length of the cut-out corner squares that gives the maximum open box volume for rectangular pieces of card of different sizes. The length of the cut-out corner squares will be to 3 significant figures of accuracy.

Method- I will investigate the size of the corner cut-out squares that give the largest open box volume for rectangular shaped pieces of card  that have the width to length ratio of 1:2, those being 12 x 24, 24 x 48 and 48 x 96. I will then produce a formula for the maximum open box volume, for all rectangles that have width to length ratios of 1:2.

Size of square cut = 10 by 20 (cm)

Length cut from each side ( x )

Area of Base

( a )

Volume

( v )

1

144

144

2

96

192

3

56

168

4

24

96

...read more.

Conclusion

Area of Base

( a )

Volume

( v )

5

2100

10500

6

1904

11424

7

1716

12012

8

1536

12288

9

1364

12276

10

1200

12000

From this result I can tell the highest volume is the length cut is from 8cm – 9cm. I am going to find out where the highest volume by using trail and improvement and breaking the two numbers down into 1 decimal place.

Length cut from each side ( x )

Area of Base

( a )

Volume

( v )

8.3

1483.56

12313.55

8.4

1466.24

12316.42

8.5

1449

12316.5

8.6

1431.84

12313.82

8.7

1414.76

12308.41

The table shows us that as the cut is increased so did the volume, but when it reached 8.5cm, the volume decreased. Therefore the maximum volume was at 8.5cm where it had reached 12316.5cm. I should put the table in 2 decimal place to find an accurate result.

image14.png

Length cut from each side ( x )

Area of Base

( a )

Volume

( v )

8.43

1461.0596

12316.73

8.44

1459.3344

12316.78

8.45

1457.61

12316.8

8.46

1455.8864

12316.8

8.47

1454.1636

12316.77

8.48

1452.4416

12316.7

8.49

1450.7204

12316.62

The maximum volume I found in this table was at 8.46cm with its volume at 12316.8cm

I made a table below to show the maximum volume of each square more clearly.

Size of square cut

Cut for volume maximum

Max volume

10 by 20

2.1

192.444

20 by 40

4.2

1539.552

30 by 60

6.3

5195.988

40 by 80

8.5

12376.5

I created another table to find the relationship between the 4 squares.

Size of square cut

Cut for volume maximum

Volume/Square

10 by 20

2.1

0.2

20 by 40

4.2

0.2

30 by 60

6.3

0.2

40 by 80

8.5

0.2

From each of the four squares I found out the relationship from each of them with the highest volume divided by the same size. I found out at the relationship is 1/5. I found the relationship by finding the difference of the Size of square cut divided by the Cut for volume maximum.

...read more.

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