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  • Level: GCSE
  • Subject: Maths
  • Word count: 3314

Open Box Problem.

Extracts from this document...

Introduction

Open Box Problem

Introduction

The open box problem is about an open box, which is to be made from a sheet of card. Identical sized squares are cut off the four corners as shown in the diagram below.

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

Squares

I am going to begin the open box investigation by investigating a square with the width and length of 24cm. The formula that needs to be used to get the volume of a box is:

Volume = Length x Width x Height

If I am to use a square of side length 24cm, then I can calculate the side lengths minus the cut out squares using the following equation.

Volume = x(24-2x)(24-2x)

x = is the square that is to be cut out.

24 = is the size of the length and the width

For this square I will make a table to show the volume of different cuts. I will use whole numbers for the different cuts to find roughly where the maximum volume occurs.

24cm by 24cm Square

For this square I will make a table to show the volume of different cuts. I will use whole numbers for the different cuts to find roughly where the maximum volume occurs.

Cut x

L (24-2x)

W (24-2x)

Volume

1

22

22

484

2

20

20

800

3

18

18

972

4

16

16

1024

5

14

14

980

6

12

12

864

7

10

10

700

8

8

8

512

9

6

6

324

10

4

4

160

11

2

2

44

12

0

0

0

...read more.

Middle

21

0

0

0

Volume = x(42-2x)(42-2x)

As you can see the table above shows that the cut of x, which will give the maximum volume for the open box is 7. This proves my prediction right again using the L/6 formula.

The scatter diagram below also shows that when the cut of x is 7, the open box will have its highest volume. Notice yet again the shape of the scatter diagram is the same.

image03.png

48cm by 48cm Square

I will now be investigating a 48cm by 48cm square. This is the last square that I will be investigating. I will use the formula L/6 again to predict the cut of x, which will give this open box its maximum volume:

Length/width of square            48

        4                                 6

                                             =      8

Therefore according to this equation, the value of the cut of x, which will give this open box its maximum volume, is 8cm.

Cut x

L (48-2x)

W (48-2x)

Volume

1

46

46

2116

2

44

44

3872

3

42

42

5292

4

40

40

6400

5

38

38

7220

6

36

36

7776

7

34

34

8092

8

32

32

8192

9

30

30

8100

10

28

28

7840

11

26

26

7436

12

24

24

6912

13

22

22

6292

14

20

20

5600

15

18

18

4860

16

16

16

4096

17

14

14

3332

18

12

12

2592

19

10

10

1900

20

8

8

1280

21

6

6

756

22

4

4

352

23

2

2

92

24

0

0

0

Volume = x(48-2x)(48-2x)

The table above proves that the maximum volume of the open box is reached when the cut x 8cm.

The scatter diagram below also shows that the maximum volume of the open box is reached when the cut of x is 8cm. Notice yet again the shape of the graph is the same.

image04.png

The reason I am went through squares with constant measurements (24cm, 30cm, 36cm, 42cm, 48cm) is because I am trying to prove that L/6 will equal the cut of x which will give the open box its maximum volume, and these numbers all easily divide by 6.

Rectangles

In this part of my coursework, I will be investigating the measurements of cut x to give a maximum volume for an open box made out of a rectangle. I will be investigating rectangles in the ratio of:

  • 1:2
  • 1:3
  • 1:4

The reason I am investigating rectangles with ratios of length by width is to be systematic. For example I might investigate a rectangle in the ratio of 1:2 with the these measurements:

This therefore means to calculate the volume of its open box the formula is now:

Volume = x(20-2x)(40-2x)

Unlike before, the values at the beginning of the brackets aren’t the same. They are the measurements of the length and width, which is in the ratio of 1:2.

Rectangles in the ratio of 1:2

10cm:20cm

Cut x

L (10-2x)

W (20-2x)

Volume

1

8

18

144

2

6

16

192

3

4

14

168

4

2

12

96

5

0

10

0

As you can see, the table above shows the volume of an open box for different measurements for cut x for a rectangle with the measurements of 10cm by 20cm. The table shows that the maximum volume is when the cut x is 2. Notice that this is 10/5, which is the length of the rectangle divided by 5. To calculate the volume for this particular open box, I used the formula that I stated above except I substituted the values in the brackets:

Volume = x(10-2x)(20-2x)

20cm:40cm

Cut x

L (20-2x)

W (40-2x)

Volume

1

18

38

684

2

16

36

1152

3

14

34

1428

4

12

32

1536

5

10

30

1500

6

8

28

1344

7

6

26

1092

8

4

24

768

9

2

22

396

10

0

20

0

The table above shows that the maximum volume occurs when the cut of x is equal to 4. Notice that this is again 20/5 which is the length divided by 5. To calculate the volume for this open box the formula was:

Volume = x(20-2x)(40-2x)

100cm:200cm

I calculated the volume for this open box using this formula:

Volume = x(100-2x)(200-2x)

The reason I am doing numbers with such big ratios is because I want to test the fact that when the rectangle has a ratio of 1:2, the length/5 will then give the cut of x which will give that certain rectangle its maximum volume. I have constructed a table in an excel spreadsheet, just like the ones before to investigate the cut of x which will give this open box its maximum volume. I will not include the table for this, as it will take up to much space.

However for this ratio I have found out that this time, the value of x, which gives this open box its maximum volume is 21, which isn’t 100/5. The answer to 100/5 is 20. Although this is close, as the values gets larger the smaller the divisor seems to be. The divisor is between 4.7 and 4.8 for this ratio.

10,000cm:20,000cm

I calculated the volume for this particular open box using the formula:

Volume = x(10,000-2x)(20,000-2x)

As you can see, the ratio of the rectangle I have chosen is a very large one. This is again because I want to test what the divisor will turn out to be. As you can see from the table on the next page, there is a very significant difference between 10,000/5, which is 2000, and the value of x, which give this open box its maximum volume, which is 2113. The difference is 113, which is quite big.

Cut x

L (10,000-2x)

W (20,000-2x)

Volume

2000

6000

16000

192000000000

2113

5774

15774

192450087588

As you can see I have only put 10,000/5 for cut x, which is 2000 and, 2113, which is the cut of x which give the open box its maximum volume. This is because the original table would have taken up a lot of space and is insignificant data.

As you would have seen in the table above, there is a big difference in the measurement of x, which gives the open box its maximum volume. This time the ratio of 10,000:20,000 is much bigger than 100:200. However, the divisor is still between 4.7 and 4.8. Therefore I have reason to believe that the limit is between 4.7 and 4.8 for rectangles in the ratio of 1:2.

Rectangles in the ratio of 1:3

20cm:60cm

Cut x

L (20-2x)

W (60-2x)

Volume

1

18

58

1044

2

16

56

1792

3

14

54

2268

4

12

52

2496

5

10

50

2500

6

8

48

2304

7

6

46

1932

8

4

44

1408

9

2

42

756

10

0

40

0

...read more.

Conclusion

Volume = x(20-2x)(80-2x)

40cm:160cm

Cut x

L (40-2x)

W (160-2x)

Volume

1

38

158

6004

2

36

156

11232

3

34

154

15708

4

32

152

19456

5

30

150

22500

6

28

148

24864

7

26

146

26572

8

24

144

27648

9

22

142

28116

10

20

140

28000

11

18

138

27324

12

16

136

26112

13

14

134

24388

14

12

132

22176

15

10

130

19500

16

8

128

16384

17

6

126

12852

18

4

124

8928

19

2

122

4636

20

0

120

0

The table above shows that the value of x, which gives the maximum volume for this particular open box, is 9cm. However, this is not the length divided by 4, which is supposed to be 40/4, which is 10. Instead the divisor is now 4.4. The formula used to calculate the volume of the open box for different cuts is:

Volume = x(40-2x)(160-2x)

100cm:400cm

Cut x

L (100-2x)

W (400-2x)

Volume

23

54

354

439668

25

50

350

437500

The table above shows that the cut of x, which gives the maximum volume, is 23cm. The divisor is between 4.3 and 4.4. Notice that the previous rectangle had the same divisor. Therefore, I believe that the limit is around 4.3-4.4. The formula used to calculate the volume for the different square cuts is:

Volume = x(100-2x)(400-2x)

1000cm:4000cm

Cut x

L (1000-2x)

W (4000-2x)

Volume

232

536

3536

439708672

250

500

3500

437500000

In the table above, I have included the cut of x, which gives the open box its maximum volume and the cut of x, which is equal to length/4. As you can see from the table above, the cut of x, which gives this open box its maximum volume, is 232cm. Yet again the divisor is also between 4.3-4.4. Therefore, one can conclude that the limit of the divisor for the ratios of 1:3, and 1:4.

From that, I can predict that for any ratio (except 1:1 and 1:2), the length divided by 4-4.5 will give the square cut of x which will give the open box its maximum volume.

...read more.

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