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  • Level: GCSE
  • Subject: Maths
  • Word count: 1334

the Open Box Problem

Extracts from this document...

Introduction

THE OPEN BOX PROBLEM

Aim:  

To find out the size of the square cut which makes the volume of the box as large as possible for any square sheet of cards.

Prediction:  

I predict that as the size of the square increases, the volume of the box should also increase.

Assumption:

Length of the square cut won’t go under 1

The length of the squares cut has to be identical

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The Square Open Box:

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In this investigation I’m first using a 20x20 square to start with, since it’s easier because the area is smaller. To get myself started, I would try to use a trail run. A trail run would help because it can give me the main idea of the investigation.

        To calculate the volume of an object, we use the formula:

Length x Width x Height

        To find the volume of the square box, we first need the length and width of the base, which is the middle square in the sheet. Then we need the height, which is the length of the identical squares (yellow squares) when folded.

...read more.

Middle

10

1000

11

704

12

432

13

208

14

56

        In the 30X30 square box, 5 give the highest volume. This shows that the ratio of 1:6 (5:30) also work on a 30X30 square box. Although it worked on both 20X20 squares and 30X30 squares, it might still be wrong. So therefore I will try a 40X40 box to make sure.

        A 40X40 square table:

Small square length (cm)

Volume (cm³)

1

1444

2

2592

3

4368

4

4096

5

4500

6

4704

7

4732

8

4608

9

4350

10

4000

11

3564

12

3072

13

2548

14

2016

15

1500

16

1024

17

612

18

288

19

76

        In a 40X40 square, the largest volume shown in the table is 7. By using the ratio 1:6, I get 6.7 as the result. Now I will do a graph in order to prove that the largest volume should lie on 6.7.

        According to my graph, the point between 6 and 7 went higher than 7. This is an evidence for the ratio of 1:6 being true.

        To prove it even further, I will use the rule of differentiation to prove this formula. Differentiation is used since it shows the rate of change of a curve, which in this case would be the volume curve: x (l-2x)(w-2x)

Multiplied out the volume formula is 4x³-2x²w-2x²l+xlw

                                                 n-1

To calculate x, newton’s law of differentiation

...read more.

Conclusion

Small square length (cm)

Volume (cm³)

1

144

2

192

3

168

4

96

5

0

        On length 5 the volume drops to 0. This happens because the width is only 10 cm long. If you cut it more than 5cm, there would not be a side to fold and to form a rectangle box. For a 20x10 rectangle, shown in the table the largest volume lies on length 2.  I didn’t find anything here yet, so I’m trying a 15X30 rectangle, which is still in ratio of 1:2.

        Here is the table for a 15X30 rectangle:

Small square length (cm)

Volume (cm³)

1

364

2

572

3

648

4

616

5

500

6

324

7

114

        For the 15X30 rectangle, the largest volume lies on length 3. I still don’t really see a pattern or ratio yet, so I’m going to do a 20X40 rectangle.

        The table for 20X40 rectangle:

Small square length (cm)

Volume (cm³)

1

684

2

1152

3

1428

4

1536

5

1500

6

1344

7

1092

8

768

9

396

        By reading the results, I see a pattern for the rectangles.

10X20: 2

15X30: 3

20X40: 4

I predict that for a 25X50 rectangle the largest volume will lay on length 5, because of the pattern I see now. So I’m going to try a 25X50 rectangle.

Table of 25X50 rectangle:

Small square length (cm)

Volume (cm³)

1

1104

2

1932

3

2508

4

2856

5

3000

6

2964

7

2772

8

2448

9

2016

10

1500

11

924

12

312

...read more.

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