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In this investigation, we have been told to find out the largest possible volume for an open ended tube made from a piece of paper.

Extracts from this document...

Introduction

Maths Coursework

In this investigation, we have been told to find out the largest possible volume for an open ended tube made from a piece of paper.

Squares and Rectangles:

I have been given a piece of card measurements 24x32cm:

                                           32cm

                        24cmimage00.png

I have to find out the largest possible volume for an open ended tube made from this piece of paper. I am considering whether the side used as a base will make a difference to the volume and what affect using different shapes will have.

I will start by using squares and rectangles for the base.

Firstly, I will use the 24cm side as the base:

These are the results I got for the above:

image07.png

                        32cm

                                                                Side y

                                        Side x

Length of side x

Length of side y

Area of base (xXy)

Volume of tube

(areax32)

1

11

11

352

2

10

20

640

3

9

27

864

4

6

32

1024

5

7

35

1120

6

6

36

1152

From this I can see that in order to get the optimum volume of the tube using a square or rectangular base, using the 24cm side as the base, it is best to use a square base.

Next, I shall see if the same applies to using the 32cm side as the base.

Here are the results for the above:

image15.png

                      24cm

                                                                        Side y

                                        Side x

Length of side x

Length of side y

Area of base (xXy)

Volume of tube (area x 24)

1

15

15

360

2

14

28

672

3

13

39

936

4

12

48

1152

5

11

55

1320

6

10

60

1440

7

9

63

1512

8

8

64

1536

...read more.

Middle

Longest Side Should be Folded

x+8 = long side, x = short side

  x+8    2 X x >     x    2 X (x+8)image21.pngimage02.pngimage01.png

  1.  4

x2+16x+64 X x  >  x2 X(x+8)image03.pngimage04.png

  1. 16                                                        

x3+16x2+64x   >   x3+8x2image05.png

  1. 16               short side

(x)

x3+16x2+64x > x3+8x2

                                                          long side (x+8)

16x2+64x > 8x2

8x2+64x > 0 which is true


Polygons:

I now wish to find out if using a different shape can increase the area. I will use an even length for all of the base sides and I will use the 24cm side as the base. Here are the results for the above:

Shape

No. of sides

Length of sides

Area of base

Volume

3

8

27.7

886.8

4

6

36

1152

image06.png

5

4.8

39.63

1268.47

image08.png

6

4

41.66

1330.215

45.84

1466.88

From this, I

...read more.

Conclusion


Formula for volume of tube:

b = longest side,  b is the base

b

h

b  2        X   a        = volume of tubeimage22.pngimage26.png

  1. p

b2a                  = volume of tube

16b

pa                 = volume of tube

16


To find the formula for any shape:

height       X side length of shape = area of 1 triangle

                        2

image27.png

        area X no. of sides = area of base

area of base X height= volume

p= length of each side of shape, base length image29.png

                                                no. of sides(n)

sin360/n/bn        = sin  180-360/nimage30.png

                            2


x = volume of shape, b= base (longest) length, n= no. of sides

xsin360  =  b/n X sin   180-360/nimage31.pngimage04.png

                                        2

x = b/n X sin   180-360/nimage32.pngimage04.png

                            2

                sin360/nimage33.png

This is a quick formula for working out the optimum volume of the tube for any polygon.

...read more.

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