# 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.

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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

24cm

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:

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:

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 |

Middle

Longest Side Should be Folded

x+8 = long side, x = short side

x+8 2 X x > x 2 X (x+8)

- 4

x2+16x+64 X x > x2 X(x+8)

- 16

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

- 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 | |

5 | 4.8 | 39.63 | 1268.47 | |

6 | 4 | 41.66 | 1330.215 | |

45.84 | 1466.88 |

From this, I

Conclusion

Formula for volume of tube:

b = longest side, b is the base

b

h

b 2 X a = volume of tube

- 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

area X no. of sides = area of base

area of base X height= volume

p= length of each side of shape, base length

no. of sides(n)

sin360/n/bn = sin 180-360/n

2

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

xsin360 = b/n X sin 180-360/n

2

x = b/n X sin 180-360/n

2

sin360/n

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

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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