# Tubes. I was given a piece of card measuring 24 cm by 32cm, and my task was to investigate and come up with as many open ended tubes by using this piece of card

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Introduction

Introduction:

For this coursework, I was given a piece of card measuring 24 cm by 32cm, and my task was to investigate and come up with as many open ended tubes by using this piece of card. I would have to use either 24cm or 32cm as my perimeter or length of prism and then I would swap the figures around. The purpose of this is to find the maximum volume for a particular shape.

I will begin this coursework by starting with a rectangle, as this is the easiest shape to start of with, and slowly work my way with the following shapes: square, triangle, pentagon, hexagon, heptagon, octagon and cylinder. This will then give me a brief outline of the shape volumes. Throughout the coursework I will be using this formula to help me find the volume:

Volume = area of cross-section x length

To start of with I will use the 24cm as my constant perimeter and the 32cm as my constant length.

Rectangle and Square

Length of rectangle | Width of rectangle | Area in cm2 | Volume in cm3 |

1 | 11 | 11 | 352 |

2 | 10 | 20 | 640 |

3 | 9 | 27 | 864 |

4 | 8 | 32 | 1024 |

5 | 7 | 35 | 1120 |

6 | 6 | 36 | 1152 |

7 | 5 | 35 | 1120 |

8 | 4 | 32 | 1024 |

9 | 3 | 27 | 864 |

10 | 2 | 20 | 640 |

11 | 1 | 11 | 352 |

For a rectangle there are many combinations for you to have the perimeter as 24cm. Therefore, I have drawn this table to list all the combinations for the perimeter adding up to 24cm. As you can clearly see, the 6cm by 6cm cross-section gives me the largest volume. This tube is not a rectangle, as the sides are the same, so it must be a square. However to justify this, I will use figures around 6 * 6 to see if I can get a volume above 1152cm3.

Length of rectangle | Width of rectangle | Area in cm2 | Volume in cm3 |

5.5 | 6.5 | 35.75 | 1144 |

5.6 | 6.4 | 35.84 | 1146.88 |

5.7 | 6.3 | 35.91 | 1149.12 |

5.8 | 6.2 | 35.96 | 1150.72 |

5.9 | 6.1 | 35.99 | 1151.68 |

6 | 6 | 36 | 1152 |

Formula for a square

y/4 * y/4 = area

(y/4 * y/4) * x = volume

Simplified version = y2/16 * x → xy2/16

Triangles

Base of triangle | Dimensions of triangle | Area in cm2 | Volume in cm3 |

10 | 7 | 24.49 | 783.68 |

9 | 7.5 | 27 | 864 |

8 | 8 | 27.71 | 886.72 |

7 | 8.5 | 27.11 | 867.52 |

6 | 9 | 25.46 | 814.72 |

5 | 9.5 | 22.91 | 733.12 |

4 | 10 | 19.6 | 627.2 |

3 | 10.5 | 15.59 | 498.88 |

2 | 11 | 10.95 | 350.4 |

1 | 11.5 | 5.74 | 183.68 |

For the triangle, I have used isosceles triangles to find out the largest volume. To do this I have used the Pythagoras theorem, a2 + b2 = c2, where a and b are the base and height, and c is the hypotenuse. I have used this theorem to find out the height of the triangles, in order to find the area and from the area I will find out the volume.

As you can see from the chart above, the 8 * 8 * 8 dimensions give me the largest volume. This is not an isosceles triangle, but it is an equilateral triangle. To justify that the 8 * 8 * 8 dimensions have the largest volume this I have made the chart below.

Base of triangle | Dimensions of triangle | Area in cm2 | Volume in cm3 |

7.5 | 8.25 | 27.56 | 881.92 |

7.6 | 8.2 | 27.61 | 883.52 |

7.7 | 8.15 | 27.66 | 885.12 |

7.8 | 8.1 | 27.69 | 886.08 |

7.9 | 8.05 | 27.7 | 886.4 |

8 | 8 | 27.71 | 886.72 |

Middle

So far, I have noticed that the regular shapes give me the largest volume, for example the square and the equilateral triangle. So therefore, I will be using regular shapes from now on

Conclusion

Conclusion:

In this piece of coursework I have found out that in part 1, a cylinder gives me the largest volume because it has an infinite amount of sides which results in it having the largest volume. Also I learnt that as the length goes down and the perimeter goes up, you get a shape which has a large volume. I then use this information and evidence for part 2 so I can make a formula helping me to find out the volume with an unknown figure for the area.

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