# Tube Maths Investigation

Extracts from this document...

Introduction

For this project we had to investigate the volumes of open-ended tubes made from a rectangular piece of card with the dimensions, 32cm by 24cm. Either of the two sides could be folded to make the perimeter, with the other side being the height. I made a couple of conjectures before I started. They were:

· the more sides, the bigger the volume

· regular polygons are the best

1. For the first question, the volumes have to be investigated from various polygon shapes as the base side.

The first shape I used was a square, with a height of 24cm, and perimeter of 32cm:

To find the area of the base, I used the formula: A = x2, where ‘x’ is the side of the square. Once the area was found it was multiplied by the height.

A = 82

V = 1536cm3

For a square with height 32cm and perimeter 24cm, it is:

A = 62

V = 1152cm3

From the method I used, a formula can be found for an open ended tubes with square bases. If we let ‘h’ be the height, and ‘p’ be the perimeter, it would be:

V = p 2 x h

4

V = p 2 x h

16

V = p2h

16

The next shape was an equilateral triangle based tube:

2 3

10h = 24

cm

1

Middle

12 6 12

V = p x v p2 x h

4 36

The next shape was a cylindrical tube.

h = 24

p = 32

A = pr2

A = 32 2 x p

2p

V = 1956cm3

For the cylindrical tube with a height of 32cm and circumference of 24cm, the volume was:

A = 24 2 x p

2p

V = 1467cm3

From these two results, a formula could be deduced, using the same method used, but turning it into an algebraic equation, by substituting in ‘p’ and ‘h’.

V = p 2 x p x h

2p

V = p2 h

4p

I collected results from other shaped bases, and they are recorded in the table below:

no. of sidesperimeter(cm)height(cm)volume(cm3)cylinder24321467cylinder32241955324328873322411824243211524322415366243213306322417738243213918322418541024321418103224189112243214331232241911All of these bases were constructed from regular polygons, as they have the optimum area. Showing the volume of a tube with height of 32cm, and a base of 4cm by 8cm can prove this.

A = 4 x 8

V = 1024cm3

This value is more then 100cm3 less when compared to the volume of a tube with height 32cm, and base of a 6cm square.

This proves that a regular polygon based tube has the optimum volume.These results show a pattern, that the more sides there are, the greater the area. However, a cylinder has the greatest volume, as it is infinity sides.

Conclusion

3. This question involves investigating the volumes of tubes made from card with an area equal to that of 24cm by 32cm, (768cm2).

These figures are for a square based tube made from a card with area 768cm2.

perimeter(cm)height(cm)volume(cm3)17684841921921448672481423041924921676813686476800.136864076,8000.013686400These results show an increase in volume when the perimeter is greater. The perimeter length and the volume are directly proportional,

eg: ‘p’ from 768 to 7680, ‘V’ from 36864 to 368640. The perimeter is 1000% bigger in one, and so is the volume.

The perimeter length and the volume are directly proportionalThis graph shows this:

The graph shows the way in which the volume increases as the perimeter increases. Similarly, the volume can get nearer and nearer to zero with the decrease in value of ‘p’.

The area of the card is equal to ‘ph’, perimeter multiplied by height. This can be substituted into the formulae, thereby removing the necessity of having ‘h’ in the equation. This leaves the final formulae to be:

V = Ap _4n x tan 180/nV = Ap_4pgeneral formula cylindrical formula

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

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