• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
• Level: GCSE
• Subject: Maths
• Word count: 1375

# Tubes Maths Investigation

Extracts from this document...

Introduction

TUBES

INVESTIGATION

BE HAPPY.

By

LEE SUMMERS

Tubes Investigation

The aim of this investigation is to make tubes out of a piece of paper 24cm by 32 cm.  The tubes have no top or bottom face and the main aim is to work out which shape of tube gives the best volume.

The first shape I will use to make a tube will be a square base.  This is because it’s an easy shape to start off with.  To work out the volume of the tube I must first work out the area of the base, then multiply this by the height of the tube.  The first tube will have the 24cm side of the paper as the base and 32cm as the height, whilst the second will have the 32cm side as the base and the 24cm side as the height.  Both of these are shown below:

V=bxh

=32(6x6)

=32x36V=volume of tube

=1152cm3b=area of base

h=height of tube

V=bxh

=24(8x8)

=24x64

=1536cm3

From this I can see that although the paper from which the tubes are made is the same there is a difference in the volumes, with the larger base giving the largest volume.

Middle

x is needed.  To work this out we need to use the length and angle we know and trigonometry.  This stage is shown below

x=tan54x2.4

=3.30(2d.p.)

The final stage is working out the area of the base.  This can now be worked out easily by using what was found in the last stage.  The height of the triangle is now simply multiplied by the base of the small triangle and the answer is multiplied by the number of sides.  So this is

V=(bxh)

=32(5(2.4x3.3))

=32(5x7.92)

=32x39.64

=1268.47cm3

The 32cm pentagonal base would, therefore be worked out as:

x=tan54x3.2

=4.40(2d.p.)

V=(bxh)

=24(5(3.2x4.4))

=24(5x14.09)

=24x70.47

=1691.30cm3

I shall now use this method to work out the volume of hexagonal tubes, then octagonal tubes.

24cm hexagonal based tube.

a=180(6-2)

=180x4

=720°

x=tan60x2

=3.46(2d.p.)

V=n (bxh)

=32(6(2x3.46))

=32(6x6.93)

=32x41.57

=1330.22cm3

32cm hexagonal based tube

a=180(6-2)

=180x4

=720°

x=tan60x2.6

=4.62(2d.p.)

V=n (bxh)

=24(6(2.6x4.62))

=24(6x12.32)

=24x73.90

=1773.62cm3

24cm based octagonal tube

a=180(8-2)

=180x6

=1080°

x=tan67.5x1.5

=3.62(2d.p.)

V=n (bxh)

=32(8(1.5x3.62))

=32(8x5.43)

=32x43.46

=1390.59cm3

32cm based octagonal tube

a=180(8-2)

=180x6

=1080°

x=tan67.5x2

=4.83(2d.p.)

V=n (bxh)

=24(8(2x4.83))

=24(8x9.66)

=24

Conclusion

I will set the value for A and modify the value of x to see what shape paper gives the most effivient tube.

The first value I will try as A will be 100.  This is because it is an easy number to divide and multiply.  I will start off with x being small and gradually make it larger until I think I have found the optimum size for the paper, and as cylinders are the best tube I will use this as a starting point.  The results for this are shown in the table below.

NOTE:x is equal to l and A/x is equal to h

From this table I can see that as the length of the base of the tube increases, so does the volume.  More importantly however is the fact that there appears to be no limit to this and the volume will continue to rise until the paper the tube is made from is the shortest and longest it can possibly be.

Therefore in order to make a tube with the largest volume it should have the following properties

• It should have a circular base
• The base should be very long and the height should be small
• If the base cannot be circular then it should be a regular polygon with as many sides as possible

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

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Fencing Problem essays

1. ## Mathematics Gcse Coursework Tubes Investigation

� 32 = 1152cm3 Triangular Prism Base � Height = Cross Section � 2 � height of tube. 24 � 3 = 8cm length of one side 8 � 2 = 4cm half on one side 4square + b square = 8square 64 - 16 = b square b square

2. ## Equable shapes Maths Investigation

9 2.5 23.1 23.1 10 2.5 25 25 Using my rule that a length or width of an rectangle cannot be less than 2 ( because the formula includes subtracting 2 from x, so unless bigger than two will be a number below zero)

1. ## Maths Investigation on Trays.

This was mainly used with the 9 by 9 tray because at the start it was looking that the hypothesis was wrong but soon after a decimal search I found it to be true. So overall the hypothesis correct also I found a easy way to predict when the max volume will come up.

2. ## Graphs of Sin x, Cos x; and Tan x

Try to split problems of this type into a series of two-dimensional questions. Confused? Look at the following example. The pyramid VABCD has a square base of length 4cm. The height of the pyramid is 3cm. Question 1 Calculate the length of VA The Answer We need to use the right-angled triangle VAO.

1. ## Borders Investigation

This is part of our formula. Therefore the remaining part of the cross must be represented by , which easily divides into two equal halves, each thus of area . The next step is to rigorously prove this formula. Geometric Near-proof One way to approach the problem is to look at it geometrically, and apply geometrical theorems and formulae.

2. ## Fencing investigation.

Quadrilaterals Squares 250m 250m 250m 250m Area = length x breadth As we already know both the length and the width of the square, it is the easiest shape to solve the area of. Area = length x breadth Area = 250m x 250m Area = 62,500m2 Quadrilaterals Rectangles 400m

1. ## Geography Investigation: Residential Areas

To back up what I see I will take photographs of eye-sores and the different types of residential areas I experience throughout my investigation. Methodology How do Basingstoke's residential areas change, improve and reflect different urban models? I have chosen this key question as the basis of my coursework because

2. ## Fencing - maths coursework

= 2474.87m 377.5m 377.5m 245m S= 0.5 (377.5m + 377.5m + 245m) = 500m 330m 330m 500 x 122.5 x 122.5 x 255 =4374.12m2 340m S= 0.5 (330m + 330m + 340m) = 500m 500 x 160 x 170 x 170 = 48083.26m2 450m 450m 100m S= 0.5 (450m + 450m + 100m)

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to