• 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
• Level: GCSE
• Subject: Maths
• Word count: 1434

# Investigation of Open Ended Tubes.

Extracts from this document...

Introduction

Jesper Pipping        Mathematics Cousework; Gepmoetry – Open Ended Tubes

Investigation of Open Ended Tubes

Introduction.

Tubes form an integral part of life in any developed country. Whether to transport fuel, water, coolant or information tubes are used in almost all appliances.

Because of their widespread use it is important to maximise efficiency; in most cases this would mean maximum volume with minimum material used. That is what I am investigating.

## Section 1

• To investigate the maximum possible volume of a tube made using a 320 by 240mm rectangular piece of card.

Task 1 is to discover which side would be most efficient to use as the length and which will form the perimeter of the cross-section. This will allow me to disregard the length of the tube in future calculations as the relationship will be linear if the length is always the same. Which will improve the ease of reading of future formulae.

Let P = Perimeter of cross-section (in section 1, either 240 or 320mm)

Let L = Length of tube

Let A = Area of cross-section

As a tube is a prism, the volume, V equals the area A multiplied by the length L.                V = AL

For the sake of simplicity, I shall assume that the cross-section is a square.  The general relationship should hold true for all shapes.

## V = (P/4)²L

V = (P/4)²L

V = (P²/16)L

V = P²L/16

Middle

.

h² = a²-(20-x)² = (300-a)²-x²

Which clearly shows that  a>h and 300-a>h. This means

that h is greatest when a = 300-a = 300/2 = 150.

If two sides should be the same size for optimum area, we can calculate the optimum

Length of the third one. First some limitations:

2a+b = 320

a+b>a

2a>b

h² = a²-(a-x)² = (320-2a)²-x²

this again shows that a>h and 320-2a>h. This means that h is

greatest when a = 320-2a = 320/3.

The triangle with the greatest area has all sides the same.

Therefore all its angles are 60°.

Therefore it is equilateral.

Task 3 is the same as task 2 but n = 4.

Left is a rhombus. The top and bottom sides are parallel and equal, call them a. The sides on the left and right are also parallel and equal. Call them b.d is the base of the triangle formed of d, b and c. c is the perpendicular of a. A is the area of the shape. θ is the angle between b and a

A = ac.

c = √(b²-d²)

when d = 0,b=c

when b = c, θ=90

b>c

Therefore, the area is greatest when θ=90 as that is when b = c

and c cannot be greater than b.

Therefore a quadrilateral with 4 90° angles has the greatest area.

Conclusion

Л.

Let r be the radius

The Perimeter is the circumference

The circumference = 2 Лr                                r = P/(2 Л)

The area is Лr²                                        A = Л P/(2 Л)²

Multiplied by length                                V= Л P/(2 Л)²L

The maximum volume of a prism made out of a piece of card 240 by 320mm is therefore 1955695.941mm³

Section 2

To ascertain how to create a tube with the greatest possible volume using a piece of card with an area (B) of 1200cm².

1200 = PL

From section one, we know that the cross-section with the greatest area is a circle. So we can make an equation for the volume like so:

P = 1200/L

Area = radius² Л

Area = ((1200/L)/2Л)² Л

V = ((1200/L)/2Л)² ЛL

This equation gives an exponential curve as shown below. L is x on this graph, V is Y, so it clearly shows that the smaller L is, the greater the volume.

As P                ∞, V                ∞

A Cylinder where P = ∞ would not be a 3 dimensional objest, and so have a Volume of 0, it would be an infinite sheet, a plane. This means that one can make a prism out of a card of any area with an unlimited volume. It is therefore impossible to give a definite answer to the task, as it is always possible to increase the volume of the piece of card simply by shortening the Length and increasing the Perimeter of the cross-section.

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. ## A length of guttering is made from a rectangular sheet of plastic, 20cm wide. ...

Number Of Sides (cm) Length Of Sides (cm) Angle Half Base Height Area of 1 Triangle Total Area 20 4 =A2/B2 =180/(B2*2) =C2/2 =E2/TAN(RADIANS(D2)) =E2*F2 =G2*B2 20 5 =A3/B3 =180/(B3*2) =C3/2 =E3/TAN(RADIANS(D3)) =E3*F3 =G3*B3 20 6 =A4/B4 =180/(B4*2) =C4/2 =E4/TAN(RADIANS(D4))

2. ## Borders Investigation

In that area formula, this should give a result which proves our original formula. To sum it, we use a standard proven result which says that i.e. that is the sum of the series to n terms. We can use this result to sum our series: As we can see,

1. ## Geography Investigation: Residential Areas

If my prediction is correct, I would then be able to find the average area rating of any street after that on the same route. Location The United Kingdom is situated in the far west of Europe but is not connected to the continent by land.

2. ## Biological Individual Investigation What Effects Have Management Had On Grasses In Rushey Plain, Epping ...

All of these factors present a tripping hazard. This hazard can also be minimized by watching where one is going, and staying to the paths whenever possible. Rushey Plain used to be quite marshy, and the clay-based soil is water retentive. It is also within half a mile of a lake, so there is the possibility of mosquitoes biting, and causing irritation, or even infection.

1. ## Geography As Environmental Investigation

Bad Fairly Good Good Excellent 5. What is your usual means of transport in this area ? Car Bus Walk Train Cycle Other 6. Do you Think that the new shopping/cinema complex has had a positive affect on the area.

2. ## Maths Investigation on Trays.

(18-2x)2 = 4(18-2x)x (18-2x)(18-2x) = 4x(18-2x) (18-2x) (18-2x) 18-2x = 4x 18= 4x + 2x 18= 6x so x must equal 3. As you can see by the results the formulae was correct in finding when the max volume will occur. I will see if this formula works for the other trays.

1. ## In this investigation, we have been told to find out the largest possible volume ...

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 From these results, I can see that it is also better to use a square base using the 32cm side as the base.

2. ## Quality of life in Leicester.

53% of the residents could not afford their homes and therefore all lived off the local authorities. On the west side we have the inner city of Westcotes. The terraced housing in Westcotes was built in the late 19th century/ early 20th century.

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