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Investigation of Open Ended Tubes.

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Jesper Pipping        Mathematics Cousework; Gepmoetry – Open Ended Tubes        

Investigation of Open Ended Tubes


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

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

...read more.



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



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

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


Therefore, the area is greatest when θ=90 as that is when b = cimage01.png

and c cannot be greater than b.

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

...read more.



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

Radius = P/2Л

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.

...read more.

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