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:

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

        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.

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Formula for a square

y/4 * y/4 = area

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

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

Triangles

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 ...

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