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# 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

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

 Length of rectangle Width of rectangle Area in cm2 Volume in cm3 1 11 11 352 2 10 20 640 3 9 27 864 4 8 32 1024 5 7 35 1120 6 6 36 1152 7 5 35 1120 8 4 32 1024 9 3 27 864 10 2 20 640 11 1 11 352

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.

 Length of rectangle Width of rectangle Area in cm2 Volume in cm3 5.5 6.5 35.75 1144 5.6 6.4 35.84 1146.88 5.7 6.3 35.91 1149.12 5.8 6.2 35.96 1150.72 5.9 6.1 35.99 1151.68 6 6 36 1152      Formula for a square y/4 * y/4 = area

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

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

Triangles

 Base of triangle Dimensions of triangle Area in cm2 Volume in cm3 10 7 24.49 783.68 9 7.5 27 864 8 8 27.71 886.72 7 8.5 27.11 867.52 6 9 25.46 814.72 5 9.5 22.91 733.12 4 10 19.6 627.2 3 10.5 15.59 498.88 2 11 10.95 350.4 1 11.5 5.74 183.68

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 used this theorem to find out the height of the triangles, in order to find the area and from the area I will find out the volume.

As you can see from the chart above, the 8 * 8 * 8 dimensions give me the largest volume. This is not an isosceles triangle, but it is an equilateral triangle. To justify that the 8 * 8 * 8 dimensions have the largest volume this I have made the chart below.

 Base of triangle Dimensions of triangle Area in cm2 Volume in cm3 7.5 8.25 27.56 881.92 7.6 8.2 27.61 883.52 7.7 8.15 27.66 885.12 7.8 8.1 27.69 886.08 7.9 8.05 27.7 886.4 8 8 27.71 886.72

Middle                                So far, I have noticed that the regular shapes give me the largest volume, for example the square and the equilateral triangle. So therefore, I will be using regular shapes from now on

Conclusion

x as the length. The constant in this formula is: A2 / 4π. This means that whatever values are placed into the formula, A2 / 4π will always remain the same in order to give a correct answer. This then leaves x as the variable, which means that when you change figures, the x value will change as with a fixed area the lengths can change. This also means that A is inversely proportional to x, so this means that if length decreases the volume increases, and this can be done in the opposite way.

Conclusion:

In this piece of coursework I have found out that in part 1, a cylinder gives me the largest volume because it has an infinite amount of sides which results in it having the largest volume. Also I learnt that as the length goes down and the perimeter goes up, you get a shape which has a large volume. I then use this information and evidence for part 2 so I can make a formula helping me to find out the volume with an unknown figure for the area.

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