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 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.
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 in order to find out the largest volume for each shape.
Regular Pentagon
From the pentagon formula, I have found a formula for an n-sided polygon. This is basically a formula for any regular sided polygon, and instead of going through all the working out, one can use the formula to make the working out much simpler and easier.
Formula for an n-sided polygon
Regular Hexagon
Cylinder
Now I will change my perimeter from 24cm to 32cm and I will change my length from 32cm to 24cm. this will then leave me with the results of which shape has the largest volume.
As you can see clearly, the chart has demonstrated that a cylinder measuring a perimeter of 32cm and a length of 24cm has the largest volume. To justify this, I will use a million sided shape to find the volume and to see if a cylinder gives me the largest volume.
Volume of a million
sided polygon
Part 2
In the second part of my coursework I have been asked to investigate as many open ended tubes I can get from a card with an area of 1200cm2. From my evidence in part 1, I have learnt that a cylinder gives a maximum volume. Also I have learnt that the smaller the length and the larger the perimeter results in me getting a higher volume. So therefore I will use the card to make a cylinder with small lengths and larger perimeters.
So far I have used my theory of using small lengths and large perimeters. As you can see I can still keep going on for the largest volume as the length can keep getting shorter and the perimeter can keep getting bigger. Instead of working out the same procedure for finding out the volume, I will make a formula to make it much easy for me to find out the volume with a formula and also to find the volume with an unknown figure for the area.
I know that y = A/x so therefore I will place it into this formula for me to complete:
Π (A/x) 2*x
→ simplifies to → π A2/x2 * x → π A2/x
Using my knowledge of fractions
Π A2/x ÷ 4 π2/1 → π A2 * 1/4 π2
The Pie’s cancel out to form: A2 / 4 πx
Volume for any area shape = A2
This formula will help me to get the volume of a shape with A as the area and 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.