Maths Coursework : Tubes
Syllabus 1385
TASK 2
TUBES
The basis of my investigation is, when I fold in various ways a piece of card like this, what will occur.
32cm
24cm
These are the prisms I will be using:
The first shape I will try to investigate will be rectangles. I will try all the various possibilities. I will start on the smaller side i.e. the 24cm. A substantial rule to remember to find the volume, is Length x Height x Width.
A = 11 x 1 = 11
V = 11 x 32 = 352cm(
A = 10 x 2 = 20
V = 20 x 32 = 640cm(
A = 9 x 3 = 27
V = 27 x 32 = 864cm(
A = 8 x 4 = 32
V = 32 x 32 = 1024cm(
A = 7 x 5 = 35
V = 35 x 32 = 1120cm(
A = 6 x 6 = 36
V = 36 x 32 = 1152cm(
A = 5 x 7 = 35
V = 35 x 32 = 1120cm(
I realised in this review that the more cubic the piece of card gets the more possibilities in maximising its volume space. To prove this notion I even tried to work out the rectangle after it, which resulted in a smaller volume.
After folding along the 24cm side, I moved on to the 32cm edge, the results are worked out in the same way except that the depth is now 24cm instead of 32cm.
A = 15 x 1 = 15
V = 15 x 24 = 360cm(
A = 14 x 2 = 28
V = 28 x 24 = 672cm(
A = 13 x 3 = 39
V = 39 x 24 = 936cm(
A = 12 x 4 = 48
V = 48 x 24 = 1152cm(
A = 11 x 5 = 55
V = 55 x 24 = 1320cm(
A = 10 x 6 = 60
V = 60 x 24 = 1440cm(
A = 9 x 7 = 63
V = 63 x 24 = 1512cm(
A = 8 x 8 = 64
V = 64 x 24 = 1536cm(
A = 7 x 9 = 63
V = 63 x 24 = 1512cm(
This examination proved my theory again, that squares have the biggest volume. This proves that shapes that are regular have a bigger volume than those that are not regular.
The biggest volume the rectangle ever reached was 1536cm³, to enforce this fact I have drawn a table:
24cm length
Length
x
Width
x
Depth
=
Volume
X
1
X
32
=
352cm(
2
X
0
X
32
=
640cm(
3
X
9
X
32
=
864cm(
4
X
8
X
32
=
024cm(
5
X
7
X
32
=
120cm(
6
X
6
X
32
=
152cm(
7
X
5
X
32
=
120cm(
8
X
4
X
32
=
024cm(
9
X
3
X
32
=
864cm(
0
X
2
X
32
=
640cm(
1
X
X
32
=
352cm(
32cm side
Length
x
Width
x
Depth
=
Volume
X
5
X
24
=
360cm(
2
X
4
X
24
=
672cm(
3
X
3
X
24
=
936cm(
4
X
2
X
24
=
152cm(
5
X
1
X
24
=
320cm(
6
X
0
X
24
=
440cm(
7
X
9
X
24
=
512cm(
8
X
8
X
24
=
536cm(
9
X
7
X
24
=
512cm(
0
X
6
X
24
=
440cm(
1
X
5
X
24
=
320cm(
2
X
4
X
24
=
152cm(
3
X
3
X
24
=
936cm(
4
X
2
X
24
=
672cm(
5
X
X
24
=
360cm(
I would like to move on to triangles, and determine what their biggest volume would be. To work out the area of a triangle the formulae needed is half base times height. In this particular case I am not given the height, so foremost I have to work out the height, this is done using what's known as Pythagerous's Theorem. This method used is, A² + B² = C ².
A is known as the base, B the height, and C the length. But in this specific case the formulae is rearranged to C² + A² = B² .
As a matter of interest another way to work out the area of a triangle is (AB Sin C, but this only works when the triangle is when you know 2 sides and the angle between them.
The triangles I will be investigating are only isosceles, as any other kind would take too long, and it makes the calculations easier.
Here again, once the area of the triangle is found like last time you have to times the area by the length of the prism to find the volume.
I will start with triangles folded on the 24cm side:
1 11
B = 11( - 1( = 121 - 1 = 120
B = 120 = 10.95
Area = 10.95 x 1 = 10.95cm(
Volume = 10.95 x 32 = 350.54cm(
2
B = 10( - 2( = 100 - 4 10 B = 96 = 9.8
Area ...
This is a preview of the whole essay
Here again, once the area of the triangle is found like last time you have to times the area by the length of the prism to find the volume.
I will start with triangles folded on the 24cm side:
1 11
B = 11( - 1( = 121 - 1 = 120
B = 120 = 10.95
Area = 10.95 x 1 = 10.95cm(
Volume = 10.95 x 32 = 350.54cm(
2
B = 10( - 2( = 100 - 4 10 B = 96 = 9.8
Area = 9.8 x 2 = 19.6cm(
Volume = 19.6 x 32 = 672.07cm(
B = 9( - 3(= 81 - 9
B = 72 = 8.49
9 9 Area = 8.49 x 3 = 25.46cm(
Volume = 25.46 x 32 = 814.5cm(
6
60?
8 8 1/2 x A x B x Sin C
= 1/2 x 8 x 8 x Sin C
Area = 27.71 cm(
Volume = 27.71 x 32 = 886.81 cm(
8
B = 72( - 52( = 49 - 25
7 7 B = 24 = 4.9
Area = 4.9 x 5 = 24.49 cm(
Volume = 24.49 x 32 = 783.84cm(
10
It is not possible to continue with this shape folding on the 24cm side, as the next shape would have a base of 12cm and its 2 sides a length of 6cm thereby resulting in a straight line. As the common rule is that the base has to be less than both opposite side added together. But in this particular case the 2 sides added together actually equals the value of the base.
Now after finishing working out the calculations to try finding the biggest volume on an open ended tube, it figured out to be all the values around the triangle the same length. With the rectangular it happened to be that the square had the biggest volume, and with the triangles it transpired the equilateral had the largest volume. Therefore with this information I can work out that for the rest of the coursework I will not work out irregular shapes for I know that the regular ones will have the bigger volume.
Conclusively the only calculation I need to make folding along the 32cm side is the equilateral triangle, as I have already concluded it has the biggest volume.
60?
10 ( 10 (
10(
1/2 x B x Sin C
= 1/2 x 10 2/3 x 10 2/3 x Sin C
Area = 49.27cm(
Volume = 49.27 x 24 = 1182.41cm(
The following tables and graph is to show the distinction between the triangles folded on the 24cm side and the 32cm side.
24 cm
Base
/2 Base
X
Height
X
Length
=
Volume
2
X
0.92
X
32
=
350.5cm(
4
2
X
9.8
X
32
=
627.07cm(
6
3
X
8.49
X
32
=
814.59cm(
8
4
X
6.93
X
32
=
886.81cm(
0
5
X
4.9
X
32
=
783.84cm(
32 cm
Base
/2 Base
X
Height
X
Length
=
Volume
2
X
4.95
X
24
=
359.2cm(
4
2
X
3.82
X
24
=
665.11cm(
6
3
X
2.65
X
24
=
910.74cm(
8
4
X
1.31
X
24
=
086.12cm(
0
5
X
9.8
X
24
=
175.76cm(
0 2/3
5 1/3
X
9.24
X
24
=
182.41cm(
2
6
X
8
X
24
=
152cm(
4
7
X
3.87
X
24
=
650.66cm(
The next shape on my agenda is a pentagon. A subsequent point I noticed that not only did the volume increase in only regular shapes, but also the volume was at its maximum on the 32cm side, so from now on I am only working values folded on the 32cm side.
A regular pentagon always has its 5 sides the same length, and its exterior angle always equals 72?. In this distinct case the lengths of the sides are 6.4cm. To work out the area of the pentagon, I need to split it into 5 separate isosceles triangles, which the length actually is the base and the top angles are the same as the exterior angles, being 72?. Once I have split up the pentagon, I will extract one isosceles triangle out of the pentagon work out its area then times it by 5 to equal the whole area of the pentagon.
72?
6.4cm
6.4cm
To find out the area of the triangles I have to find out the height first and foremost. Previously I used Pythagerous's theorem because I had the lengths of the other 2 sides, but this time I only have the length of one side and one angle and therefore I have to use trigonometry.
If I cut the triangle in half, I will be able to find the height instantly. Now the length I obtained is 3.2cm and the angle is 36?.
36?
x cm
3.2
The length x, is the adjacent. I have the opposite, which is 3.2cm. Therefore I have to use 'Tan' to work out side x. The formula that accords with 'Tan' is;
O
T x A
This can be rearranged to read as Tan 36 = 3.2/x, where x is the adjacent. Since I don't know what x is, I can rewrite this formula as x = 3.2/Tan 36
x = 4.404cm
Now that I have the height of the triangle I can simply work out the area of the isosceles as a whole, by multiplying 1/2 its base by its height, which is 3.2 x 4.404 = 14.09 then times it by 5 to ascertain the area of the whole pentagon, which is 70.47cm , and finally to get the volume I have to times it by 24cm (the length), which equals 1691.3cm(.
The area of the hexagon is worked out the same as we did with the pentagon. The distinction is a pentagon has 5 sides and I therefore had to divide it into 5 separate isosceles triangles whereas a hexagon has 6 sides indicating I have to divide it into 6 separate equilateral's triangles.
I will do the same as I did with the pentagon i.e. I will use a regular shape as the perimeter of 32cm.
The length of each side of the hexagon is 32 ? 6 = 5 ( which also consists as the base of the isosceles triangle and thereby the top angle of each triangle is 360 ? 6 = 60?.
60?
5 (
5 (
Again, to find out the area of the triangle I found the height, and to find the height I have to use repeat trigonometry. Like last I cut the triangle in half to find its height using trigonometry's method.
30?
Adj x cm Hyp
2 (
Opp
Using the Tan rule, I can work out the height.
Tan 30? =2 ( ? Adj
Adj = 2 ( ? Tan 30º
Adj =4.188cm
The formula for the area of the triangle is Half Base x Height, which is:
2 ( x 4.188 = 12.3168 cm(
The formula for the area of the hexagon is the area of the triangle x 6:
2.3168 x 6 = 73.9008 cm(
The last stage is to find the area of the hexagon as a prism, I have to times the area of the hexagon which we worked out above x the length of the prism i.e. 24:
73.9008 x 24 = 1773.62 cm(.
It is clearly patent from my survey that, in whatever shape I used that the bigger the area of the cross section was, the bigger was the volume. To make it simpler I have constructed a graph:
SHAPE
SIDES
VOLUME CM(
Triangle
3
82.41cm(
Square
4
536cm(
Pentagon
5
691.3cm(
Hexagon
6
773.62cm(
This because all the edges of the regular shapes in simple terms are as though they are touching the outline of an imaginary circle. So a triangle which has 3 sides touches the outline of the circle in an only minimal space, whereas a hexagon for e.g. has more sides touching the circle and that is why it has the bigger volume. The perimeter need not be the same.
Triangle inside a circle Hexagon inside a circle
From this it is conclusive to deduct that the only shape that has infinite sides touching this imaginary circle would be a circle itself!
I know that the distance around the circle (the circumference) is 32 but the essential element that I need to know is what is the radius, to be able to work out the area. The formula for the circumference of the circle is 2?r.
Then all that is left, is to take out the 2? and you're left with the radius.
32 ? 2? = 5.093
The formula for the area of the circle is ?r².
?r² = ? x 5.093
Area = 81.4873 cm(
Multiply the area by the length to find the volume of the cylinder.
Volume = 81.4873 x 24 = 1955.7cm(.
The result to find the biggest volume of an open-ended tube using the card 32cm by 24cm is 1955.7cm(, the shape was a circle folding on the 32cm length.
This next question needs more work in the mathematical sense. I will try finding 3 general formulas from the following shapes;
> Squares and rectangles.
> Pentagons.
> Circles.
These general formulas will give me a precise accurate evaluation when working out any form of a square/rectangle/pentagon and circle. In annexation to the formula I will provide an example incorporating the formula.
FORMULA FOR SQUARE AND RECTANGULAR PRISMS.
I have all ready deduced that the biggest volume occurs when folding on the 32cm side and also that the cross sectional area is the same length, consequently each side of the square will be a ( so each side the length would be divided by 4. 4 = L/4
H
L
To obtain the maximum volume I will fold along the longest side.
H
L/4
L/4
I will use the letter 'V' to represent the volume an 'H' to represent the height.
It is insufficient to use letters in formulas, but if the letters are used to create formulas for certain types of shapes than it is accepted. As I have stated earlier to work out the volume of a square/rectangle prism is Length x width x height. All the sides of the shape are equal so each side can be represented as L/4. So V = L/4 (Height) x L/4 (Width) x Length.
L/4 x L/4 x H
L²H
16
This formula should work when evaluating any square or rectangle. To prove this I will invent a shape and apply this formula to it.
40
0 10
40
So L²H 40²10
6 16
= 1000, and it does when working the same shape out in the conventional way.
FORMULA FOR THE PENTAGON PRISM
With the pentagon, I have to divide it into 5 separate isosceles triangles in order to workout the height of the triangle. I will then only concentrate on 1 triangle and use the trigonometry method.
L/5 L/5
L/5 L/5
H L/5
Like last time I will calculate the angle on top of each triangle to be 36°. 360° ? 5 equals a triangle to make it into a right angled triangle I have to divide the answer by 2
= 36° 36°
Tan
As this is common place in all pentagons I can include this number in my over all formula.
Furthermore 'Tan' is needed to find the height of the triangle, so that can likewise be included in the formula.
The base of a triangle is its length so if the base is 5(L/5) then I can split the triangle into 2 right-angled triangles, inferring the base would be
L/5 ? 2 = L/10.
Therefore the height of the right-angled triangle would be
Tan 36° x L/10 = H.
Now this is established I have to find the height.
L x L = L² = L² x Tan 36° x H
0 x 10 = 100 100
FORMULA FOR THE CIRCLE PRISM
I will now try finding a formula for an open-ended cylinder.
First I will investigate a tube with a height of 24cm and a circumference of 32cm.
24cm
Radius: Circumference = 2?r
Radius: 32 = 2?r
Radius: r = 32
2?
Radius: r = 16
?
Radius: r = 5.0929cm
Volume of cylinder: ?r²h
Volume of cylinder: ? x 5.0929² x 24
Volume of cylinder: = 1960cm³ (to 3 s.f.)
Now I will do the same as I have done except that I will use a cylinder with a height 32cm and a perimeter of 24cm.
Radius: Circumference = 2?r
Radius: r = 24
2?
Radius: r = 12
?
Radius: r = 3.8197cm
Volume of cylinder: ?r²h
Volume of cylinder: ? x 3.8197² x 32
Volume of cylinder: = 1470cm³ (to 3 s.f.)
VOLUME FOR ANY CYLINDER
I will now work out a theory to find out the volume of the largest shape, the cylinder.
Acm²
L
If the area of this piece of card is Acm² and the Length is Lcm then it is easy to determine the height will be A
L
If I then turn this card into a cylinder
The height of this cylinder will be A
L
And the circumference will be L.
So: L = 2?r
Which is the same as: r = L
2?
Area of the circle: ?r²
So: ? L ²
2?
Which is the same as: ? x L x L
2? 2?
Which equals: L²
4?
Volume of cylinder: L² x height
4?
Volume of cylinder: L² x A
4? L
4