From this I can see that in order to get the optimum volume of the tube using a square or rectangular base, using the 24cm side as the base, it is best to use a square base.
Next, I shall see if the same applies to using the 32cm side as the base.
Here are the results for the above:
24cm
Side y
Side x
From these results, I can see that it is also better to use a square base using the 32cm side as the base. In addition, I can see that it is better to use the 32cm side as the base and the 24cm one as the height.
Proof
Using a Square Base Over a Rectangular One
L= the length of the base/4. T= the amount added or taken away from each side to make the shape not regular.
(L+T)(L-T) < LXL
L2+LT-LT-T2 < L2
L2-T2 < L2 which is true
To help show this more clearly I will substitute for numbers, using the values for a square 32cm base:
(8+2)(8+2)<8X8
64+16-16-4<64
64-4<64
60 < 64
Longest Side Should be Folded
x+8 = long side, x = short side
x+8 2 X x > x 2 X (x+8)
- 4
x2+16x+64 X x > x2 X(x+8)
- 16
x3+16x2+64x > x3+8x2
- 16 short side
(x)
x3+16x2+64x > x3+8x2
long side (x+8)
16x2+64x > 8x2
8x2+64x > 0 which is true
Polygons:
I now wish to find out if using a different shape can increase the area. I will use an even length for all of the base sides and I will use the 24cm side as the base. Here are the results for the above:
From this, I can see that it is better to use a shape of more sides to make a larger volume.
Proof
I can see that each of the shapes can be split up into small triangles as follows:
X4 = 36
8/sin120 = x/sin30
sin54/36 = sin72/4.8
= 1330.2
Using a square base:
b is the longest side
b = base, h = height
b 2 X h = volume of tube
4
b 2 X 1200 = volume of tube
-
b
b2 X 1200 = volume of tube
- b
1200b2 = volume of tube = 75b
16b
b = 1200 X 75 = 90,000 = volume
The above shows that when b is large the volume of the tube will be larger. This proves that the longer side should be folded, and this side should be as long as possible, infinitely long.
Formula for volume of tube:
b = longest side, b is the base
b
h
b 2 X a = volume of tube
- p
b2a = volume of tube
16b
pa = volume of tube
16
To find the formula for any shape:
height X side length of shape = area of 1 triangle
2
area X no. of sides = area of base
area of base X height= volume
p= length of each side of shape, base length
no. of sides(n)
sin360/n/bn = sin 180-360/n
2
x = volume of shape, b= base (longest) length, n= no. of sides
xsin360 = b/n X sin 180-360/n
2
x = b/n X sin 180-360/n
2
sin360/n
This is a quick formula for working out the optimum volume of the tube for any polygon.
