The Fencing problem
Introduction
A farmer has 1000m of fencing. Wit this, he wants to enclose a field,
of the maximum area possible, of any shape, which ill use all of his
fencing, and have the greatest area possible. This means, any shaper
can be used, but it must be flat, so with no height inside the
perimeter. The task of this project is to investigate which shape of
field would give the maximum area, using only 1000m of fencing as a
perimeter. I will now outline some hypotheses, to give a structure to
my investigation. Firstly, I believe that shapes of a regular nature,
with sides of equal length will provide the greatest area. If I then
prove this to be correct, I further believe that shapes with a greater
number of sides will have the greatest area, and this leads to the
idea of a circle enclosing the greatest area, as it has infinite
sides. These three hypotheses rest heavily on each other, but the
evidence for one will help to prove the next.
Regular shapes have the largest area
The first hypothesis is stated was that areas with a
regular shape, and sides of equal lengths would produce the greatest
area. I will attempt to prove this correct, using diagrams and
calculations.
Rectangles:
This is the regular shape for a 4-side polygon.
[image001.gif]
Side Length = 1000/4 = 250
Area = 250²
Area of the shape = 62 500m²
This is a rectangle, with sides of unequal length.
[image002.gif] Side length = 400 and 100
Area = 400 x 100
Area = 40 000m²
[image003.gif]
Area =Side Length * Side Length
Area = 300 * 200
Area = 60 000m²
Here are some more rectangles I have worked out:
RECTANGLES
Perimeter = 1000 M
Area = 250 x 250
= 62500 m^2
[image004.gif]
250 M
250 M
Perimeter = 1000 M
Area = 300 x 200
= 60000 m^2
[image005.gif]
300 M
200 M
[image006.gif]
[image007.gif]
150 M
[image008.gif]
Perimeter = 1000M
Area = 400 x 100
= 40000 ^2
400M
100M
[image009.gif]
Perimeter = 1000M
Area = 450 x 50
= 22500 m^2^
450M [image011.gif]
50M
Rectangles Line Graph
[image013.gif]
Text Box: Area (000 m2)
Width of Rectangle (m)
This graph shows that the longer the two sides of the polygon area,
the smaller the area is, but it tends towards a limiting case as it
begins to flatten out. This shows that the length of the sides is
inversely proportional to the area of the polygon, proving that the
square has the largest volume.
There are large discrepancies in the area of the two various, but they
have the same perimeter. I have used a spreadsheet to avoid long
...
This is a preview of the whole essay
Text Box: Area (000 m2)
Width of Rectangle (m)
This graph shows that the longer the two sides of the polygon area,
the smaller the area is, but it tends towards a limiting case as it
begins to flatten out. This shows that the length of the sides is
inversely proportional to the area of the polygon, proving that the
square has the largest volume.
There are large discrepancies in the area of the two various, but they
have the same perimeter. I have used a spreadsheet to avoid long
calculations, and get accurate results. Below are the results:
Side 1
Side 2
Side 3
Side 4
Perimeter
Area
50
450
50
450
1000
22,500
100
400
100
400
1000
40,000
150
350
150
350
1000
52,500
200
300
200
300
1000
60,000
250
250
250
250
1000
62,500
300
200
300
200
1000
60,000
350
150
350
150
1000
52,500
400
100
400
100
1000
40,000
450
50
450
50
1000
22,500
I plotted this data on a graph, and it shows a parabola peaking at
62,500m². This is the square, with sides of equal length. I have now
proved my first hypothesis for 4 sided shapes.
[image016.jpg]
These results show a clear answer, that the regular shape for a for
sided shape has the largest area, and shown below are my results for
triangles.
Triangles
For triangles, I will use Pythagoras theorem to work out the area for
each of the triangles I do.
[image017.gif] [image018.gif] Side Lengths = 400m, 400m, 200m
According to Pythagoras theorem the height is 387.3m, and the area
38729.8m².
[image017.gif] [image019.gif] This is an equilateral triangle, and
because there are 3 sides, the length of side will be 333.33'.. The
lengths are 333.33 '.. The height should also be 333.33'.m, so I can
now work out the area. The ½ base x Height formula gives us an area of
55555.56m².
Here are some more triangles I worked out, that help to give evidence
for my hypotheses:
[image020.gif]
H^2 = 450^2 - 50^2
H^2 = 200000 m
H = 447.2136 m
447.2136 / 2 = 223.6068
223.6068 x 100 = 22360.68
Area = 22360.68 m^2^
H^2 = 300^2 - 200^2
H^2 = 50000 m
H = 223.6068 m
223.6068 / 2 = 111.8034
111.8034 x 400 = 44721.36
Area = 44721.36
Isosceles Triangles Graph
[image022.gif]
Text Box: This graph is very similar to the rectangles one, and leads
me to the same conclusion. A further observation is that the
difference between the areas versus the length each time decreases, so
the change in area each time decreases. Text Box: Base (m) Text Box:
Area (000 m2)
This pattern for the two results gives me a conclusion, which is that
the regular shapes such as a square and an equilateral triangle. This
proves my hypothesis correct but now I have to find which shape gibes
the largest area with a 1000m perimeter.
Below are some other shapes which also display a trend that shows that
the regular type of shape has the largest area.
[image026.gif] [image027.gif] PENTAGONS
200 m^
200 m^
[image028.gif]
200 m^
[image029.gif]
[image030.gif]
H
100 m
200 m
OCTAGON
125 m^
[image031.gif]
Tan 45^0 = 62.5/H
H = 62.5/ tan45^0 = 62.5
62.5 x 125 = 7812.5
Area = 7812.5 m^2
7812.5 x 5 = 39062.5 m^2
62.5 m^
[image032.gif]
[image033.gif]
H
H
125 m
62.5 m^
[image034.gif]
22.5^0
H
[image035.gif]
45^0
HEXAGON
166.7 m
[image036.gif]
11.1 m
The Area problem
To find out which regular shape has the largest area, I need to go
through and find a formula to help work the area of an N sided shape,
so I can find out which shape has the largest area with a 1000m
perimeter.
All the shapes above a triangle and square can be split up into a
number of triangles, which I can work out the area for and then
multiply by the number of sides. This will make working out all the
areas of the regular shapes above a square much easier.
To build up a formula, I need to decide what I will refer to different
units as. The number of sides, and therefore the number of triangles
will be known as N.
[image037.gif]
[image038.gif]
[image039.gif]
[image040.gif]
I need an example to help me work out the area, so I have chosen a
pentagon. I can work out the length of each side, by dividing 1000 by
N, giving me the length of sides for the shape. The length of side for
this regular pentagon must be 200m, because the pentagon has five
sides:
[image041.gif]
I have shown below one of the triangles from the pentagon, and have
shown the working out I have used to find its area. This information
will help me to build up the formula I need to work out the area of a
regular polygon.
The interior angle of the bottom two corners is 54'0,because exterior
angles are 540'0 for a pentagon, and 54'0 is one half of one fifth of
the number, which gives me the interior angles.
Below is the working out for the area of the triangle, using tangent.
Tan = Opposite/Adjacent
[image042.gif] Tan 54'0 = Opposite/100m
Tan 54'0x 100 = Opposite
(The Opposite is the Height, which
will enable me to work out the area)
Height = 137.64m
½ Base x Height= Area
Area = ½ x 200 x 137.64
Area = 13 763.82m²
The area of the whole shape:
Area = 13 763.82m² x N , N=5
Area = 13 763.82m² x 5
Area = 68 819.10m²
The formula can be found using the example above, but is has to
include these parts to be correct: N, the number of sides, the
perimeter (1000m), and the exterior angle of the shape. This is the
formula:
Area = N ( ½ * 1000/ N ((500/ N)tan180/ N)
Area = N (500/ N ((500/ N) tan180/ N)
Here is an example, using the square that I have already found the
area for, which is 62 500m², so this will allow me to check if I got
the answer correctly.
Area = N (500/ N ((500/ N) tan180/ N)
Area = 4(500/4 ((500/4) /tan180/4)
Area = 4(125((125)/tan45)
Area = 4 * 125 * (125/1)
Area = 62 500m²
[image044.jpg] This proves my formula correct. I still have to find
out the shape that has the greatest area, but the formula I created
means I have been able to create a table for shapes up to 30 sides, 1
and 2 excluded, as they will not work. Below is the table I have
created, and it shows a definite trend towards increasing area with
increasing sides.
Another way to write the formula, so it is easier to understand is:
Text Box: N x 1000 x 500/n 2 n tan(180/n)
The correlation is obvious, and when I put it into graph form, the
correlation is even more obvious. It begins to plateau, so this shows
that there is a definite trend to the increase in area decreasing each
time, so a final shape with the largest area will be found.
[image047.jpg]
The trend shows a line tending towards a plateau, so there must be a
limiting case. According to my hypothesis about the shape that has the
maximum area, the limiting case must be p r ², where r is the radius
of the circle with a perimeter of 1000m. p r ² works out the area of
a circle, but this will not be reached until the sides reach infinity,
which is a circle. This is not possible, but the type of graph is
called an asymptote, which is a graph with a final number that can
never quite be reached.
[image048.gif]
Perimeter = 2pr
1000 = 2pr
r = 1000 / 2 p
r = 500 / p
r = 159.15494
Area = pr^2
Area = p(159.15494)^2
Area = p(25330.296)
Area = 79577.472 m^2
I have worked out the area of a circle with perimeter of 1000m, which
is 79577.47155m². In other words as N tends to the limiting case
which is, the maximum area available tends towards p r ². p r ² can
be drawn as a straight line, showing the limiting case, although this
will not be entirely accurate, as the place the line, you would have
to have to have infinity on the scale, which is impossible.
The gradient of the graph decreases further up the scale, but the
liming case of infinity can never be reached, so the horizontal line
cannot be drawn.
Conclusion
The conclusion I draw from this is that a regular shape with an
infinite number of sides has the maximum area while maintaining a
perimeter of 1000m. This shape with infinite sides, also known as a
circle has the greatest area for a 1000m perimeter according to the
area. This leads me to the conclusion that a circular field would give
the largest possible area for fencing 1000m long.
