The Fencing Problem - Math Coursework


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The Fencing Problem - Math


The task
--------

A farmer has exactly 1000m of fencing; with it she wishes to fence off
a level area of land. She is not concerned about the shape of the plot
but it must have perimeter of 1000m.

What she does wish to do is to fence off the plot of land which
contains the maximun area.

Investigate the shape/s of the plot of land that have the maximum
area.


Solution
--------


Firstly I will look at 3 common shapes. These will be:
------------------------------------------------------

[IMAGE]

A regular triangle for this task will have the following area:

1/2 b x h

1000m / 3 - 333.33

333.33 / 2 = 166.66

333.33² - 166.66² = 83331.11

Square root of 83331.11 = 288.67

288.67 x 166.66 = 48112.52²

[IMAGE]A regular square for this task will have the following area:

Each side = 250m

250m x 250m = 62500m²

[IMAGE] A regular circle with a circumference of 1000m would give an
area of:

Pi x 2 x r = circumference

Pi x 2 = circumference / r

Circumference / (Pi x 2) = r

Area = Pi x r²

Area = Pi x (Circumference / (Pi x 2)) ²

Pi x (1000m / (pi x 2)) ² = 79577.45m²

I predict that for regular shapes the more sides the shape has the
higher the area is. A circle has infinite sides in theory so I will
expect this to be of the highest area.

The above only tells us about regular shapes I still haven't worked
out what the ideal shape is.

Width (m)

Length (m)

Perimeter (m)

Area (m²)

500

0

1000

0

490

10

1000

4900

480

20

1000

9600

470

30

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This table shows the results of every rectangle possible for the 1000m
perimeter with a 10m gap between each reading.

The results increase and then decrease symmetrically proving that the
SQUARE 250 x 250 is the rectangle with the highest area because it is
in the middle. This would produce a curve on a graph with the square
at the top and in the middle.

A square is a regular shape.

The graph below is of the table on the previous page.

[IMAGE]

The graph has the curve that I explained on the previous page.

Now I have completed the rectangles the next step I will choose is to
try the same for triangles.

I decided to use isosceles triangles:

The base length (m)

The 2 equal lengths (m)

Perimeter (m)

Area (m²)

0.00

500.00

1000.00

0.00

20.00

490.00

1000.00

4898.98

40.00

480.00

1000.00

9591.66

60.00

470.00

1000.00

14071.25

80.00

460.00

1000.00

18330.30

100.00

450.00

1000.00

22360.68

120.00

440.00

1000.00

26153.39

140.00

430.00

1000.00

29698.48

160.00

420.00

1000.00

32984.85

180.00

410.00

1000.00

36000.00

200.00

400.00

1000.00

38729.83

220.00

390.00

1000.00

41158.23

240.00

380.00

1000.00

43266.62

260.00

370.00

1000.00

45033.32

280.00

360.00

1000.00

46432.75

300.00

350.00

1000.00

47434.16

320.00

340.00

1000.00

48105.35

333.33

333.33

1000.00

48000.00

340.00

330.00

1000.00

48083.26

360.00

320.00

1000.00

47623.52

380.00

310.00

1000.00

45731.55

400.00

300.00

1000.00

43487.07

420.00

290.00

1000.00

40222.82

440.00

280.00

1000.00

35575.62

460.00

270.00

1000.00

28781.50

480.00

260.00

1000.00

24000.00

500.00

250.00

1000.00

0.00

[IMAGE]

This also shows that the shape with the largest area is the regular
one.

The graph isn't symmetrical this is because the isosceles triangles
with the given perimeter can only have a base length from 0m - 500m
otherwise the triangle would be flat.

This shows us that regular shapes are better. The triangle, square and
circle tested make it appear that the amount of sides has something to
do with the area. This is proved by the square having a greater area
than that of the triangle but less than the circle.

To prove this I will investigate regular shapes with more sides.

A regular pentagon:

[IMAGE]

A regular pentagon has 5 sides and with the perimeter of 1000m it will
therefore have 200m for each side.

The way I will use to find the area of a pentagon will be too split it
into triangles.

To find the area we need to know the height, the values we know are:

54Ù’

[IMAGE]

72Ù’

The angle at the top of the triangle can be calculated by dividing
360Ù’ by 5 (the number of sides and triangles) this gives 72Ù’. Then to
work out the other 2 angles 72Ù’ must be taken from 180Ù’ (the sum of
angles in a triangle) to give 108Ù’. In an isosceles triangle the
bottom two angles are the same so they will both be 108 / 2 = 54Ù’

200m

To then work out the area of the triangle we need to split it in half
and use trigonometry.

[IMAGE][IMAGE]

1

1 = Base: 100m

1 = Top angle: 36Ù’

1 = Bottom angle: 54Ù’

To work out this are we will use TAN because we need the opposite and
we have the adjacent.

TAN 54 = opposite / 100

TAN 54 x 100 = opposite

TAN 54 = 1.3764

Height of triangle = 137.64m

Area of triangle = 137.64m x 100 = 13763.8

Area of pentagon = 13763.8 x 5 = 68819.10m²

This regular pentagon is as expected, the area is more than the
triangle and the square but less than the circle. This once again
proving that the more sides a regular shape has the more area there
is.

To work out the area for a regular shape I did the following:

B = Base

N = Number of sides

P = Perimeter

H = height

Firstly I worked out the formula for the area of a triangle. This is =
b x h / 2

Then I worked out the area for any shape = b x h x n / 2

This shows that the formula to work out the area of any regular shape
is =

1/2 x P x P .

n x 2 tan(180/n)

This simplified is A = P² .

4n Tan(180/n)


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