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• Level: GCSE
• Subject: Maths
• Word count: 2015

The Fencing Problem.

Extracts from this document...

Introduction

Maths Coursework - The Fencing Problem

Maths Coursework

The Fencing Problem

Aim:

A farmer has brought 1000 meters of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the biggest size. I will investigate different shapes the fencing can make to achieve the largest area.

I am going to start investigating different shape rectangles because they are the easiest shapes to work with, the perimeter all of these shapes will have to be 1000 meters. Below are two rectangles (to scale) showing how different shapes with the same perimeter can have different areas. I will use a scale of 1cm: 100m.

1) 2)

400m (4cm)

Height 300m (3cm)

200m (2cm)

100m (1cm)

Width

I will work out the area of both rectangles by using the formula below. Both rectangles have a perimeter of 1000m

1)

Area of rectangle = height * width

Area of rectangle = 400m * 100m

Area of rectangle = 40000 m²

2)

Area of rectangle = height * width

Area of rectangle = 300m * 200m

Area of rectangle = 60000m²

As you can notice the areas of the rectangles 1 and 2 are different though the perimeters of both are 1000m.

Now I will put the areas widths and lengths of rectangles.

Middle

360

140

220

1000

50400

370

130

240

1000

48100

380

120

260

1000

45600

390

110

280

1000

42900

400

100

300

1000

40000

410

90

320

1000

36900

420

80

340

1000

33600

430

70

360

1000

30100

440

60

380

1000

26400

450

50

400

1000

22500

460

40

420

1000

18400

470

30

440

1000

14100

480

20

460

1000

9600

490

10

480

1000

4900

500

0

500

1000

0

The highlighted row gives the biggest area; after I go past this row I start to repeat my self.

A square with the perimeter 1000m gives me the largest area. I will further investigate this because I was going up in increments of 10. I will go into the decimal widths and lengths.

Below is a table of results.

 Width Length Area 249 251 62499 249.5 250.5 62499.8 24975 250.25 6249994 250 250 62500 250.25 249.75 62499.9 250.5 249.5 62499.8 251 249 62499

From this table I can see that the perfect square with the perimeter 1000m produces the largest area.

From this graph I can see that the highest point is on the 250m mark. The graph is a parabola and is symmetrical. At the centre point of the graph I reach the highest value for the area, after this I just repeat my self.

In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m * 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side.

Conclusion

To find the length of the base of a segment I would divide 1000 by the number of sides, so I could use 1000/s (s=number of sides), but because I want to find half of one segment I use (1000/s)/2. To find the interior angle of the n sided polygon I can use, (360/s)/2. I am finding half of the interior angle of a segment. Then I have to find the length of the perpendicular in the segment. To do this I use the tangent formula. Adjacent = opposite/ Tangent

A = [(1000/n)/2]/ Tan (360/s)/2

Then to find the area of the regular polygon I have to multiply the value of the line A by half of the base (1000/s)/2 then I have to multiply that by the number of sided the regular polygon has. So basically it’s A * base *0.5 * number of sides.

As the table above shows as the number of sides go up the area goes up. Now I am going to investigate the area of a perfect circle. I predict that this will ultimately have the largest area because it has an infinite amount of sides.

Diameter = 318.47m

pR²

3.14 * 159.235 (2 d.p) * 159.235 (2 d.p) = 79617.16561m²

I can see that this produces the largest area.

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