# The Fencing Problem

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Introduction

The Fencing Problem

A farmer has exactly 1000 metres of fencing and wants to fence off a plot of level land.

She is not concerned about the shape of the plot, but it must have a perimeter (or circumference) of 1000m.

She wishes to fence off the plot of land, which contains the maximum area.

I am going to investigate the shape or, shapes, that have the maximum area with a perimeter of a 1000 metres. I am going to first investigate triangles, because they are polygons with the least number of sides. Then I will progress to other polygons with increasing number of sides. Finally, I will work out the area of a polygon with infinite number of sides, i.e. a circle, with the circumference of 1000m.

Triangles

I will investigate isosceles triangles because triangles such as scalene have more than one different variable so there are lots of possible combinations. If I know the length of the base of an isosceles triangle, I can work out the lengths of the other two sides because they are the equal. For example, if the base is 200m, the sides would be 400m long. To work this out I used the following formula:

length of side = (1000-base )/2

This is a diagram showing the lengths of a triangle.

b = base of triangle

Middle

333.4

333.30

288.617

48112.520

333.5

333.25

288.531

48112.504

333.6

333.20

288.444

48112.476

333.7

333.15

288.357

48112.435

333.8

333.10

288.271

48112.381

333.9

333.05

288.184

48112.314

334.0

333.00

288.097

48112.233

Using this results table I can draw a final graph of area against base of triangle. From the graph, I can see that the maximum area of a triangle with a range of bases, ranging from 333m to 334m, is a triangle with a base of 333.3m, with sides of 333.3m long and a perimeter of 1000m.The area of this triangle is 48112.522m². The triangle is an equilateral triangle.

Quadrilaterals

I will investigate if and how the interior angle of a parallelogram effects the area of the parallelogram and see if the parallelogram has a bigger area than a rectangle with the same perimeter of 1000m. To work out the area of a parallelogram, I will use the following formula:

l = length of parallelogram

w = width of parallelogram

h = perpendicular height of parallelogram

a = area of parallelogram

a = l * h

However, because I only know the length and width of the parallelograms, I will need to work out the perpendicular height. For example, if the length of a parallelogram was 300m and the width of a parallelogram was 200m, I wouldn't be able to use the formula shown above. The formula to work out the perpendicular height is:

h = w * sinx°

I used sine because I know the hypotenuse (width) and I want to work out the opposite (perpendicular height). I will now work out the area of the following parallelogram.

h = 200m * sin45°

= 141.4m

a = 300m * 141.4m

= 42420m²

I will change the angle x but keep the length and width the same to find the interior angle needed to create the maximum area. The results were as followed:

x° | height/m | area/m² |

10 | 34.73 | 10418.89 |

20 | 68.40 | 20521.21 |

30 | 100.00 | 30000.00 |

40 | 128.56 | 38567.26 |

50 | 153.21 | 45962.67 |

60 | 173.21 | 51961.52 |

70 | 187.94 | 56381.56 |

80 | 196.96 | 59088.47 |

90 | 200.00 | 60000.00 |

100 | 196.96 | 59088.47 |

110 | 187.94 | 56381.56 |

120 | 173.21 | 51961.52 |

130 | 153.21 | 45962.67 |

140 | 128.56 | 38567.26 |

150 | 100.00 | 30000.00 |

160 | 68.40 | 20521.21 |

170 | 34.73 | 10418.89 |

These results show that a parallelogram has the greatest area when angle x is 90° i.e. when it is a rectangle. Therefore, a rectangle has a greater area than a parallelogram.

I will investigate the areas of different rectangles, that each have the same perimeter of 1000 metres, because a rectangle is an easier shape for which to work out the area for than any other quadrilaterals. To work out the area, I will use the following formula:

l = length of rectangle

w = width of rectangle

a = area of rectangle

a = l * w

The two examples I worked out are shown below.

The area of the rectangles above are as follows.

- area = length * width

area = 150m * 350m

area = 525000m²

- area = length * width

area = 200m * 300m

area = 60000m²

The examples above show that not all rectangles with the same perimeter have the same area.

The formula to work out the area of any rectangle, with a perimeter of 1000m for any given length is:

width = 500 - length

area = length * width

Substituting the width in the formulas above creates this formula:

area = length * (500 - length)

area = (500 * length) - length²

I'm going to use the formula shown above to find the maximum area of a rectangle with a perimeter of 1000 metres. To save time and go through all the possible measurements by a calculator, I'm going to make a spreadsheet. I will change the length of the rectangle in steps of 10m. The formulas I will use in the spreadsheet are:

width = 500 - length

area = length * width

The results are as followed:

length/m | width/m | area/m² |

0 | 500 | 0 |

10 | 490 | 4900 |

20 | 480 | 9600 |

30 | 470 | 14100 |

40 | 460 | 18400 |

50 | 450 | 22500 |

60 | 440 | 26400 |

70 | 430 | 30100 |

80 | 420 | 33600 |

90 | 410 | 36900 |

100 | 400 | 40000 |

110 | 390 | 42900 |

120 | 380 | 45600 |

130 | 370 | 48100 |

140 | 360 | 50400 |

150 | 350 | 52500 |

160 | 340 | 54400 |

170 | 330 | 56100 |

180 | 320 | 57600 |

190 | 310 | 58900 |

200 | 300 | 60000 |

210 | 290 | 60900 |

220 | 280 | 61600 |

230 | 270 | 62100 |

240 | 260 | 62400 |

250 | 250 | 62500 |

260 | 240 | 62400 |

270 | 230 | 62100 |

280 | 220 | 61600 |

290 | 210 | 60900 |

300 | 200 | 60000 |

310 | 190 | 58900 |

320 | 180 | 57600 |

330 | 170 | 56100 |

340 | 160 | 54400 |

350 | 150 | 52500 |

360 | 140 | 50400 |

370 | 130 | 48100 |

380 | 120 | 45600 |

390 | 110 | 42900 |

400 | 100 | 40000 |

410 | 90 | 36900 |

420 | 80 | 33600 |

430 | 70 | 30100 |

440 | 60 | 26400 |

450 | 50 | 22500 |

460 | 40 | 18400 |

470 | 30 | 14100 |

480 | 20 | 9600 |

490 | 10 | 4900 |

500 | 0 | 0 |

Conclusion

a = n * ((500/n) * ((500/n) * tan(90-(180/n)))

Simplified:

a = 500 * ((500/n) * tan(90-(180/n)))

Multiplied out brackets:

a = 250000/n * tan(90-(180/n)))

I used the formulas above to see what happens as you increase the number of sides. The results are as followed:

Number of sides | area/m² |

10 | 76942.09 |

20 | 78921.89 |

30 | 79286.37 |

40 | 79413.78 |

50 | 79472.72 |

60 | 79504.74 |

70 | 79524.04 |

80 | 79536.56 |

90 | 79545.15 |

100 | 79551.29 |

110 | 79555.83 |

120 | 79559.29 |

130 | 79561.98 |

140 | 79564.11 |

150 | 79565.84 |

160 | 79567.24 |

170 | 79568.41 |

180 | 79569.39 |

190 | 79570.22 |

200 | 79570.93 |

210 | 79571.53 |

220 | 79572.06 |

230 | 79572.52 |

240 | 79572.93 |

250 | 79573.28 |

260 | 79573.60 |

270 | 79573.88 |

280 | 79574.13 |

290 | 79574.36 |

300 | 79574.56 |

From this results table, I can produce a graph of area against number of sides. From the graph and table, I can see that as you increase the number of sides, the area increase. Therefore, the shape that has the maximum area, with a perimeter of 1000m, is a circle because it has infinite sides. As n increases and approaches infinite, the shape should become a circle. Now I am going to investigate a circle and see if this is true.

Circle

a = area of circle

c = circumference of circle

r = radius of circle

a = r²

c = 2r

If the circumference was 1000m:

1000m = 2r

1000/2 = r

Simplified:

500/ = r

Substitute r in the formula to work out the area of a circle:

a = *(500/)²

a = *(500/)*(500/)

a = *(250000/²)

Simplified:

a = 250000/

a = 79577.47m²

The area of the circle is larger than the 30 side shape because a circle has infinite sides.

Conclusion

For any given n-sided shape, regular shapes have the maximum area and as the number of sides of a shape increase, the area increases. A circle has the maximum area because it has infinite sides. Therefore the farmer should fence off a circle with a perimeter 1000m because it would create an area of 79577.47m².

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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