# The Fencing Problem

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Introduction

The fencing problem

There is a need to make a fence that is 1000m long. The area inside the fence has to have the maximum area. I am investigating which shape would give this.

I am going to start investigating different shape rectangles, all which have a perimeter of 1000m. Below are 2 rectangles (not to scale) showing how different shapes with the same perimeter can have different areas.

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 X 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. To work out the area of a rectangle with a bas length of 200m, I subtract 200 from 500, giving 300 and then times 200 by 300. I can put this into an equation form.

1000 = (500 – )

Below is a table of results, worked out using the formula. I have gone down by taking 10m off the base every time.

## Height (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 |

Middle

All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas.

Now that I have found that a square has the greatest area of the rectangles group, I am going to find the triangle with the largest area. Because in any scalene or eight angled triangle, there is more than 1 variable, there are countless combinations, so I am only going to use isosceles triangles. This is because if know the base length, then I can work out the other 2 lengths, because they are the same. If the base is 200m long then I can subtract that from 1000 and divide it by 2. This means that I can say that

Side = (1000 – 200) ÷ 2 = 400. This can be rearranged to the following formula.

Side = . Where X is the base length.

Conclusion

20, 50, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000.

Below is a table showing the results that I got.

No. of sides | Area (m2) |

20 | 78921.894 |

50 | 79472.724 |

100 | 79551.290 |

200 | 79570.926 |

500 | 79576.424 |

1000 | 79577.210 |

2000 | 79577.406 |

5000 | 79577.461 |

10000 | 79577.469 |

20000 | 79577.471 |

50000 | 79577.471 |

100000 | 79577.471 |

On the next page is a graph showing the No. of sides against the Area.

As you can see form the graphs, the line straightens out as the number of side’s increases. Because I am increasing the number of sides by large amounts and they are not changing I am going to see what the result is for a circle. Circles have an infinite amount of sides, so I cannot find the area by using the equation that I have used to find the other amount of sides out. I can find out the area of a circle by using . To work out the circumference of a circle the equations is d. I can rearrange this so that the diameter is circumference/. From that I can work out the area using the r2 equation.

1000/ = 318.310

318.310/2 = 159.155

X 159.1552 = 79577.472m2

If I place this point on my graph it is at the same place, area wise, as the last results on my graph were. From this I conclude that a circle has the largest area when using a similar circumference.

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

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