# The Fencing Problem

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Introduction

The Fencing Problem

Aim

A farmer wants to fence off a plot of level land with exactly 1000m of fencing. The farmer is not worried about the shape of the plot but the perimeter has to be 1000m. So it could be a shape with as many sides or as little sides as possible. The aim is to find the plot, which contains the maximum area. I am going to investigate different shapes that could be used to fence the plot using exactly 1000 metres of fencing each time to obtain the maximum area.

Hypothesis

My prediction is that the shape that will give me the maximum area with a 1000m perimeter/circumference is a regular shape. I also think that the shape will be a circle. I believe this the shape will be a regular because if the shape isn’t regular, the angles are not equal and this would mean that some angles would be bigger and some smaller. The angles that are bigger would gain the shape some area but the smaller angle would lose more area than it has gained. In my investigation, I will prove to you that this is true.

Method

- Firstly, I am going to start with rectangular shape because they are the easiest to calculate seeing that to find the area is a simple multiplication of any two adjacent sides.
- I will then look at other four-sided shapes like trapeziums and parallelograms and see if they have a larger area than the rectangle with the largest area.

Middle

## Rectangles - 10metre pitch

The next thing that I am going to do is to decrease the pitch to 10m. This is so I can see which length give me the largest area with sides 10m apart. The reason I decreased the pitch is because the shape that gives me the maximum area could have a length between the maximum and one of the other two points next to it. So decreasing the pitch would give me more reliable results. As you can still see, when the length is 250m, the area is still the largest.

Length (m) | Width (m) | Perimeter | Area (m^2) |

200 | 300 | 1000m | 60000 |

210 | 290 | 1000m | 60900 |

220 | 280 | 1000m | 61600 |

230 | 270 | 1000m | 62100 |

240 | 260 | 1000m | 62400 |

250 | 250 | 1000m | 62500 |

260 | 240 | 1000m | 62400 |

270 | 230 | 1000m | 62100 |

280 | 220 | 1000m | 61600 |

290 | 210 | 1000m | 60900 |

300 | 200 | 1000m | 60000 |

As you can see in this graph, the points form a curve that is very similar to the curve of the previous graph. This is because the points I have chosen are all between the three highest points on the previous graph.

## Rectangles - 1metre pitch

Now I have decreased the pitch even further to 1metre. As you can see from this table and graph, it shows that the length that gives us the biggest area is still 250m, which is a regular four-sided shape or a square. On the other hand, it shows that the two points next to the peak have very close values to the peak value, they are only 1cm^2 away. Although all three tests show me the same result, I will do one more with an even smaller pitch to confirm this result.

## Other 4-sided shapes

## Parallelogram

Conclusion

## Conclusion of Quadrilaterals and Triangles

After the tests that I have done, I have found out that regular shapes give me the largest area for a perimeter of 1000m. I also found out that a regular four-sided shape (62500m^2) gives me a larger area than a regular three-sided shape (48112.52243m^2) with equal perimeters. This may mean that the more sides a shape has, the bigger the area would be with a fixed perimeter. I will test this out now by using regular polygons with five or more sides.

## Conclusion

In my investigation, I have found out that whatever number of sides you have in a shape, regular shapes always give the maximum area. I know this because for three and four sided shapes, the regular shapes always gave me the maximum area. I tried to show a counter example but I couldn’t find one. For example, for quadrilaterals, I drew a parallelogram and trapezium to see if they could be a counter example but it didn’t work. Also, in triangles, only the regular equilateral triangle gave the maximum area. So for the rest of the investigation, I used regular shapes because I knew that these gave the maximum area for a shape with any number of sides.

When I tested polygons, I found a reliable method to find the area of the shapes. For every polygon, I used the same method or formula but changing the number of sides for each shape. Now, I can write out the formula to show you what I did.

This is the most simplified version of the general formula.

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

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