The Fencing Problem.

Charles Watson

The Fencing Problem

Introduction

I have been given 1000 meters of fencing and my aim is to find out the maximum area inside.

Prediction

I would predict that the more sides the shape has, then possibly the bigger the area it will have, although I have nothing to base this on, it will be what I am about to investigate.

Shapes:

I am going to start with the rectangle, I think this is a good starting block because I am able to vary the widths and lengths to see which has the bigger area. If I discover that the rectangles with equal sides i.e. square bring me the best result, then I will try to direct my investigation into furthering that particular theory.

Rectangles

Area = 40 000 m2

Area = 60 000 m2

Area = 62 500 m2

It appears that the square shape has a bigger area, I would possibly say that this is because the square has two bigger numbers, which are multiplied together to give a greater number than when a big number is multiplied with a smaller number.

However, I cannot take this for granted and I think using one more shape will be useful in order to back up my theory.

Area = 52 500m

This proves my theory regarding squares and I shall now put my results into a graph to show what I have found.

Length (m) Width (m) Area (m)

400 100 40 000

300 200 60 000

250 250 62 500

150 350 52 500

I will now further my investigation by looking at shapes of a different nature:

Regular Pentagon

The regular pentagon has 5 sides, and as we get 1000m of fencing, this means each side will be 200m (1000÷5=200).

I will now work out a triangle inside the pentagon and divide it by 6 later on. I have ‘halved’ the triangle in the pentagon and will be using Pythagoras, as it is a right angled triangle.

100 + 100 ...