The Fencing Problem.

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The Problem

A farmer has exactly 1000 meters 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 of 1000 meters. She wishes to fence off the plot of land, which contains the maximum area.

Investigate the shape or shapes, which could be used to fence in the maximum area using exactly 100 meters of fencing each time. In this investigation I am going to experiment with different shapes and I will find out which shape will give here the maximum area

Investigation

Squares

There are several shapes that I will be able to investigate for this coursework and to start with I will find the area of a square as this is the easiest shape to work out.

Area of a Square = Length x Width

Firstly to find the length of each side I will divide 1000m by 4 which equals 250m. Then to find the area we must square this number.

  • 1000 / 4 = 250 meters

  • 250 x 250 = 62500 meters ²

Rectangle

The next shape that I will work out is a rectangle as this is almost as easy as a square to work out.

Area of Rectangle = Width x Length

In this table in the following page I will work out the area for a large number of rectangles. I will put it in a table as this will be the easiest way to work out many different areas.

Each time I will increase the length by 10 and decrease the width by 10 to see what different area I will get.

From the results above it is clear that a square (250m x 250m) will give a bigger area than any other rectangle that you could have. So immediately I can discount a rectangle as a possible area that the farmer will want to have.

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Triangles

The next kind of shape I am going to investigate will be the triangle.

If the base is ‘B’ then we can work out the length of the two other sides using the following formula:

Length of x = (1000- B) / 2

 

Now that we know the length of the sides we can then work out the height which we will need to find the area of the triangle. To do this we use Pythagoras.

Pythagoras’s Theorem is Z² = X² + Y²

 

Having found the base (B) ...

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