A farmer has exactly 1000m of fencing and wants to fence off a plot of level land. She is not concerned about the shape of the plot. She wishes to fence off the plot of land which contains the maximum area. I will be investigating the shape

The maximum value in scalene is identical to an isosceles! This could mean that the more irregular shapes are the smaller the area, or this could be put as the more regular the shape and the more sides it has the larger the area. I do not have enough evidence to prove this theory so I will carry on.

I’ll now test the values nearest the maximum to make absolutely sure that my maximum value is correct.

The graph of the scalene triangles, has a line of symmetry shown

As you can see the maximum value highlighted below, show’s the shape of an isosceles triangle.  Therefore I will be doing isosceles triangles next.

I’ll start my investigation of isosceles triangles by working out the area of a few triangles. Then I will show my results in the form of a table and analyse them.

        

Answer = 11858.54cm2

        

Answer = 22360.68cm2

        

Answer = 31374.75cm2

        

Answer = 38729.83cm2

        

Answer =44194.17cm2

        

Answer =47434.16cm2

        

Answer =47925.72cm2

        

Answer =44721.36cm2

        

Answer =35575.62cm2

I will be using hero’s formula to find the area of isosceles triangles. Here is my table.

Although this table shows the rough maximum value it does not pinpoint it, just

gives a general sketch. My next table will attempt to zoom in on the max area.

Look at the highlighted value; this is the largest area for isosceles triangles and is extremely close to the value of an equilateral triangle to make sure this is

just not chance I will zoom in from 332-334 in 0.1’s.

This excerpt is the maximum value; it is very near to the sides and area of an equilateral triangle which is next.

There is only 1 possible equilateral triangle with a perimeter of 1000. I have drawn it and the area below.

Answer =48112.52cm2

I will now be moving on to the next part of my investigation which is polygons, although I will not be investigating irregular polygons as the other shapes I have investigated have shown that the highest area yield shapes are regular, and so I will do only regular polygons.

I will be using pentagons as an example as to how I will work out the area of polygons.

I will take the same steps when working out the rest of the polygons.

Using this simpler means of comparing the areas I can show that the circle is the largest area, and the farmer should shape her field in the shape of one, the circle is the largest shape, as it is rigid and unhindered by angles. Although it would not be a good shape to tessellate with other shapes.

I’ll be using a couple of abbreviations when trying to explain and map out how I got to the general formula.

A = 1000 / 2n           - This is the length of the base

B = 360 / 2n             - This is the angle of the small triangle used in my formula

C = Height x Base     - This is the area of the smaller triangles I have taken out of the shape

                2

D = C x 2n                 - This is the area of the full shape

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                

I’ll now substitute the abbreviations for the real formula; I will attempt to simplify some of it straight away as it can get a bit messy.

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Now I can present the formula not in its finished state, but a usable one, the penultimate stage. I’ve cancelled out the like terms and I’m left with only a few more things to simplify.

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                

I will now do the finished version of the formula and then test it out on 3 n-gon shapes.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 

Now to test the general formula, I will at first use the shapes which areas I’ve found out.

Octagons – 25000/8TAN(180/8)=  75444.174cm2

Nonagons – 25000/9TAN(180/9)=  76318.817cm2

Decagons (10 sided) – 250000/10TAN(180/10)= 76942.088cm2

50 sided shape -  250000/50TAN(180/50)= 79472.724cm2

Now here is a table containing the areas

My graph show’s a general increase as the sides increase reaffirming my prediction that the area would increase as the sides went up. Although to show this theory working in a larger scope here is a table of areas increasing in 50’s.

The graph of the shapes going up in 50’s shows the same trend as before

After my research, I’ve concluded that the circle with a perimeter of 1000m is the largest area, and the farmer, if she wishes to have the largest area, should use this shape, although since circles do not tessellate, nor is it easy to make this shape on a large scale, she should use a simpler shape.