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This is an example of how an investigation into area optimisation could progress. The problem is this:

A farmer has 40m of fencing. What is the maximum area he can enclose?

**Case 1: The rectangle:**

Reflection – the rectangle turns out to be a square, with sides 10m by 10m. Therefore the area enclosed is 100 metres squared.

**Case 2: The circle:**

Reflection: The area enclosed is greater than that of the square – this time we have around 127 metres squared enclosed.

**Case 3: The isosceles triangle:**

Reflection – our isosceles triangle turns out to be an equilateral triangle, and it only encloses an area of around 77 metres squared.

**Case 4, the n sided regular polygon**

Reflection: Given that we found the cases for a 3 sided and 4 sided shape gave us the regular shapes, it made sense to look for the n-sided regular polygon case. If we try to plot the graph of the area against n we can see that for n ≥3 the graph has no maximum but gets gets closer to an asymptote. By looking at the limit of this area (using Wolfram Alpha) as n gets large we can see that the limiting case is the circle. This makes sense as regular polygons become closer to circles the more sides they have.

**Proof of the limit using L’Hospital’s Rule**

Here we can prove that the limit is indeed 400/pi by using L’Hospital’s rule. We have to use it twice and also use a trig identity for sin(2x) – but pleasingly it agrees with Wolfram Alpha.

So, a simple example of how an investigation can develop – from a simple case, getting progressively more complex and finishing with some HL Calculus Option mathematics.