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.