Tuesday, 29 March 2011

Maths Puzzle 6: Maximum area for a perimeter

Here's a puzzle I devised, pondered and then worked on solving:  imagine you have a piece of string, tied so that it forms a closed loop exactly 12 cm long.  What's the maximum area that could be contained by the string if the shape that was formed was always a right-angled triangle?


Okay, so it's not a difficult question to describe, but taking it apart leads to some interesting questions.  At one extreme, we have a triangle that's almost exactly 6cm tall with almost no width, and at the other extreme, we have a triangle that's almost exactly 6cm wide with no height.  Somewhere in between we have a triangle with maximum area.


The main formulae that we need to use to help us are the area of a triangle, the perimeter of a triangle, and the relationship between the sides of a right-angled triangle.


The area of a triangle is half times the base times the height.  In our case, let's call the base x and the height y.


A = xy / 2


The perimeter of a triangle is the sum of the three sides, x, y and the third side (which will be the side opposite the right angle), the hypotenuse h.


x + y + h = 12


Finally, the relationship between the sides of a right-angled triangle are set by Pythagoras' Theorem, which says that x2 + y2 = h2


I'm sure there's a complicated expression that connects x, y, h, A and the perimeter, but that's not how I solved this problem.  In other words, I solved the problem by testing various values of x, and using a spreadsheet to determine y and h, and therefore A.


The relationship between x (the height) and A (the area) is shown in the graph and table below.




A few points I noticed about my results:

As I had anticipated, the maximum area is obtained by having an isoceles triangle, with two sides of 3.51, which is 12/ (2 + sqrt 2)), and it isn't 3 1/2.

What surprised me, though, is that the graph is not symmetrical.  I suppose, on reflection, there's no way it can be symmetrical:  if having length x at 0 cm leads to zero area, and having length x at 6 cm also leads to zero area, then for the graph to be symmetrical, the area would have to peak at exactly x=3, and this wouldn't be an isoceles triangle.

I also assumed it would be symmetrical because I was too busy looking at the symmetry of x and y.  For each value of x, there's a corresponding value of y, and the two can be swapped around (so that (x,y) is a valid pair, and so is (y,x).  However, the relationship between x and A, the area, is much more complicated, and it is not symmetrical.

This second graph is perhaps a little confusing, but it represents the way that the two sides x and y, and the area A are connected.  The area of the bubbles or circles represents the area of the triangle which is produced when the sides of the triangle are x (along the bottom) and y (up the side).  I've highlighted the maximum area with a lighter blue.

For those who are interested - no, I haven't calculated A in terms of x alone, and I haven't differentiated either. I've just used Excel to 'goal seek' values of y and h for a given value of x, and worked from there! Some things to note: there are better ways of enclosing area other than using a triangle. For example, by using a square with three sides of 3cm each, we could enclose an area of 9 cm2, which is almost 50% more than with the right-angled triangle. The optimum approach would be to have a circle. For a circumference of 12 cm, the radius of the circle would be 1.91 cm (12 divided by 2 pi) and the area would be 11.46 cm2, which is approaching double the area enclosed by the triangle, and 27% more than the square.

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