Sheep must eat; the isoperimetric inequality

You have some sheep and access to some nice grassland for them to graze. You have some wiring lying around with which you can build a pen (corral). You of course remember the isoperimetric inequality which says that the maximum area for a given perimeter length is a circle. So you set up your circular corral and let your sheep graze. So far so good.

But as you watch your sheep graze you wonder: can we prove that for a given length of perimeter wire the circle does indeed maximize the area of enclosed grass? What about a square or a rectangle? Or what about other regular polygons with additional sides: a pentagon, hexagon, octagon, etc?

Square vs. uneven rectangle first

You decide to address the square vs. (uneven) rectangle problem first.

the area A of any rectangle equals width w × length l:

A = wl

Below (left) we have plotted the function for area f(w) = wl for a given perimeter (50). Notice how the area reaches its maximum when w equals P/4; i.e., when the rectangle is a square.

(You can play with the perimeter but you will see that the function always tops at P/4)

This gives us some strong evidence that indeed, an uneven rectangle can never yield more area than a square for a given perimeter.

However, we have not proven(!) that this holds for any and all rectangles and all perimeters. After all, we have only plotted the function for one perimeter, and perhaps you have tried a few more.

To see if the relationship holds for all(!) perimeters and rectangles we need a more complete proof.

Our approach: write the function for area A based on perimeter P and width w alone, then take the derivative of that function, set it to zero and solve for w.

We know that any rectangle's perimeter P equals the sum of twice the width and twice the length:

P = 2w + 2l

Thus:
l = P/2 - w

And therefore;
A = w(P/2 - w)

or
A = (P/2)w - w²

Next, we take the derivative, set it to equal 0 and solve for w.
f(w) = (P/2)w - w²

f'(w) = P/2 - 2w

P/2 - 2w = 0

P/2 = 2w

Aha! Since one half of the perimeter P equals twice the width w, it therefore also must equal twice the length l, and hence, width and length are equal and thus we have a square.

So now we know that for a given perimeter length P, a square with width and length P/4 will always give a larger area than a rectangle where w ≠ l.

Next, n-gon or circle?

Next question: given the isoperimetric inequality we know that the circle gives the maximum area for a given perimeter. But can we prove it and how does a circle's area compare to that of other regular polygons (n-gons)?

The formula for the area A of an n-gon with n sides and perimeter P is:

\[ A = \frac{P^2}{4n \tan\left(\frac{\pi}{n}\right)} \] Below (right) we have plotted this function for P = 100 and 3 ≤ n ≤ 50. Comparing it with the area of the circle with that same perimeter, you quickly see that as the polygon gets more sides, its area becomes closer to that of the circle... but will, in fact, never quite get there.

To prove this we follow the same approach as with the area of the rectangle. We take the derivative of the function, set it to zero and solve for n.

\[ f(x) = \frac{P^2}{4n \tan\left(\frac{\pi}{n}\right)} \] The derivative of this function is: \[ f'(n) = \frac{P^2}{4n^2 \tan\left(\frac{\pi}{n}\right)} \left( \frac{\pi}{n \sin\left(\frac{\pi}{n}\right) \cos\left(\frac{\pi}{n}\right)} - 1 \right) \] When we set this to zero and solve for n we find that n approximates infinity (∞).^*

And how about an n-gon with perimeter P and an infinite number of sides?

You can play with these area functions below.

perimeter:      max number of n-gon sides n (3 ≤ n ≤ 100):      Plot!

^*At first glance it seems that the π/n terms in the denominators will go to 0 as n approaches ∞. However, since tan(θ) and sin(θ) equal θ for very small values of θ, we can replace tan(π/n) and sin(π/n) with π/n for very large values of n and we see that the n's cancel out.