# How was the isoperimetric inequality formulated?

I'm tyring to understand how the isoperimetric inequality came into existence. It seems like finding the region which yields maximum area when enclosed by a curve of fixed length is an old problem. Queen Dido seems to have figured out the solution a long time ago: that the region should be a circle.

I'm currently reading how Steiner's method of characterizing how certain regions cannot yield maximum area.

My question however is that how was the following inequality formulated:

$4\pi A\leq L^2$

where $A$ is the area of the region which we will be considering and $L$ is the length of the curve which encloses it.

I understand we are trying to find a relationship between the area of the region and the length of the curve. I expect the area to be a function of the length perhaps ($A$~$f(L)$ where ~ represents some relation). In this case it happens that we have $A \leq \frac{1}{4\pi}L^2$

My question is simply -- why?

EDIT

Given a region $\mathbb{L}$ with $A=Area(\mathbb{L})$ and a curve with a fixed length of $L$

why do we have that $4\pi A\leq L^2$ ?

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If you believe that the optimal curve is a circle then surely you can figure out the relationship between its area and circumference yourself. – Rahul Mar 3 '13 at 13:10
Are you saying that inequality is purely based on the circle? If so, what you're saying doesn't make sense as the isoperimetric inequality only says that the region is a circle if $A=\frac{1}{4\pi}L^2$ – Adeeb Mar 3 '13 at 13:23
For fixed length the circle has the maximum area $\iff$ for fixed length no curve has more area than the circle $\iff$ for fixed length no curve has area more than $\frac1{4\pi}L^2$ $\iff$ $A\le\frac1{4\pi}L^2$. Of course, to prove the premise that the circle has the maximum area, you have the whole rest of the proof of the isoperimetric inequality. – Rahul Mar 3 '13 at 20:19