TITLE: Ratio of diameter to area of a set
QUESTION [5 upvotes]: It may be quite a basic and common thing but I haven't found much after a while of searching and I failed to figure that myself...
Let's have a (connected) set $M$ and let $\text{diam}(M)$ be its diameter.
How big can his area be?
Or in other words - evaluate this expression:
$$\sup_{M}\left\{\frac{\text{area}(M)}{\text{diam}^{2}(M)}\right\}$$
For example let $M$ be a square with side $a$. Then:
$$\frac{\text{area}(\text{square})}{\text{diam}^{2}(\text{square})} = \frac{a^{2}}{\left( a\sqrt{2} \right)^{2}} = \frac{1}{2}$$
I'm really looking forward to see a (sketch of a) proof of such thing because no matter how simple it looks I just don't even know where to start...

REPLY [3 votes]: I found a great one-page proof at http://www.math.ntnu.no/~hanche/blog/isodiametric.pdf.
They assume that one point on the boundary is at the origin and that the shape lies in the upper half plane. They then use polar coordinates for the area:
$\int_0^{\pi} 1/2 (f(\theta))^2 d\theta$ and rewrite it as 
$1/2\int_0^{\pi/2} (f(\theta))^2 +(f(\theta+\pi/2))^2 d\theta$  
The integrand is the squared length between the points at $\theta$ and $\theta+\pi/2$ on the boundary, which is at most $d^2$. So the greatest the area can be is $(\pi d^2)/4$.
The link has a great picture.