Question: > 
> **Problem:**
> 
> 
> Given a sequence of $n$ lengths, $L\_1,L\_1,...,L\_n$ where each $L\_k$ is the length of the $k$th side, I need to find a sequence of $n$ points, where $p\_k=(x\_k,y\_k)$, such that $\operatorname{dist}(p\_k,k\_{k+1})=L\_k$ and where $\operatorname{dist}(a,b)$ is the Euclidean distance between $a$ and $b$. So they way the sequence goes, $p\_1=(0,0)$ then $p\_2=(0,L\_1)$. The points must correspond to the ordered clockwise vertices of the simple polygon having the maximum possible area for the given side lengths.
> 
> 
> 

For example, if we were given the lengths $1,1,1$, then obviously this is an equilateral triangle. And the maximum area is given by the coordinates, $(0,0),(0,1),(\sqrt{1-0.25},0.5)$.

Another example, if we were given the lengths $1,2,1,2$, then we would get a rectangle with coordinates $(0,0),(0,1),(2,1),(2,0)$.

How would I work this out for large $n$ or in general for every $n$? Because I cannot think of a way past the case of a square/rectangle.

Answer: 
This proof was given [here](http://www.drking.org.uk/hexagons/misc/polymax.html).

Take the polygon and arrange it so that all the vertices lie on a circle:
[![enter image description here](https://i.stack.imgur.com/U4u4q.png)](https://i.stack.imgur.com/U4u4q.png)

Now, "glue" the red caps onto the side of the polygon:

[![enter image description here](https://i.stack.imgur.com/eUOjA.png)](https://i.stack.imgur.com/eUOjA.png)

Note that the perimeter around the red caps stays constant as I move the vertices. The area of the red shaded sections also stay the same. According to the isoperimetric inequality, the area of the total shape (both the red and white sections) is maximized when the polygon is cyclic. Ergo, the area of the polygon is maximized when the polygon is cyclic.
In general, if $A\_1 A\_2\ldots A\_n$ is a $n$-sided polygon and the lengths $l\_1,l\_2,\ldots,l\_{n-1},l\_n$ of $A\_1 A\_2,A\_2 A\_3,\ldots,A\_{n-1} A\_n, A\_n A\_1$ are fixed, the maximum area is achieved by the cyclic polygon, i.e. the polygon having all its vertices on a circle. You may easily prove this fact through [Lagrange multipliers](https://en.wikipedia.org/wiki/Lagrange_multiplier) or Steiner's [symmetrization method](https://en.wikipedia.org/wiki/Symmetrization_methods). It is an istance of the [isoperimetric inequality for polygons](https://mathproblems123.wordpress.com/2012/04/27/2523/), that can be used to prove the usual [isoperimetric inequality in the plane](https://en.wikipedia.org/wiki/Isoperimetric_inequality) through a limit process.