Linear transformation of a polygon maximizing its area with respect to its perimeter. Given a polygon $P$ on the plane, is there a rigorous method or algorithm to compute or approximate a linear transformation $T$ which maximizes the following ratio?
$$\frac{\mathrm{Area}[T(P)]}{\mathrm{Perimeter}[T(P)]^2}$$
 A: Comment 1: In general, there is no closed form solution to find the linear transformation, you are seeking. 
Comment 2: This is good news. You can reformulate your problem as a convex optimization problem and efficiently solve it using readily available convex packages. 
SHORT ANSWER
Please refer to Brain Rushton's answer. Once you understand his explanation and reach at the end. It is very easy to solve it. This is the well-known unconstrained minimization of sum of norms. This can be converted into a second order cone program (SOCP). 
\begin{align}
\min_{t_i,z}&\sum_{i=1}^{n}t_i \\
subject~to~&\lvert\lvert F_iz\rvert\rvert \leq t_i,~~\forall i=1,2,\dots,n
\end{align}
This can be efficiently solved using convex optimization packages like Stephen Boyd's CVX. Please read about minimization of sum of norms here (section 2.2)
LONG ANSWER
Here I will explain how to reach that formulation.
Notation: I assume the co-ordinates of the polygon are given and they are $z_1$,$z_2$,...,$z_n$. The co-ordinates are counted in one direction (clockwise or anticlockwise). So $z_1$ is adjacent to $z_2$, $z_2$ to $z_3$,.... and finally $z_n$ to $z_1$. Also $T$ denote the transformation. After applying this transformation, let $a_i=Tz_i, \forall i$. Note that all this vectors are $2 \times 1 $ and $T$ is $2 \times 2$. Thus 
SKETCH OF APPROACH
step 1: Here I will prove that the problem is independent of area. It is solely dependent only on perimeter.
step 2: The problem can be converted as a convex problem. 
STEP 1
Define $n$ matrices $A_i$ which are $2 \times 2$ as
\begin{align}
A_1=[z_1,z_2] \\
A_2=[z_2,z_3] \\
A_3=[z_3,z_4] \\
\dots \\
\dots \\
A_n=[z_n,z_1]
\end{align}
Then the area of the polygon before transformation is given by 
\begin{align}
P=(1/2)*(~\det(A_1)+\det(A_2)+\dots+\det(A_N)~)
\end{align}
It is easy to see that the area after transformation is given by 
\begin{align}
Q &=(1/2)*(~\det(TA_1)+\det(TA_2)+\dots+\det(TA_N)~) \\
 &=(1/2)*(\det(T))*(~\det(TA_1)+\det(TA_2)+\dots+\det(TA_N)~) \\
&=\det(T)P
\end{align}
where I used the fact $\det(AB)=\det(A)\det(B)$. Thus the area of the polygon just gets multiplied by the determinant of the linear transformation. As user Brian Rushton has pointed out in another answer. There is no less of generality, if you take this determinant as one, as the ratio is always preserved. Thus following his arguments, we can say the optimization problem is nothing but to minimize the perimeter of the new linear transformation.
STEP 2
Thus the problem becomes 
\begin{align}
\min_{T\in \mathbb{R}^{2\times 2}}\mathrm{Perimeter}[T(P)]
\end{align}
Perimeter before transformation is given by
\begin{align}
Peri &=\lvert\lvert z_2-z_1\rvert\rvert+\lvert\lvert z_3-z_2\rvert\rvert \dots +\lvert\lvert z_n-z_1\rvert\rvert \\
&=\lvert\lvert d_1\rvert\rvert+\lvert\lvert d_2\rvert\rvert \dots +\lvert\lvert d_n\rvert\rvert 
\end{align}
where $d_1=z_2-z_1$ and so on. After transformation, perimeter is given by 
\begin{align}
T(Peri)&=\lvert\lvert Td_1\rvert\rvert+\lvert\lvert Td_2\rvert\rvert \dots +\lvert\lvert Td_n\rvert\rvert 
\end{align}
Now comes the reformulation part.  Define the $4 \times 1$ vector
\begin{align}
z=\begin{bmatrix}T_{11} \\ T_{12} \\T_{21} \\T_{22}  \end{bmatrix}
\end{align}
where $T_{ij}$ is the $(i,j)$ entry of $T$.
Then make the important observation
\begin{align}
\begin{bmatrix}T_{11} & T_{12} \\T_{21} & T_{22}  \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}x & y & 0 & 0 \\ 0 & 0 & x & y\end{bmatrix}\begin{bmatrix}T_{11} \\ T_{12} \\T_{21} \\T_{22}  \end{bmatrix}
\end{align}
Thus, let $d_i=(x_i,y_i)$ for all $i$. Also, for all $i$ define $2\times 4$ matrix 
\begin{align}
F_i=\begin{bmatrix}x_i & y_i & 0 & 0 \\ 0 & 0 & x_i & y_i\end{bmatrix}
\end{align}
Thus you have 
\begin{align}
\lvert\lvert Td_i\rvert\rvert=\lvert\lvert F_iz\rvert\rvert
\end{align}
Substituting you have, 
\begin{align}
\min_{T\in \mathbb{R}^{2\times 2}}\mathrm{Perimeter}[T(P)]~=~\min_{z\in \mathbb{R}^{4\times 1}} \lvert\lvert F_1z\rvert\rvert+\lvert\lvert F_2z\rvert\rvert \dots +\lvert\lvert F_nz\rvert\rvert 
\end{align}
This is the well-known unconstrained minimization of sum of norms. This can be converted into a second order cone program (SOCP). Please read about minimization of sum of norms here (section 2.2) 
\begin{align}
\min_{t_i,z}&\sum_{i=1}^{n}t_i \\
subject~to~&\lvert\lvert F_iz\rvert\rvert \leq t_i,~~\forall i=1,2,\dots,n
\end{align}
Essentially, you restrict each term $\lvert\lvert F_iz\rvert\rvert$ below $t_i$ and minimize the sum of all $t_i$. This is same as minimizing sum of $\lvert\lvert F_iz\rvert\rvert$
A: This is not an answer, but may be a helpful reformulation. We can scale any linear transformation to make it have determinant 1 (so that it preserves area) without changing the isoperimetric ratio. This changes the problem to,

Given a polygon, what matrix $A$ in $SL_2(\mathbb{R})$ maximizes $\frac{1}{(Perimeter(A(P)))^2}$?

Clearly, this is the same as minimizing the perimeter. Also, we can regard each edge as a vector $a_i$. Since translation does not change the length distortion under $A$, your question reduces to,

Given a collection of $n$ 2-dimensional vectors $a_1,...,a_n$, what matrix $A$ in $SL_2(\mathbb{R})$ minimizes $\sum |Aa_i|$?

I don't know the answer to this question, either, but it seems like something that must have been studied before. Hopefully someone can finish this answer!
Edit: I realized that if you consider a 2x2 matrix as a function of four variables, you can solve this last reformukation using Lagrange multipliers (where the constraint is det A=1).
Edit: Because the perimeter and area are also invariant under rotation, we can choose a final rotation that simplifies the form of the matrix.  For instance, we can force $(1,0)$ to be an eigenvector; this is equivalent to making the transformation matrix upper triangular.  The additional constraint on the determinant leaves a two-parameter matrix:
$$
A = \left(
\begin{matrix}s & k \\ 0 & 1/s
\end{matrix}\right).
$$
