(The same system with slightly different questions has been asked in MSE.)
Let $\Omega\subset \mathbb{R}^2$ be some simply connected planar domain. We seek for a mapping $\mathbf{r}:\Omega\rightarrow \mathbb{R}^3$ such that
a) The lengths of all axes-aligned segments remain unchanged,
b) The boundary remains planar, i.e, $\mathbf{r}|_{\partial \Omega}\subset \mathbb{R}^2$,
c) The volume between the plane and the surface is maximal.
Writing $\mathbf{r}=\left(u(x,y),v(x,y),w(x,y) \right)$, being $(x,y)$ rectangular coordinates on $\Omega$, the 1st condition is expressed as
\begin{align} (u_x)^2+(v_x)^2+(w_x)^2&=1\\ (u_y)^2+(v_y)^2+(w_y)^2&=1, \tag{*} \end{align}
while
\begin{align} w\left(\partial \Omega\right)=0 \tag{**} \end{align}
for the 2nd condition, and the functional
\begin{align} V\left[u(x,y),v(x,y),w(x,y)\right]=\int_\Omega dxdy\:\left(u_x v_y-u_y v_x\right)w, \tag{***} \end{align}
is to be maximized to satisfy the 3rd condition. Introducing the multipliers $\lambda_1=\lambda_1(x,y)$ and $\lambda_2=\lambda_2(x,y)$ to enforce the constraints, we find the Euler-Lagrange equations
\begin{align} 0&=u_x v_y-u_y v_x+\left(\lambda_1 w_x \right)_x+\left(\lambda_2 w_y \right)_y\\ 0&=-\left(wv_y-\lambda_1 u_x\right)_x+\left(wv_x+\lambda_2 u_y\right)_y\\ 0&=\left(wu_y+\lambda_1 v_x\right)_x-\left(w u_x-\lambda_2 v_y\right)_y. \end{align}
The boundary conditions of $u$ and $v$ are free, meaning that among all the possible conditions we seek for the ones that maximize $V$ (for $\lambda_1$ and $\lambda_2$ as well).
Is the problem well-posed? What are the conditions for the existence of a (unique?) maximizer? Is it possible to come up with the maximizing boundary conditions of $u,\:v,\: \lambda_1,\:\lambda_2$? Is there a general protocol to eliminate the multipliers in this kind of problems?
I suspect the existence of a maximizer can be safely established. The picture below shows a numerical simulation for $\Omega$ a unit square.
