# Optimal contour shape for variational problem over captured area

Let's assume we have a continuous and finite scalar function $f(x,y)$ over the $xy$ plane ($\mathbb{R}^{2}$) and this function is to be integrated over a bounded area (surface) $A\subset\mathbb{R}^{2}$, which represents a subdomain on the same $xy$ plane, as $\int_{A} f(x,y) dA$. And let the area be nonzero, or perhaps even a constant area $A=A_{0}$. Then, it is clear that the shape of this area, as defined by the contour enclosing it (call it $g(x,y)=0)$, will influence the result of the integral.

The question now is what is the contour shape for this area that would extremize this integral? How to express the variational problem correctly (well posed) to be in terms of the contour shape?

This question is the core behind a variation problem I am trying to formulate. My attempted formulation (reported here: Formulation of contour variational problem) is having difficulty producing a reasonable answer in the form of independent Euler-Lagrange equations. Probably it is ill posed somewhere, but I cannot find why/how. Any insight or advice on how to formulate it would be helpful.

(Note, this bears some similarity to Queen Dido's classical variational problem, but here we are are not restricting the contour perimeter -- rather we require nonzero (or fixed) area, and we have a function $f$ as the integrand in the area integral).

• Do you mean $A = \{(x,y) : f(x,y) > m\}$ for an appropriate $m$? – Mateusz Kwaśnicki Sep 25 '17 at 11:21
• @MateuszKwaśnicki I think you are saying that $f$ is being restricted in some way over $A$, but this is not the case. $f(x,y)$ has no such restrictions, and can take any variation, as long as it is continuous. $A$ is an area in the physical sense, and therefore is non-negative. However, the problem will become trivial if we allow $A=0$, therefore, we added the restriction that $A>0$, or that $A$ is a fixed constant (call it $A_{0}$). – user135626 Sep 25 '17 at 11:29
• If you don't impose some condition on the boundary (such as fixed length, etc.) then, obviously, the functional will have no critical points (as long as $f$ is positive), so there won't be an associated Euler-Lagrange equation. – Robert Bryant Sep 25 '17 at 11:37
• @user135626: If your question is "for what kind of set $A$ with a given area the integral $\iint_A f$ is maximal", then the answer is: super-level set $A = \{(x,y) : f(x,y) > m\}$ for an appropriate $m$. I apologise if I misunderstand the problem. – Mateusz Kwaśnicki Sep 25 '17 at 12:08
• @user135626: Well, there is no algebraic expression unless you know $f$ explicitly. And if you ask about numerical methods, any standard method (bisection, Newton's iteration) will do, depending on regularity of $f$. – Mateusz Kwaśnicki Sep 25 '17 at 16:00