# Maximizing perimeter to area ratio of a function

I was trying to find the curve, $$y = f(x)$$, from $$x = x_i$$ to $$x = x_f$$, constrained by $$y(x_i) = 0$$ and $$y(x_f) = 0$$, such that the ratio between the arc length of the curve and the area below the curve was minimized.

This setup cannot produce all possible shapes, for example a square, since the square would be a vertical line followed by a horizontal line then another vertical line, which is not smooth. However, it can produce a semi-circle which is the answer I'm trying to arrive at ie. $$y = \sqrt{r^2 - x^2}$$

My first approach was to use calculus of variations. However, I'm not sure where the constraints of $$y(x_i) = 0$$ and $$y(x_f) = 0$$ can fit in (Perhaps I'm required to use parametric equations). These constraints are so that I get a "closed" curve.

I can define arc length as $$S = \int_{x_i}^{x_f}\sqrt{1 + (dy/dx)^2}dx$$

I can then define area as $$A = \int_{x_i}^{x_f}{ydx}$$

Thus, the function I want to minimize is $$(S/A)^2$$. I have to square it so that the function doesn't "cheat" and go below the x axis, creating an unbounded negative ratio.

The Euler–Lagrange equation gives us: $${\displaystyle {\frac {\partial L}{\partial f}}-{\frac {d}{dx}}{\frac {\partial L}{\partial f'}}=0}$$

If I was simply trying to minimize arc length, I could have $$L = \int_{0}^{1}{\sqrt{1 + (dy/dx)^2}}$$

However, my $$L$$ is not a simple function, its actually the division of two integrals. Perhaps I can simplify the division into a single integral, but I haven't been able to figure out how. I would appreciate any advice on how to proceed, thanks!

This the oldest known problem in Calculus of Variations called the Dido's Problem. In modern times, this is known as an Isoperimetric Problem. I understand that you want to minimize the ratio between the perimeter and the area. The standard way I know to solve this is by making the area constant$$(A)$$ and then try to find the shortest curve that has this constant area $$A$$.
\begin{align} &\int_{x_i}^{x_f}\sqrt{1 + \dot{y}^2}dx\rightarrow min\\ &\int_{x_i}^{x_f}{ydx}=A \end{align} With end constrains $$y(x_i) = 0$$ and $$y(x_f) = 0$$. Then you can solve it as a bolza problem as follows.
\begin{align} &L=\lambda _0\sqrt{1 + \dot{y}^2}+\lambda_1y\\ &\dot{y}L_{\dot{y}}-L=constant \end{align} Where the second equation is given by the Euler-Lagrange equation and $$\lambda_0$$ is positive for a minimization problem and $$\lambda$$'s are unique upto a multiple. So it would be sufficient to check for $$\lambda_0=0,1$$.