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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!

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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 $\lambda_0=0,1$.

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