Consider the smooth function $f$ on interval $[0,1]$ with the constraint that $f(0)=f(1)=0$. My question is that, if the arc-length of $f$ is fixed, and assume that the area of the figure enclosed by the $f$ and $x$-axis is the largest within all the admissible function (that is all the smooth function $g$ satisfying that $g(0)=g(1)=0$ and the length of curve $(x,g(x))$, $x\in (0,1)$ equals to $l$), must $f$ be a half ellipse?
Intuitively, I guess this conjecture is valid combining some physical intuition, but I can't know how to prove it.
Indeed, I can "translate" this question into the following:
Define two functionals on space $F=\{f\in C^\infty([0,1]):f(0)=f(1)=0\}$: $$ L(f)=\int_0^1\sqrt{1+[f'(x)]^2}\,dx,\\ E(f)=\int_0^1 f(x)\,dx. $$
Denote $$ \lambda=\sup_{f\in F, L(f)=l} E(f), $$ where $l$ is a fixed positive constant. Suppose $f\in F$, $L(f)=l$ and $E(f)=\lambda$, does $f$ satisfies $$ \frac{(x-1/2)^2}{a^2}+\frac{f(x)^2}{b^2}=1 $$ for some constants $a,b\in\mathbb{R}$? Notice that this question doesn't possess a variational structure, so I have no idea how to solve it.
Any help will be appreciated a lot!