What function maximizes area for a constant arc length? Suppose I have a continuous function $f$, such that $f(0) = f(1) = 0$. Given the length $l$ of the curve between $0$ and $1$, which function maximizes the area under the curve? I know that if $l \leq \pi/2$ the curve is a circular arc: but if it's greater, then it cannot still be so because the curve wouldn't be a function. What would be the function for $ l = 2 $, for example?
 A: [Significant credit goes to @san for their observation on how to prove this, see the comments on the question].
For a curve of length $2l + \pi/2$, $l > 0$, no maximum exists: the area is always strictly less than $l + \pi/8$, but you can get arbitrarily close to that bound.
Preliminary lemma: if $g: [a, b] \to \mathbb R$, $g$ continuous, $0 \leq a < b \leq 1$, and the length of the curve defined by $g$ is $l < \pi/2$, then the area under $g$ is $A < \pi/8$.
Proof: reflect in the $x$-axis; we then have a closed curve of length $< \pi$, so by the isoperimetric inequality $2A \leq (2L)^2/4\pi < \pi/4$, i.e. $A < \pi/8$.
Now let $f: [0, 1] \to \mathbb R$, $f$ continuous, and suppose the length of the curve defined by $f$ is $2l + \pi/2$.
Let $x_0 = \min\{ x \in [0,1]: f(x) = l\}$ and $x_1 = \max\{ x \in [0, 1]: f(x) = l \}$.
Then the distance along the curve between $x = 0$ and $x = x_0$ is strictly greater than $l$, similiarly the distance along the curve between $x = x_1$ and $x = 1$ is strictly greater than $l$. Thus the distance along the curve between $x = x_0$ and $x = x_1$ is less than $\pi/2$, and so by the lemma, the area between this section of the curve and the line $y = l$ is $< \pi/8$. Combining this with the area between the line $y = l$ and the line $y = 0$, we get an area of size $< l + \pi/8$ which contains the area under the curve defined by $f$, and so the area under the curve defined by $f$ is $< l + \pi/8$.
Now for the second part of the claim, that is, that we can get arbitrarily close:
Let $\min\{1/2, l\} > \epsilon > 0$. Define $f: [0, 1] \to \mathbb R$ as follows:
$f(t) = (l-\epsilon)t/\epsilon \text{ if } 0 \leq t \leq \epsilon $
$f(t) = (l-\epsilon) + \sqrt{(\frac12 - \epsilon)^2 - (x-\frac12)^2}  \text{ if } \epsilon \leq t \leq 1-\epsilon $
$f(t) = (l-\epsilon)(1-t)/\epsilon \text{ if } 1-\epsilon \leq t \leq 1$
That is, we draw a straight line from $(0,0)$ to $(\epsilon, l-\epsilon)$, then a semicircle of radius $\frac{1}{2} - \epsilon$ around $(\frac{1}{2}, l-\epsilon)$, then a straight line from $(1-\epsilon, l-\epsilon)$ to $(1, 0)$.
The lengths of these three parts are less than $l$, $\pi/2$, and $l$ respectively, so the length of this curve is $< 2l + \pi/2$. Meanwhile the area under it is
$$\frac{\epsilon(l - \epsilon)}{2} + (1-2\epsilon)(l-\epsilon) + \frac{\pi(1/2-\epsilon)^2}{2} + \frac{\epsilon(l - \epsilon)}{2}$$
As $\epsilon \to 0$, this area $\to l + \pi/8$, so we have continuous curves of length less than $2l + \pi/2$ that get arbitrarily close to enclosing an area of $l+ \pi/8$ (and we can obviously extend these to curves of length $2l + \pi/2$ without decreasing the area under them).
