We are given two real positive numbers $A$ and $l$, such that $4\pi A \leqslant l^2$ (so the Isoperimetric inequality holds). Is it true we can always find a plane, simple, closed curve so that curve has length $l$ and region that it encloses has area $A$ ? If not, give a counterexample.
Intuitively, I think it is true, but I'm not sure have to define the curve with given properties.
On
Let $s = \frac12\sqrt{\ell^2-4\pi A}$ and $r = \frac1{2\pi}(\ell - 2s)$.
Start from a circle of radius $r$, cut it along a diameter into two semicircles. Insert a $(2r) \times s$ rectangle between the "cut" and glue the two semi-circles to the two sides of rectangle with length $2r$. You obtain a "pill"-like figure of area $A$ and perimeter $\ell$.
The algebra is left as an exercise.
The area of an ellipse with radii $a$ and $\frac A{\pi a}$ is $A$. If we let $a$ vary continuously over the interval $[\sqrt{\frac A\pi},\infty)$, i.e. from a circle to an arbitrarily stretched almost-line, the perimeter varies accordingly (and as a continuous function of $a$) over $[2\sqrt{\pi A},\infty)$.
If you want something more constructive, as shape like below (with a part of a circle pushed up by some distance $a$), can give you curves with area $A=\pi r^2$ and perimeter $\ell = 2\pi r+4a$ where you can easily pick $a$ arbitrarily in $[0,2r)$, i.e., this covers the cases $$ 4\pi \le \frac{\ell^2}{A}<\frac{(2\pi+8)^2}{\pi}=64.9382\ldots$$ So for $\frac{\ell^2}A$ in that range, simple pick $r$, compute the desired $a$, and construct the figure accordingly - with the middle part thin enough to avoid problems, namely if the width of the middle part is $b$, you want $a^2+b^2<4r^2$.
On the other hand, $x\times y$ rectangles with $x\le y$ cover the cases $\frac{\ell^2}A\ge 16$. for given $\frac{\ell^2}A$, it is straightforward to compute $y$ for given $x$, say, from a quadratic equaiton.