I have recently learnt some Calculus of Variations and was trying to apply this to a question I made:
Over all functions $f: [0, 1] \to \mathbb{R}$ satisfying $f(0) = f(1) = 0$ with fixed curve length $\ell \geq 1$ (i.e. $\int_0^1 \sqrt{1 + (f'(x))^2} \ \mathrm{d}x = \ell$), find $f$ which maximise and minimise \begin{align*} \int_0^1 f(x) f(1 - x) \ \mathrm{d}x. \end{align*}
Ordinarily, I would proceed by Lagrange Multipliers and use Euler-Lagrange equations to solve for $f$, but I'm not sure how this would work with $f$ being shifted above. I considered rederiving the Euler-Lagrange equation for this as well, but the fact that it is a shifted argument makes me think this would likely not be nice to work with.
Any help would be appreciated, thanks!
Hint: The extended functional reads
$$ J[f]~=~\int_0^1\!\mathrm{d}x\left( f(x)f(1-x)+\lambda \sqrt{1+f^{\prime}(x)^2} \right) -\lambda \ell, $$
where $\lambda$ is a Lagrange multiplier. The EL equation becomes non-local:
$$ 2f(1-x)~=~\lambda\frac{d}{dx}\frac{f^{\prime}(x)}{\sqrt{1+f^{\prime}(x)^2}}.$$