Past exam of the short options course of the University of Oxford / 2015 / 6 / c / Solution

When an endpoint's horizontal coordinate is fixed but its height is free, the natural boundary conditions for a free endpoint require $F_{y^{\prime }}=0$ there. In this problem,Every admissible extremal is an upper circular arc with centre on $y=0$, by part (b). Its tangent is horizontal precisely above its centre. Thus imposing the natural boundary conditions for a free endpoint at $x=1$ forces $h=1$. The fixed height at $x=-1$ gives $R^2=4+4c^2$, soIts free endpoint height is $y(1)=2\sqrt{1+c^2}$. Positivity on the whole interval follows because the radicand is at least $4c^2$.

There can be no solution satisfying the natural boundary conditions for a free endpoint at both ends: an upper circular arc has only one point with a horizontal tangent, whereas the two endpoints have distinct horizontal coordinates. Therefore