= Solution
For the upper <function graph>, the vertical cross-section of the enclosed region has length $y(x)$, so its <area> is $J[y]=\int_0^{2a}y(x)\,dx$. A maximizing nonnegative arch therefore uses $F=y$ and $G=\sqrt{1+y'^2}$ in the normal <Euler-Lagrange equation>. It becomes
$$
1+\lambda\frac d{dx}\left(\frac{y'}{\sqrt{1+y'^2}}\right)=0.
$$
Integrating once gives $y'/\sqrt{1+y'^2}=(c-x)/\lambda$. The concave upper branch has $\lambda>0$ and integrates to $y=\sqrt{\lambda^2-(x-c)^2}+d$. The equal zero endpoint heights force $c=a$ and $d=-\sqrt{\lambda^2-a^2}$. Hence
$$
\boxed{y(x)=\sqrt{\lambda^2-(x-a)^2}-\sqrt{\lambda^2-a^2}.}
$$
It is a <circular arc>. To find its parameter, substitute into the length <integral>: $\sqrt{1+y'^2}=\lambda/\sqrt{\lambda^2-(x-a)^2}$, giving
$$
\boxed{l=2\lambda\arcsin(a/\lambda).}
$$
This scalar equation determines $\lambda$ numerically when needed; no explicit solution is required.
The arch is a <global maximum> in its smooth <function graph> regime. For $\lambda>0$, the integrand $y-\lambda\sqrt{1+y'^2}$ is concave and <strictly concave> in its slope. Its supporting-tangent inequality, integrated against any competing <function graph> with the same endpoint values, has <first-order term> zero by the <Euler-Lagrange equation>. Thus $J[y]-\lambda I[y]\le J[y_*]-\lambda I[y_*]$. On the fixed-length class this is $J[y]\le J[y_*]$; strict slope <concavity> also gives uniqueness.
The source leaves the length range implicit. A twice <continuously differentiable> <function graph> of this form needs $\lambda>a$, giving \b[$2a<l<\pi a$]. At $l=2a$ the straight <function graph> is the degenerate solution. The semicircle limit $l=\pi a$ has vertical endpoint tangents and is outside that smooth <function graph> class. For $l>\pi a$, no arc in the claimed smooth <function graph> family exists, so the stated formula cannot describe a twice <continuously differentiable> maximizing <function graph> under those unrestricted data. Allowing overhanging <curves> changes the admissible class.
\Image[/past-exam-of-the-options-ii-course-of-the-university-of-oxford/2005-circular-arch.png]
{title=The circular arch and its smooth graph length range}
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