Past exam of the options II course of the University of Oxford / 2005 / G / 2 / ii / 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 $G=\sqrt{1+y^{\prime 2}}$ in the normal Euler-Lagrange equation. It becomesIntegrating once gives $y^{\prime }/\sqrt{1+y^{\prime 2}}=(c-x)/\lambda$. 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 $d=-\sqrt{{\lambda }^2-a^2}$. HenceIt is a circular arc. To find its parameter, substitute into the length integral: $\sqrt{1+y^{\prime 2}}=\lambda /\sqrt{{\lambda }^2-(x-a{)}^2}$, givingThis 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 $y-\lambda \sqrt{1+y^{\prime 2}}$ 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 $J[y]-\lambda I[y]\le J[y_{\ast }]-\lambda I[y_{\ast }]$. On the fixed-length class this is $J[y]\le J[y_{\ast }]$; 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 $2a<l<\pi a$. 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.