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Along an extremal, differentiate $f-y^{\prime }{f}_{y^{\prime }}$. The chain rule givesWhen $f_x=0$, the Euler-Lagrange equation makes the right side zero. This proves the Beltrami identity, $f-y^{\prime }{f}_{y^{\prime }}=k$.

Place the wall on $y=0$ with attachments at $x=\pm a$, and write the fence above it as $y(x)\ge 0$. Maximize area $A={\int }_{-a}^{a}ydx$ subject to length ${\int }_{-a}^a\sqrt{1+y^{\prime 2}}dx=2l$. The isoperimetric constraint gives the augmented integrand $f=y-R\sqrt{1+y^{\prime 2}}$, where $R$ is a constant multiplier. Its Euler-Lagrange equation is $y^{\prime \prime }/(1+y^{\prime 2}{)}^{3/2}=-1/R$, so the fence has constant negative curvature. Equivalently the Beltrami identity is $y-R/\sqrt{1+y^{\prime 2}}=k$.

The resulting maximum-area circular segment with fixed arc length and chord hasHere $0<\theta <\pi /2$ is the half-angle subtended by the arc. Its half-length is $R\theta =l$ and half-chord is $R\sin \theta =a$. The function $\theta /\sin \theta$ increases from one to $\pi /2$, so the given range $a<l<\pi a/2$ determines a unique minor arc. Its enclosed area is $A_{\max }=R^2(\theta -\sin \theta \cos \theta )$.

There is also a global maximality proof for simple fences in the chosen half-plane, without assuming competitors are graphs. Complete the candidate minor arc with the complementary major arc of the same circle, which lies below the wall. Attach this same fixed major arc to any competing fence of length $2l$. The completed closed curve always has perimeter $2\pi R$, and its area is the competitor's fenced area plus the same fixed area below the wall. The planar isoperimetric inequality bounds that total by $\pi R^2$, attained by the candidate circle. Hence no competitor encloses more area, and equality requires the circular arc.