Consider a simple closed curve $\gamma$ in $\mathbb R^3$. Suppose that $\gamma$ has length $\ell$ and contains a line segment $s$ of length $k<\ell/2$. Let $\Sigma$ be a surface with boundary $\gamma$ of minimum area. I'm interested in knowing what such curve $\gamma$ causes the area of $\Sigma$ to be largest possible.


I suspect that the area of $\Sigma$ is largest possible when $\gamma$ is the union of $s$ and a circular arc of length $\ell-k$, in which case $\Sigma$ is a *circular segment*. Is this known, or does anyone have an idea how it can be (dis)proved?


If $\gamma$ is not required to contain a line segment (i.e., the case $k=0$), it is well known that $\Sigma$ has maximum area when $\gamma$ is a circle, so that agrees with my claim.



Isoperimetric inequality for minimal surfaces bounded by space curves containing a line segment
The area minimizing disc $\Sigma$ with boundary $\gamma$ has nonpositive curvature in the sense of Alexandrov.
Applying Reshetnyak majorization theorem (see [9.56](https://arxiv.org/pdf/1903.08539v5.pdf)), we get a convex plane figure $F$ and a length-nonincreasing map $m\colon F\to \Sigma$ such that the restriction $m|\_{\partial F}$ is an arc-length parametrization of $\gamma$.
In particular,
$$\textrm{area}\,F\geqslant\textrm{area}\,\Sigma.$$
Observe that the line segment $s$ corresponds to a line segment in $\partial F$.


This construction reduces you question to the convex plane figures, where you know the answer already.
In addition one need to use that for a plane curve area minimizing surface is a disc.


