Stretch

The soap film will be a surface of revolution such that $y(-\tfrac12L) = y(\tfrac12L) = 1$ and such that the surface area $\int_{-\frac12L}^{\frac12L} 2\pi y\sqrt{1 + (y')^2}\,dx$ is minimized. Standard Calculus of Variations arguments tell us that $\frac{y}{\sqrt{1 + (y')^2}}$ is constant. Solving the equation $a^2\big(1 + (y')^2\big) = y^2$ we obtain $y \; = \; a\cosh\left(\tfrac{x+u}{a}\right)$ for some constant $u$ . Since $y(-\tfrac12L) = y(\tfrac12L) = 1$ we deduce that $u=0$ and that $y(x) = a\cosh\big(\tfrac{x}{a}\big)$ where $1 = a\cosh\big(\tfrac{L}{2a}\big)$ , and hence $L \; = \; 2a\cosh^{-1}a^{-1} \hspace{2cm} 0 < a < 1$ Plotting the graph of $y = 2x \cosh^{-1}x^{-1}$ , we solve numerically to find that the maximum value of $L$ is $\boxed{1.32549}$ , achieved when $a = 0.552434$ . Note that, for smaller values of $L$ , there are two different values of $a$ , and therefore two different surfaces that satisfy the equation. It turns out that the "small $a$ " solution is an unstable minimum.