satisfy the Euler-Lagrange equation Two circles of unit radius, each normal to the line through their centers are a distance d apart.  A soap film is formed between themas shown below; energetic considerations require the filem to assume a shape of minimum surface area.
a) show that for any value of d less than a critical value $d_{c}$, there are two surfaces satisfying the appropriate Euler-Lagrange equation, while for d > $d_{c}$ there is no surface.  evaluate $d_{c}$ for a typical d < $d_{c}$, sketch the 2 surfaces.  which has the lesser area?
b) show that in a certain range $d_{o}$ < d < $d_{c}$, the minimum surface as given by the Euler-Lagrange equation has an area larger than 2$\pi$, so hat the surface is "metastable"; that is, the surface is stable against small perturbations but not against arbitrary perturbation.  evaluate $d_{o}$.
c) what happens when d is in creased beyond $d_{c}$? 
Here is an drawing of the surface.
Here is what i have so far.
 A: The free energy of the film is to be minimized by surface area minimization. If we assume that d is along x-axis due to axial-symmetry minimal surface is a surface of revolution given by
$$A[y]=2\pi \int_{x_1}^{x_2} y \sqrt{1+y'^2}\ dx$$
Correspending Euler Lagrange equation
$$\frac{\partial g}{\partial y}-\frac{d}{dx}\bigg(\frac{\partial g}{\partial y'}\bigg)=0$$
where ommiting constants
$$\frac{\partial g}{\partial y}=\sqrt{1+y'^2}$$ 
$$\frac{d}{dx}\bigg(\frac{\partial g}{\partial y'}\bigg)=\frac{d}{dx}\bigg(\frac{y\ y'}{\sqrt{1+y'^2}}\bigg)=\frac{y'^2}{\sqrt{1+y'^2}}+\frac{y''}{\sqrt{1+y'^2}}-\frac{y\ y'^2\ y''}{\big(\sqrt{1+y'^2}\big)^3}$$
If you replace the partials and collect the terms you have
$$\frac{1}{\sqrt{1+y'^2}}-\frac{y\ y''}{\big(\sqrt{1+y'^2}\big)^3}=0$$
Further simplification by multiplying $y'$ and collecting terms
$$\frac{y'}{\sqrt{1+y'^2}}-\frac{y\ y'\  y''}{\big(\sqrt{1+y'^2}\big)^3}=\frac{d}{dx}\bigg(\frac{y}{\sqrt{1+y'^2}}\bigg)=0$$
The minimization problem reduces to find a solution to a first-order differential equation solution.
$$\frac{y}{\sqrt{1+y'^2}}=\alpha$$
which can be recast as
$$\frac{dy}{dx}=\sqrt{\frac{y^2}{\alpha^2}-1}$$
and separation of variables
$$\int dx=\int \frac{dy}{\sqrt{\frac{y^2}{\alpha^2}-1}}$$
Necessary substitution as $y=\alpha\ \cosh t$ and
$$\int dx=\alpha \int dt\Rightarrow x+\beta=\alpha\ t$$
and by back-substitution
$$y=\alpha \cosh \frac{x+\beta}{\alpha}$$
The boundary conditions must be integrated to the solution. To be able to use symmetry I assume origin lies in the middle of length d. The curve must pass through points $(\frac{-d}{2},1)$ and $(\frac{d}{2},1)$. Due to symmetry we can say that $\beta=0$. The following condition must be satisfied
$$1=\alpha \cosh \frac{d}{2\alpha}$$
I guess from this point you can analyze required points.
A: Here's a very helpful link which I believe answers this question:
http://ocw.mit.edu/courses/mathematics/18-086-mathematical-methods-for-engineers-ii-spring-2006/readings/am72.pdf
