Surface of constant mean curvature From PDE Evans, 2nd edition: Chapter 8, Exercise 12:

Assume $u$ is a smooth minimizer of the area integral $$I[w]=\int_U (1+|Dw|^2)^{1/2} \, dx,$$ subject to given boundary conditions $w=g$ on $\partial U$ and the constraint $$J[w] = \int_U w \, dx = 1.$$ Prove that graph of $u$ is a surface of constant mean curvature.
(Hint: Recall Example 4 in §8.1.2.)

First, I will print Example 4 from the textbook (page 457) below:

Example 4 (Minimal surfaces). Let $$L(p,z,x)=(1+|p|^2)^{1/2},$$ so that $$I[w] = \int_U (1+|Dw|^2)^{1/2} \, dx$$ is the area of the graph of the function $w : U \to \mathbb{R}$. The associated Euler-Lagrange equation is $$\sum_{i=1}^n \left(\frac{u_{x_i}}{(1+|Du|^2)^{1/2}} \right)_{x_i}=0 \quad \text{in }U.$$
$\quad$ This partial differential equation is the minimal surface equation. The expression $\operatorname{div} \left(\frac{Du}{(1+|Du|^2)^{1/2}} \right)$ on the left side of $(10)$ is $n$ times the mean curvature of the graph of $u$. Thus a minimal surface has zero mean curvature.

I don't have much work started on this, but this question looks interesting. (This is not a homework assignment, as with all my other PDE Evans questions.) But I am asking here because I do not understand fundamentally how mean curvature is applied to Euler-Lagrange equations. In particular, this question uses concepts of differential geometry, which I have not taken any courses in yet in my academic career.
Now, what I do know so far, is that maybe I should show that the graph of $u$ is a minimal surface. This would mean $u$ has zero mean curvature, and hence a constant mean curvature (zero is constant, obviously).
How should I start about this problem?
 A: you don't need to know any differential geometry to confirm the PDE. 
The minimal case is this: replace $w$ by $w + t \phi.$ Here $\phi$ refers to a $C^\infty$ function with compact support, the support (closure of the set where $\phi$ is nonzero) being contained in the interior of $\Omega.$ In order to have $I(w)$ a minimizer (or any critical point) it is necessary that
$$ \frac{d}{dt} \; I(w+t \phi)   $$
be zero when $t=0.$ You write this out carefully. The main lemma you need is this: if a continuous function $F,$ which will be some combination of $w$ and first and second partials of $w,$ satisfies $\int_\Omega F\phi =0$ for all such $\phi,$ then $F$ is constant zero. 
For the constant mean curvature case, replace the $\phi$ by functions $\psi$ with compact support in the interior of $\Omega,$ with the additional constraint that  $\int_\Omega \psi =0.$ Vary with $w + t \psi.$ This time, the main lemma is that, if continuous $F$  satisfies $\int_\Omega F\psi =0$ for all such $\phi,$ then $F$ is constant, but the constant is allowed to be nonzero if that is how it works out. 
The calculations in the above descriptions are simply not difficult; I recommend getting your hands dirty. It will help when later you take differential geometry. See if you can find proofs of the two lemmas. I cannot tell what you know about test functions, so it may be a matter of looking things up, or asking your professor or TA.
A: May be I am seeing the trees but here the wood also.. by means of these geometry examples underlying the question leading to the same result from same simpler situation. I see this pattern in the ODEs:


*

*In Dido's problem ... Maximum area enclosed for given boundary length ( also same as given area of minimum  boundary length),the optimal situation and Lagrangian in the plane are


$$\int w \,dx - \lambda \int \sqrt{1+w^2}dx ,\, F =  w - \lambda \sqrt{1+w^2} ; $$
Using Euler-Lagrange equation of calculus of variations/ Beltrami special case where Lagrangian appears independent of $x$ explicitly.
We are led to a constant value of curvature:
$$ \dfrac{w^{''}}{(1+w^2)^{\frac32} } = \dfrac{1}{\lambda} $$


*

*For the surface of revolution that  maximizes volume for given surface area ( or  for given volume contained within minimum surface area )  


the optimal situation Lagrangian in $\mathbb R^3$are
$$\int \pi w^2 dx - \lambda \int 2 \pi w \sqrt{1+w^2}dx ; \,F =  w^2 - 2 \lambda\, w \sqrt{1+w^2}; $$
we are led to another constant mean curvature $H = 1/ \lambda^{'} $
$$ \dfrac{w^{''}}{(1+w^{'2})^{\frac32} } + \dfrac{1} {w\, \sqrt{1+w^{'2}}}= \dfrac{1}{\lambda ^{'}}  $$
So the minimal surfaces have  $$ \kappa = const.  or \,  H= const. $$
