Euler-Lagrange equations of the Lagrangian related to Maxwell's equations Clarification on Lagrangian mechanics would be much appreciated:
Suppose 
$$L(\phi,\,\,\phi_{,i},\,\,A_i, \dot A_i)=|\dot A+\nabla\phi|^2-|\nabla \times A|^2-c\phi+d\cdot A$$
Are the corresponding Euler-Lagrange equations then:
$$c=0$$ by considering $\phi$, and
$$2(\dot\phi_{,i}+\ddot A_i)+d_i=0$$
by considering $A_i$?
I am confused by the dependent variables in this Lagrangian -- they are differentiated wrt to different variables, namely $\phi$ wrt spatial elements, whereas $A_i$ wrt time. Moreover, shouldn't $L$ also be a function of $A_{i,j}$?
Help would be much appreciated!
 A: It might help you to understand the physical significance of the lagrangian here. This is the lagrangian for Maxwell's equations in terms of the potentials. $ \phi $ and $ A $ are the scalar and vector potentials, and $ c $ and $ d$ are the charge and current distributions. The first term $ |\dot A+\nabla\phi|^2 $ is the electric field, second term magnetic, and the remaining terms the coupling between charges and fields.
It is just a physical fact that lagrangian does not explicitly depend on the other partial derivatives.
A: Even though the answer is already accepted, I feel that a proper derivation in the language of multivariate calculus fits here. 
All relevant notes in physics use index notation for curved space-time, I still prefer the vector calculus so I will give a presentation using mathematical languages. Equations (2),(3),(4),(5),(6), and (7) in blue color are the final results.


Suppose $L(\phi,\,\,\phi_{,i},\,\,A_i, \dot A_i)=|\dot A+\nabla\phi|^2-|\nabla \times A|^2-c\phi+d\cdot A$

The Lagrangian for Maxwell's equations should have $1/2$ factor in front of the first two terms (as the one in the Physics.SE question, the derivation will be almost the same with your Lagrangian):
$$
L = \frac{1}{2}|\partial_t\mathbf{A} + \nabla \phi|^2 - \frac{1}{2}|\nabla \times \mathbf{A}|^2 - \rho\phi + \mathbf{J}\cdot \mathbf{A},
$$
where I switched the notation using $c$ to $d$ to industry standard $\rho$ and $\mathbf{J}$. Notice the $(\phi,\mathbf{A})$ pair are the electric and magnetic potential pair.


Are the corresponding Euler-Lagrange equations then: $c=0$ by considering $\phi$, and
  $2(\dot\phi_{,i}+\ddot A_i)+d_i=0$ by considering $A_i$?

There must be something wrong with the derivation. I myself always prefer to use calculus of variations to derive Euler-Lagrange equation, aka, principle of least action for the action functional:
$$
\mathcal{S}[(\phi,\mathbf{A})]:= \int^{t}_0\int_{\Omega} L \,d\mathbf{x}dt.
$$
For some smooth and simply-connected $\Omega\subset \mathbb{R}^3$. The first variation of this action functional must vanish:
$$
\lim_{\epsilon \to 0}\frac{d}{d\epsilon} \mathcal{S}[(\phi,\mathbf{A}) + \epsilon(\psi,\mathbf{v})] = 0, \tag{$\star$}
$$
for any test pair $(\psi,\mathbf{v})$. The test pair satisfy:
$$ \color{red}{\text{The test pair do not change the boundary value of }  (\phi,\mathbf{A})  \text{ in both space and time.}}
\tag{$\dagger$}$$ 
i.e. the test pair satisfy homogeneous boundary conditions, for example, $\psi = 0$ or $\mathbf{v}\times \mathbf{n}_{\partial \Omega} = 0$ on boundary.
By the arbitrariness of the test pair $(\psi,\mathbf{v})$, first we can let $\psi = 0$ firstly, plugging the expression of Lagrangian into $(\star)$:
$$
0= \lim_{\epsilon \to 0}\frac{d}{d\epsilon} \mathcal{S}[(\phi,\mathbf{A}) + \epsilon(0,\mathbf{v})] \\
= \int^{t}_0\int_{\Omega} \Big((\partial_t \mathbf{A} + \nabla\phi)
\cdot\partial_t \mathbf{v} 
-\nabla \times \mathbf{A} \cdot \nabla \times \mathbf{v}
+\mathbf{J}\cdot \mathbf{v} \Big) 
\,d\mathbf{x}dt.\tag{1}
$$
The first two equations for Maxwell's system is obtained by substitution (assuming everything is smooth): 
$$\text{Magnetic field:}\quad \mathbf{B}= \nabla \times \mathbf{A},
\\
\text{Electric field:} \quad \mathbf{E} = -\partial_t \mathbf{A} - \nabla\phi.$$
So that they implies Gauss's law for magnetism, and Faraday's law respectively:
\begin{align}
\color{blue}{\nabla \cdot \mathbf{B} = 0},\tag{2} \\
\color{blue}{\nabla \times \mathbf{E} = -\partial_t \mathbf{B}}.\tag{3}
\end{align}
Back to equation (1), we have:
$$
0 = \int^{t}_0\int_{\Omega} \Big( -\mathbf{E}\cdot\partial_t \mathbf{v} 
-\mathbf{B} \cdot \nabla \times \mathbf{v}
+\mathbf{J}\cdot \mathbf{v} \Big) 
\,d\mathbf{x}dt,
$$
integrating by parts in both space and time, using the fact of $(\dagger)$, we have for any $\mathbf{v}$:
$$
0 = \int^{t}_0\int_{\Omega} \Big( \partial_t \mathbf{E}\cdot\mathbf{v} 
-\nabla \times \mathbf{B} \cdot \mathbf{v}
+ \mathbf{J}\cdot \mathbf{v} \Big) 
\,d\mathbf{x}dt.
$$
This yields Ampère's law:
$$
\color{blue}{\nabla \times \mathbf{B} = \partial_t \mathbf{E} + \mathbf{J}}.\tag{4}
$$
Lastly, let $\mathbf{v} =\mathbf{0}$ in the test pair, integrating by parts one more time and using $\psi$'s boundary condition $(\dagger)$:
$$
0= \lim_{\epsilon \to 0}\frac{d}{d\epsilon} \mathcal{S}[(\phi,\mathbf{A}) + \epsilon(\psi,\mathbf{0})] \\
= \int^{t}_0\int_{\Omega} \Big((\partial_t \mathbf{A} + \nabla\phi)
\cdot \nabla \psi -\rho \psi\Big) 
\,d\mathbf{x}dt
\\
= \int^{t}_0\int_{\Omega} \Big(-\mathbf{E} \cdot \nabla \psi -\rho \psi\Big) 
\,d\mathbf{x}dt
\\
= \int^{t}_0\int_{\Omega} \Big(\nabla \cdot\mathbf{E}  \, \psi -\rho \psi\Big) 
\,d\mathbf{x}dt.
$$
We have reached the last piece in Maxwell's equations, Gauss's law for electric field:
$$
\color{blue}{\nabla \cdot\mathbf{E} = \rho}. \tag{5}
$$
Combining (2),(3),(4), and (5) yields Maxwell's equations. 
Remark 1: here all medium-related constants are set to $1$, also neglecting the speed of light $c$. Adding speed of light the Lagrangian should be
$$
L' = \frac{1}{2}\left|\frac{1}{c}\partial_t\mathbf{A} + \nabla \phi\right|^2 - \frac{1}{2}|\nabla \times \mathbf{A}|^2 - \rho\phi + \frac{1}{c}\mathbf{J}\cdot \mathbf{A}.
$$
Remark 2: Aside from substitution, the first one of the Euler-Lagrange equations obtained (equation (4)) in index notation should be:
$$
\underbrace{\color{blue}{\varepsilon_{nmi}(\varepsilon_{ijk} A_{k,j})_{,m}}}_{\nabla\times(\nabla \times \mathbf{A})} \color{blue}{+ }
\underbrace{ \color{blue}{\ddot{A}_n + \dot{\phi}_{,n} }}_{\partial_{tt}\mathbf{A} + \partial_t(\nabla\phi)} \color{blue}{= }\underbrace{\color{blue}{J_n}}_{\mathbf{J}}.\tag{6}
$$
Also if the Lagrangian you gave is w/o the factor of $1/2$, the equation above should have a factor of $2$ on the left side. 
If we assume the magnetic potential $\mathbf{A}$ is smooth (so we can interchange the derivative w.r.t space and time) and divergence free, then the second one of the Euler-Lagrange equations should be simply the Poisson equation: 
$$
\color{blue}{-\Delta \phi = \rho}.\tag{7}
$$
In (6) and (7), $\rho$ and $\mathbf{J}$ correspond to OP's $c$ and $d$ respectively.


I am confused by the dependent variables in this Lagrangian -- they are differentiated wrt to different variables, namely $\phi$ wrt spatial elements, whereas $A_i$ wrt time. 

The electric and magnetic potentials can be functions in space and/or time. In not-so-strict term, magnetic (electric) field changing along with time generating rotational electric (magnetic) field. 

Moreover, shouldn't $L$ also be a function of $A_{i,j}$?

$L$ is already a function $A_{i,j} := \partial_j A_i$, for it has a term of $\nabla \times \mathbf{A} $ whose $k$-th component is $\varepsilon_{kji} \partial_j A_{i}$.
