Answer
We show that there is a constant $\mu $ such that the maximum of $S$ subject to the constraints occurs for ${x_i} = {A^{ - 1}}{{\rm{e}}^{\mu {E_i}}}$ for $i = 1,2,...n$,
where $A = {N^{ - 1}}\left( {{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}} \right)$.
Work Step by Step
Our task is to maximize
$S\left( {{x_1},...,{x_n}} \right) = {x_1}\ln {x_1} + {x_2}\ln {x_2} + \cdot\cdot\cdot + {x_n}\ln {x_n}$
subject to two constraints:
$g\left( {{x_1},...,{x_n}} \right) = {x_1} + {x_2} + \cdot\cdot\cdot + {x_n} - N = 0$
$h\left( {{x_1},...,{x_n}} \right) = {E_1}{x_1} + {E_2}{x_2} + \cdot\cdot\cdot + {E_n}{x_n} - E = 0$,
where $E$, ${E_1}$, ${E_2}$, ..., ${E_n}$ and $N$ are constants.
The Lagrange condition is
$\nabla S = \lambda \nabla g + \mu \nabla h$
So,
$\left( {\ln {x_1} + 1,\ln {x_2} + 1,...,\ln {x_n} + 1} \right) = \lambda \left( {1,1,...,1} \right) + \mu \left( {{E_1},{E_2},...,{E_n}} \right)$
From the Lagrange condition we obtain $n$ equations:
$\ln {x_1} + 1 = \lambda + \mu {E_1}$,
$\ln {x_2} + 1 = \lambda + \mu {E_2}$,
$\cdot\cdot\cdot$,
$\ln {x_n} + 1 = \lambda + \mu {E_n}$
Or,
$\lambda = 1 + \ln {x_1} - \mu {E_1}$,
$\lambda = 1 + \ln {x_2} - \mu {E_2}$,
$\cdot\cdot\cdot$,
$\lambda = 1 + \ln {x_n} - \mu {E_n}$
So,
$\lambda = 1 + \ln {x_1} - \mu {E_1} = 1 + \ln {x_2} - \mu {E_2} = \cdot\cdot\cdot = 1 + \ln {x_n} - \mu {E_n}$
$\ln {x_1} - \mu {E_1} = \ln {x_2} - \mu {E_2} = \cdot\cdot\cdot = \ln {x_n} - \mu {E_n}$
It follows that
$\begin{array}{*{20}{c}}
{\ln {x_1} - \mu {E_1} = \ln {x_2} - \mu {E_2}}\\
{\ln {x_1} - \mu {E_1} = \ln {x_3} - \mu {E_3}}\\
{\cdot\cdot\cdot}\\
{\ln {x_1} - \mu {E_1} = \ln {x_n} - \mu {E_n}}
\end{array}$ $ \Rightarrow $ $\begin{array}{*{20}{c}}
{\ln {x_1} - \ln {x_2} = \mu \left( {{E_1} - {E_2}} \right)}\\
{\ln {x_1} - \ln {x_3} = \mu \left( {{E_1} - {E_3}} \right)}\\
{\cdot\cdot\cdot}\\
{\ln {x_1} - \ln {x_n} = \mu \left( {{E_1} - {E_n}} \right)}
\end{array}$
So,
$\begin{array}{*{20}{c}}
{\ln \frac{{{x_1}}}{{{x_2}}} = \mu \left( {{E_1} - {E_2}} \right)}\\
{\ln \frac{{{x_1}}}{{{x_3}}} = \mu \left( {{E_1} - {E_3}} \right)}\\
{\cdot\cdot\cdot}\\
{\ln \frac{{{x_1}}}{{{x_n}}} = \mu \left( {{E_1} - {E_n}} \right)}
\end{array}$ $ \Rightarrow $ $\begin{array}{*{20}{c}}
{\frac{{{x_1}}}{{{x_2}}} = {{\rm{e}}^{\mu \left( {{E_1} - {E_2}} \right)}}}\\
{\frac{{{x_1}}}{{{x_3}}} = {{\rm{e}}^{\mu \left( {{E_1} - {E_3}} \right)}}}\\
{\cdot\cdot\cdot}\\
{\frac{{{x_1}}}{{{x_n}}} = {{\rm{e}}^{\mu \left( {{E_1} - {E_n}} \right)}}}
\end{array}$
From the $2$nd, the $3$rd, ... and the $n$th equation above we obtain
(1) ${\ \ \ \ }$ ${x_2} = {x_1}{{\rm{e}}^{\mu \left( {{E_2} - {E_1}} \right)}}$, ${\ \ \ }$ ${x_3} = {x_1}{{\rm{e}}^{\mu \left( {{E_3} - {E_1}} \right)}}$,
$\cdot\cdot\cdot$,
${x_n} = {x_1}{{\rm{e}}^{\mu \left( {{E_n} - {E_1}} \right)}}$
Substituting ${x_2} = {x_1}{{\rm{e}}^{\mu \left( {{E_2} - {E_1}} \right)}}$, ${x_3} = {x_1}{{\rm{e}}^{\mu \left( {{E_3} - {E_1}} \right)}}$, ..., and ${x_n} = {x_1}{{\rm{e}}^{\mu \left( {{E_n} - {E_1}} \right)}}$ in the constraint $g$ gives
${x_1} + {x_1}{{\rm{e}}^{\mu \left( {{E_2} - {E_1}} \right)}} + {x_1}{{\rm{e}}^{\mu \left( {{E_3} - {E_1}} \right)}} + \cdot\cdot\cdot + {x_1}{{\rm{e}}^{\mu \left( {{E_n} - {E_1}} \right)}} - N = 0$
${x_1}\left( {1 + {{\rm{e}}^{\mu \left( {{E_2} - {E_1}} \right)}} + {{\rm{e}}^{\mu \left( {{E_3} - {E_1}} \right)}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu \left( {{E_n} - {E_1}} \right)}}} \right) = N$
${x_1} = \frac{N}{{1 + {{\rm{e}}^{\mu \left( {{E_2} - {E_1}} \right)}} + {{\rm{e}}^{\mu \left( {{E_3} - {E_1}} \right)}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu \left( {{E_n} - {E_1}} \right)}}}}$
Multiplying the right-hand side with the identity $\frac{{{{\rm{e}}^{\mu {E_1}}}}}{{{{\rm{e}}^{\mu {E_1}}}}} = 1$ gives
${x_1} = \frac{{N{{\rm{e}}^{\mu {E_1}}}}}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}}$
Using ${x_2} = {x_1}{{\rm{e}}^{\mu \left( {{E_2} - {E_1}} \right)}}$, ${x_3} = {x_1}{{\rm{e}}^{\mu \left( {{E_3} - {E_1}} \right)}}$, ..., and ${x_n} = {x_1}{{\rm{e}}^{\mu \left( {{E_n} - {E_1}} \right)}}$ we obtain
${x_2} = \frac{{N{{\rm{e}}^{\mu {E_1}}}}}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}}{{\rm{e}}^{\mu \left( {{E_2} - {E_1}} \right)}} = \frac{{N{{\rm{e}}^{\mu {E_2}}}}}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}}$
${x_3} = \frac{{N{{\rm{e}}^{\mu {E_1}}}}}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}}{{\rm{e}}^{\mu \left( {{E_3} - {E_1}} \right)}} = \frac{{N{{\rm{e}}^{\mu {E_3}}}}}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}}$
$\cdot\cdot\cdot$
${x_n} = \frac{{N{{\rm{e}}^{\mu {E_1}}}}}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}}{{\rm{e}}^{\mu \left( {{E_n} - {E_1}} \right)}} = \frac{{N{{\rm{e}}^{\mu {E_n}}}}}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}}$
So, the critical point is
$\left( {{x_1},{x_2},...,{x_n}} \right) = $
$\left( {\frac{{N{{\rm{e}}^{\mu {E_1}}}}}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}},\frac{{N{{\rm{e}}^{\mu {E_2}}}}}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}},...,\frac{{N{{\rm{e}}^{\mu {E_n}}}}}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}}} \right)$
Write $A = \frac{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}}{N}$. So, ${A^{ - 1}} = \frac{N}{{{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}}}$.
Thus,
$\left( {{x_1},{x_2},...,{x_n}} \right) = \left( {{A^{ - 1}}{{\rm{e}}^{\mu {E_1}}},{A^{ - 1}}{{\rm{e}}^{\mu {E_2}}},...,{A^{ - 1}}{{\rm{e}}^{\mu {E_n}}}} \right)$
Hence, there is a constant $\mu $ such that ${x_i} = {A^{ - 1}}{{\rm{e}}^{\mu {E_i}}}$ for $i = 1,2,...n$, where $A = {N^{ - 1}}\left( {{{\rm{e}}^{\mu {E_1}}} + {{\rm{e}}^{\mu {E_2}}} + \cdot\cdot\cdot + {{\rm{e}}^{\mu {E_n}}}} \right)$.
Since the domain of logarithmic function are positive numbers, the function $S$ is increasing. Thus, we conclude that the maximum of $S\left( {{x_1},{x_2},...,{x_n}} \right)$ subject to the two constraints occurs at the critical point:
$\left( {{x_1},{x_2},...,{x_n}} \right) = \left( {{A^{ - 1}}{{\rm{e}}^{\mu {E_1}}},{A^{ - 1}}{{\rm{e}}^{\mu {E_2}}},...,{A^{ - 1}}{{\rm{e}}^{\mu {E_n}}}} \right)$
