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