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%TCIDATA{Created=Tuesday, September 11, 2007 11:02:45}
%TCIDATA{LastRevised=Friday, September 14, 2007 12:21:49}
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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}


\begin{eqnarray*}
\frac{1}{c_{1}} &=&\lambda \left( 1+\tau _{1}\right)  \\
\frac{1}{l_{1}} &=&\lambda w\left( 1-\tau _{l}\right)  \\
\beta \frac{1}{c_{2}} &=&\lambda \frac{1+\tau _{2}}{1+r} \\
\beta \frac{1}{l_{2}} &=&\lambda w\frac{1-\tau _{l}}{1+r} \\
c_{1}\left( 1+\tau _{1}\right) +\frac{c_{2}\left( 1+\tau _{1}\right) }{1+r}
&=&w\left( 1-l_{1}\right) \left( 1-\tau _{1}\right) +\frac{w\left(
1-l_{2}\right) \left( 1-\tau _{1}\right) }{1+r}
\end{eqnarray*}%
, Solution is: 

\begin{eqnarray*}
\frac{l_{1}}{c_{1}}w &=&\frac{1+\tau _{1}}{1-\tau _{l}} \\
\frac{c_{2}}{c_{1}\beta \left( 1+r\right) } &=&\frac{1+\tau _{1}}{1+\tau _{2}%
} \\
\frac{l_{2}}{c_{2}}w &=&\frac{1+\tau _{2}}{1-\tau _{l}} \\
c_{1}+\frac{c_{2}}{1+r}\frac{1+\tau _{2}}{\left( 1+\tau _{1}\right) }
&=&w\left( 1-l_{1}\right) \frac{1-\tau _{l}}{1+\tau _{1}}+\frac{w\left(
1-l_{2}\right) }{1+r}\frac{1-\tau _{l}}{1+\tau _{2}}\frac{1+\tau _{2}}{%
\left( 1+\tau _{1}\right) } \\
&&
\end{eqnarray*}%
, Solution is: $\left\{ 
\begin{array}{c}
c_{1}=w\frac{\left( 1-\tau _{l}\right) \left( 2+r\right) }{2\left( 1+\beta
\right) \left( 1+r\right) \left( 1+\tau _{1}\right) } \\ 
c_{2}=w\frac{\beta \left( 1-\tau _{l}\right) \left( 2+r\right) }{2\left(
1+\tau _{2}\right) \left( 1+\beta \right) } \\ 
l_{2}=\frac{1}{2}\frac{\beta \left( 2+r\right) }{\beta +1} \\ 
l_{1}=\frac{1}{2}\frac{2+r}{\left( 1+\beta \right) \left( 1+r\right) }%
\end{array}%
\right\} \allowbreak $ 

\[
\left[ c_{1}+\frac{c_{2}}{1+r}+G=w\left( 1-l_{1}\right) +\frac{w\left(
1-l_{2}\right) }{1+r}\right] _{c_{1}=w\frac{\left( 1-\tau _{l}\right) \left(
2+r\right) }{2\left( 1+\beta \right) \left( 1+r\right) \left( 1+\tau
_{1}\right) },c_{2}=w\frac{\beta \left( 1-\tau _{l}\right) \left( 2+r\right) 
}{2\left( 1+\tau _{2}\right) \left( 1+\beta \right) },l_{2}=\frac{1}{2}\frac{%
\beta \left( 2+r\right) }{\beta +1},l_{1}=\frac{1}{2}\frac{2+r}{\left(
1+\beta \right) \left( 1+r\right) }}
\]%
\[
\frac{d\left( -\frac{\left( 1-\tau _{l}\right) }{\left( 1+\tau _{1}\right) }%
\frac{w\left( 2+r\right) }{2\left( 1+\beta \right) \left( 1+r\right) }-\frac{%
\beta \frac{\left( 1-\tau _{l}\right) }{\left( 1+\tau _{2}\right) }\frac{%
w\left( 2+r\right) }{2\left( 1+\beta \right) }}{1+r}\frac{1+\tau _{2}}{%
1+\tau _{1}}+w\left( 1-\frac{2+r}{2\left( 1+r\right) \left( 1+\beta \right) }%
\right) \frac{1-\tau _{l}}{1+\tau _{1}}+\frac{w\left( 1-\frac{\beta \left(
2+r\right) }{2\left( 1+\beta \right) }\right) }{1+r}\frac{1-\tau _{l}}{%
1+\tau _{2}}\frac{1+\tau _{2}}{1+\tau _{1}}\right) }{d\tau _{1}}
\]

\end{document}
