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I am using wxMaxima for doing homework. I have extremely long mathematical equations. I exported my work to a .tex file. But now I have several overfull bad box warnings. I tried to use geometry package to reduce margin to almost nothing and even changed the paper to landscape. Still I am not able to make the equations fit. I am not able to figure out which LaTeX tag is preventing hyphenation from working in the math environment. Can someone suggest a solution smarter than manually fixing each of the overfull boxes?

My file looks like this:

\documentclass{article}

%% Created with wxMaxima 13.04.2

\setlength{\parskip}{\medskipamount}
\setlength{\parindent}{0pt}
\usepackage[utf8x]{inputenc}
\usepackage{graphicx}
\usepackage{color}
\usepackage{amsmath}
\usepackage{geometry}
\geometry{legalpaper, landscape, margin=0.1in}

\definecolor{labelcolor}{RGB}{100,0,0}

\begin{document}

\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i12) 
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
kill(all)$
u : A + (u0 - A)*exp(-x/%lambda);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o1) }
A+{e}^{−\frac{x}{\lambda}}\,\left( u0−A\right) 
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i2) 
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
assume(L > 0)$
I : (k/2)*(u/l)^2 + (%tau/2)*(diff(u,x))^2 + (%tau)*(u/l);
E : integrate(I,x,0,L);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o3) }
\frac{k\,{\left( A+{e}^{−\frac{x}{\lambda}}\,\left( u0−A\right) \right) }^{2}}{2\,{l}^{2}}+\frac{\tau\,\left( A+{e}^{−\frac{x}{\lambda}}\,\left( u0−A\right) \right) }{l}+\frac{\tau\,{e}^{−\frac{2\,x}{\lambda}}\,{\left( u0−A\right) }^{2}}{2\,{\lambda}^{2}}
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o4) }
({e}^{−\frac{2\,L}{\lambda}}\,(\left( \left( 2\,\lambda\,k\,{A}^{2}+4\,\lambda\,\tau\,l\,A\right) \,L+\left( \tau\,{l}^{2}−3\,{\lambda}^{2}\,k\right) \,{A}^{2}+\left( \left( 2\,{\lambda}^{2}\,k−2\,\tau\,{l}^{2}\right) \,u0−4\,{\lambda}^{2}\,\tau\,l\right) \,A+\left( \tau\,{l}^{2}+{\lambda}^{2}\,k\right) \,{u0}^{2}+4\,{\lambda}^{2}\,\tau\,l\,u0\right) \,{e}^{\frac{2\,L}{\lambda}}+\left( 4\,{\lambda}^{2}\,k\,{A}^{2}+\left( 4\,{\lambda}^{2}\,\tau\,l−4\,{\lambda}^{2}\,k\,u0\right) \,A−4\,{\lambda}^{2}\,\tau\,l\,u0\right) \,{e}^{\frac{L}{\lambda}}+\left( −\tau\,{l}^{2}−{\lambda}^{2}\,k\right) \,{A}^{2}+\left( 2\,\tau\,{l}^{2}+2\,{\lambda}^{2}\,k\right) \,u0\,A+\left( −\tau\,{l}^{2}−{\lambda}^{2}\,k\right) \,{u0}^{2}))/(4\,\lambda\,{l}^{2})
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i5) 
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
eq1 : diff(E,A) = 0;
eq2 : diff(E,%lambda) = 0;
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o5) }
\frac{{e}^{−\frac{2\,L}{\lambda}}\,\left( \left( \left( 4\,\lambda\,k\,A+4\,\lambda\,\tau\,l\right) \,L+2\,\left( \tau\,{l}^{2}−3\,{\lambda}^{2}\,k\right) \,A+\left( 2\,{\lambda}^{2}\,k−2\,\tau\,{l}^{2}\right) \,u0−4\,{\lambda}^{2}\,\tau\,l\right) \,{e}^{\frac{2\,L}{\lambda}}+\left( 8\,{\lambda}^{2}\,k\,A−4\,{\lambda}^{2}\,k\,u0+4\,{\lambda}^{2}\,\tau\,l\right) \,{e}^{\frac{L}{\lambda}}+2\,\left( −\tau\,{l}^{2}−{\lambda}^{2}\,k\right) \,A+\left( 2\,\tau\,{l}^{2}+2\,{\lambda}^{2}\,k\right) \,u0\right) }{4\,\lambda\,{l}^{2}}=0
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o6) }
({e}^{−\frac{2\,L}{\lambda}}\,(−\frac{2\,L\,\left( \left( 2\,\lambda\,k\,{A}^{2}+4\,\lambda\,\tau\,l\,A\right) \,L+\left( \tau\,{l}^{2}−3\,{\lambda}^{2}\,k\right) \,{A}^{2}+\left( \left( 2\,{\lambda}^{2}\,k−2\,\tau\,{l}^{2}\right) \,u0−4\,{\lambda}^{2}\,\tau\,l\right) \,A+\left( \tau\,{l}^{2}+{\lambda}^{2}\,k\right) \,{u0}^{2}+4\,{\lambda}^{2}\,\tau\,l\,u0\right) \,{e}^{\frac{2\,L}{\lambda}}}{{\lambda}^{2}}+\left( \left( 2\,k\,{A}^{2}+4\,\tau\,l\,A\right) \,L−6\,\lambda\,k\,{A}^{2}+\left( 4\,\lambda\,k\,u0−8\,\lambda\,\tau\,l\right) \,A+2\,\lambda\,k\,{u0}^{2}+8\,\lambda\,\tau\,l\,u0\right) \,{e}^{\frac{2\,L}{\lambda}}−\frac{\left( 4\,{\lambda}^{2}\,k\,{A}^{2}+\left( 4\,{\lambda}^{2}\,\tau\,l−4\,{\lambda}^{2}\,k\,u0\right) \,A−4\,{\lambda}^{2}\,\tau\,l\,u0\right) \,L\,{e}^{\frac{L}{\lambda}}}{{\lambda}^{2}}+\left( 8\,\lambda\,k\,{A}^{2}+\left( 8\,\lambda\,\tau\,l−8\,\lambda\,k\,u0\right) \,A−8\,\lambda\,\tau\,l\,u0\right) \,{e}^{\frac{L}{\lambda}}−2\,\lambda\,k\,{A}^{2}+4\,\lambda\,k\,u0\,A−2\,\lambda\,k\,{u0}^{2}))/(4\,\lambda\,{l}^{2})+(L\,{e}^{−\frac{2\,L}{\lambda}}\,(\left( \left( 2\,\lambda\,k\,{A}^{2}+4\,\lambda\,\tau\,l\,A\right) \,L+\left( \tau\,{l}^{2}−3\,{\lambda}^{2}\,k\right) \,{A}^{2}+\left( \left( 2\,{\lambda}^{2}\,k−2\,\tau\,{l}^{2}\right) \,u0−4\,{\lambda}^{2}\,\tau\,l\right) \,A+\left( \tau\,{l}^{2}+{\lambda}^{2}\,k\right) \,{u0}^{2}+4\,{\lambda}^{2}\,\tau\,l\,u0\right) \,{e}^{\frac{2\,L}{\lambda}}+\left( 4\,{\lambda}^{2}\,k\,{A}^{2}+\left( 4\,{\lambda}^{2}\,\tau\,l−4\,{\lambda}^{2}\,k\,u0\right) \,A−4\,{\lambda}^{2}\,\tau\,l\,u0\right) \,{e}^{\frac{L}{\lambda}}+\left( −\tau\,{l}^{2}−{\lambda}^{2}\,k\right) \,{A}^{2}+\left( 2\,\tau\,{l}^{2}+2\,{\lambda}^{2}\,k\right) \,u0\,A+\left( −\tau\,{l}^{2}−{\lambda}^{2}\,k\right) \,{u0}^{2}))/(2\,{\lambda}^{3}\,{l}^{2})−({e}^{−\frac{2\,L}{\lambda}}\,(\left( \left( 2\,\lambda\,k\,{A}^{2}+4\,\lambda\,\tau\,l\,A\right) \,L+\left( \tau\,{l}^{2}−3\,{\lambda}^{2}\,k\right) \,{A}^{2}+\left( \left( 2\,{\lambda}^{2}\,k−2\,\tau\,{l}^{2}\right) \,u0−4\,{\lambda}^{2}\,\tau\,l\right) \,A+\left( \tau\,{l}^{2}+{\lambda}^{2}\,k\right) \,{u0}^{2}+4\,{\lambda}^{2}\,\tau\,l\,u0\right) \,{e}^{\frac{2\,L}{\lambda}}+\left( 4\,{\lambda}^{2}\,k\,{A}^{2}+\left( 4\,{\lambda}^{2}\,\tau\,l−4\,{\lambda}^{2}\,k\,u0\right) \,A−4\,{\lambda}^{2}\,\tau\,l\,u0\right) \,{e}^{\frac{L}{\lambda}}+\left( −\tau\,{l}^{2}−{\lambda}^{2}\,k\right) \,{A}^{2}+\left( 2\,\tau\,{l}^{2}+2\,{\lambda}^{2}\,k\right) \,u0\,A+\left( −\tau\,{l}^{2}−{\lambda}^{2}\,k\right) \,{u0}^{2}))/(4\,{\lambda}^{2}\,{l}^{2})=0
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i7) 
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
eq1 : expand(eq1/L);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o7) }
\frac{2\,\lambda\,k\,A\,{e}^{−\frac{L}{\lambda}}}{{l}^{2}\,L}−\frac{\lambda\,k\,u0\,{e}^{−\frac{L}{\lambda}}}{{l}^{2}\,L}+\frac{\lambda\,\tau\,{e}^{−\frac{L}{\lambda}}}{l\,L}−\frac{\lambda\,k\,A\,{e}^{−\frac{2\,L}{\lambda}}}{2\,{l}^{2}\,L}−\frac{\tau\,A\,{e}^{−\frac{2\,L}{\lambda}}}{2\,\lambda\,L}+\frac{\lambda\,k\,u0\,{e}^{−\frac{2\,L}{\lambda}}}{2\,{l}^{2}\,L}+\frac{\tau\,u0\,{e}^{−\frac{2\,L}{\lambda}}}{2\,\lambda\,L}−\frac{3\,\lambda\,k\,A}{2\,{l}^{2}\,L}+\frac{\tau\,A}{2\,\lambda\,L}+\frac{\lambda\,k\,u0}{2\,{l}^{2}\,L}−\frac{\tau\,u0}{2\,\lambda\,L}−\frac{\lambda\,\tau}{l\,L}+\frac{k\,A}{{l}^{2}}+\frac{\tau}{l}=0
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i8) 
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
assume(%lambda > 0)$
eq1 : limit(eq1,L,inf);
eq2 : limit(eq2,L,inf);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o9) }
\frac{k\,A}{{l}^{2}}+\frac{\tau}{l}=0
\end{math}

\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o10) }
−\frac{\left( \tau\,{l}^{2}+3\,{\lambda}^{2}\,k\right) \,{A}^{2}+\left( \left( −2\,\tau\,{l}^{2}−2\,{\lambda}^{2}\,k\right) \,u0+4\,{\lambda}^{2}\,\tau\,l\right) \,A+\left( \tau\,{l}^{2}−{\lambda}^{2}\,k\right) \,{u0}^{2}−4\,{\lambda}^{2}\,\tau\,l\,u0}{4\,{\lambda}^{2}\,{l}^{2}}=0
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i11) 
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
eq2 : expand(eq2);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o11) }
−\frac{3\,k\,{A}^{2}}{4\,{l}^{2}}−\frac{\tau\,{A}^{2}}{4\,{\lambda}^{2}}+\frac{k\,u0\,A}{2\,{l}^{2}}+\frac{\tau\,u0\,A}{2\,{\lambda}^{2}}−\frac{\tau\,A}{l}+\frac{k\,{u0}^{2}}{4\,{l}^{2}}−\frac{\tau\,{u0}^{2}}{4\,{\lambda}^{2}}+\frac{\tau\,u0}{l}=0
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i12) 
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
facsum_combine : false$
eq2 : facsum(eq2,A,%lambda);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o13) }
−\frac{3\,k\,{A}^{2}}{4\,{l}^{2}}−\frac{\tau\,{A}^{2}}{4\,{\lambda}^{2}}+\frac{\left( k\,u0−2\,\tau\,l\right) \,A}{2\,{l}^{2}}+\frac{\tau\,u0\,A}{2\,{\lambda}^{2}}−\frac{\tau\,{u0}^{2}}{4\,{\lambda}^{2}}+\frac{u0\,\left( k\,u0+4\,\tau\,l\right) }{4\,{l}^{2}}=0
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i14) 
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
solve(eq1,A);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o14) }
[A=−\frac{\tau\,l}{k}]
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i15) 
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
eq2 : subst(−(%tau*l)/k,A,eq2);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o15) }
−\frac{\tau\,{u0}^{2}}{4\,{\lambda}^{2}}+\frac{u0\,\left( k\,u0+4\,\tau\,l\right) }{4\,{l}^{2}}−\frac{\tau\,\left( k\,u0−2\,\tau\,l\right) }{2\,k\,l}−\frac{{\tau}^{2}\,l\,u0}{2\,{\lambda}^{2}\,k}−\frac{{\tau}^{3}\,{l}^{2}}{4\,{\lambda}^{2}\,{k}^{2}}−\frac{3\,{\tau}^{2}}{4\,k}=0
\end{math}
%%%%%%%%%%%%%%%


\noindent
%%%%%%%%%%%%%%%
%%% INPUT:
\begin{minipage}[t]{8ex}{\color{red}\bf
\begin{verbatim}
(%i16) 
\end{verbatim}}
\end{minipage}
\begin{minipage}[t]{\textwidth}{\color{blue}
\begin{verbatim}
solve(eq2,%lambda);
\end{verbatim}}
\end{minipage}
%%% OUTPUT:
\begin{math}\displaystyle
\parbox{8ex}{\color{labelcolor}(\%o16) }
[\lambda=−\sqrt{\frac{\tau}{k}}\,l,\lambda=\sqrt{\frac{\tau}{k}}\,l]
\end{math}
%%%%%%%%%%%%%%%

\end{document}
share|improve this question
    
You either need to split the content manually (and use some of the alignment techniques/environments provided by amsmath), or you could consider looking into breqn. –  Werner Apr 24 '14 at 19:24
    
@Werner: I tried breqn. I got an "memory exceeded" fatal error. Seems I cannot avoid the hard work of manually splitting the lines. I am worried because the \left and \right will not work properly if I insert a newline in between –  Amit Singh Apr 24 '14 at 23:02
    

1 Answer 1

up vote 3 down vote accepted

You either need to split the content manually (and use some of the alignment techniques/environments provided by amsmath), or you could consider looking into breqn. Ultimately the former will give you more control over the alignment and spacing. As examples of this and the possibilities that exist, see

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