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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Max-min problems}

\begin{example}
The fence
\end{example}
Farmer Egbert wants to fence off some of his farmyard for a pen for his chickens.
He has 100 feet of fencing, and plans to place one side of the pen against the
side of his barn. How should he design his pen? Like most pens, this one is
supposed to be rectangular. He wants to give his chickens as much room as possible. 

\vspace{0.3cm}
{\par\centering \includegraphics{pen.ps} \par}
\vspace{0.3cm}


\subsubsection*{Solution}

The first thing to do is to identify the quantities that are variable, and label
them as variables. In this case we can take the length of the side of the pen
extending from the barn outward as \( x \), and the length of the side of the
pen parallel to the side of the barn to be \( y \). 

\vspace{0.3cm}
{\par\centering \includegraphics{pen-measurements.ps} \par}
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The problem is to maximize the area \( A=xy \). There is a constraint, in that
the length of fencing is at most 100 feet (might as well use all the fencing,
so the length is 100 feet). So, \[
2x+y=100,\]
which is a constraint between the variables. If you then solve the constraint
equation for one of the variables (tradition suggests you solve for \( y \)
in terms of \( x \)), then substitute that formula for \( y \) into the area,
turning the area into a function of \( x \), we get \[
y=100-2x,\]
 and so \begin{eqnarray*}
A(x) & = & A\\
 & = & x(100-2x)\\
 & = & 100x-2x^{2}.
\end{eqnarray*}
Then, we find the maximum by looking for critical points:\begin{eqnarray*}
0 & = & A'(x)\\
 & = & 100-4x,
\end{eqnarray*}
or \( x=25. \)  

The maximum is either at this critical point, or at an endpoint of the interval,
so \[
x=0,\: x=25,\: \textrm{or}\: x=50.\]
The other two points are where \( x=0 \) (where the whole length of fencing
is against the barn wall --- not much of a pen, but these extreme cases are
useful in the analysis of possible extreme points), and \( x=50, \) where the
pen is \( 0 \) feet wide, again not having any area.

Since the maximum has to exist, and has to be at one of these points, the only
reasonable point is when \( x=25 \). 

Now, answer the original question. The question was to find how to design the
pen, that is, find the dimensions. Since the width \( x=25, \) and \( 2x+y=100 \),
\( y=50 \), so the pen extends 50 feet parallel to the barn wall, and comes
25 out from the side of the barn. 
\end{document}
