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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}
A cardboard box 
\end{example}
Find the dimensions of the largest cardboard box with a square base which uses
only 12 square feet of cardboard. For a cardboard box, remember that the top
and bottom have two layers of cardboard (for the flaps), while the sides are
a single layer. 

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

\begin{rem}
This is typical of real problems, in that the problem is poorly phrased. Largest
how? Largest in \textit{volume}. 
\end{rem}

\subsubsection*{Solution }

Begin by drawing a sketch of the box, and then label the sides which might vary,
as say \( x \), \( y \), and \( z \), where \( z \) is the height. 

\vspace{0.3cm}
{\par\centering \includegraphics{box-measurements.ps} \par}
\vspace{0.3cm}

We have to maximize the \textit{volume} \( V=xyz \). But, there are in this
example 2 constraints, to get us down to a function of one variable. The simpler
constraint is that \( x=y \), since the base is square. The other is in terms
of the amount of cardboard to be used. If \( S \) is the amount of cardboard,\[
S=4xy+2xz+2yx=4x^{2}+4xz,\]
 where the fact that the top and bottom have a double layer is included in the
\( 4xy \) term, and the right and left sides, and front and back, account for
the other terms. The last equation comes from setting \( y=x \). 

Now, you are given that \( S=12 \), so\[
12=4x^{2}+4xz,\]
or, solving for \( z \), \( z=(3-x^{2})/x=3/x-x \). Also from that constraint
comes the fact that \( 0\leq x\leq 3 \). \( x \) is a physical length, so
can't be negative, and \( x\leq 3 \) since, if \( x>3 \), the area of cardboard
required just for the flaps would be larger than the amount alloted. 

Substitute these constraints in the original thing we wanted to optimize, and\[
V=xyz=xx(3/x-x)=3x-x^{3}=f(x)\]
 is the function you have to maximize. Differentiate \( f \), and set it to
0, and:\[
0=f'(x)=3-3x^{2},\]
so \( x=1 \). Not \( \pm 1 \), since \( -1 \) is not a physical length.  

This value of \( x \), \( x=1 \), is the place where the maximum occurs. Why?
Because the only place where there \emph{can} be a maximum is where the derivative
is 0, where the derivative doesn't exist, or the endpoints. Also, if there are
endpoints of the interval that make even halfway sense, there \textit{must be}
a maximum. In this case the endpoints, \( x=0 \) and \( x=3 \), both give
\( V=0 \). So, \( x=1 \), giving \( V=2 \), must be the maximum. You find
out which of the possible points is the optimum you want by evaluating the function
at all the possible points. 

By the way, you then should re-read the question to find out what is asked for.
In this case, it asks for the dimensions of the box: \( x=y=1 \) you already
have. You then solve for \( z \), using the constraint, \( z=3/1-1=2 \). 
\end{document}
