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\begin{frame}
\frametitle{Optimization}

\vspace{-1ex}
\begin{exampleblock}{}
A farmer has 2400ft of fencing and wants to fence a rectangular field
that borders a straight river. No fence needed along river.
\pause\medskip

What are the dimensions of the field with the largest area?
\pause\medskip

\begin{minipage}{.35\textwidth}
\begin{center}
\begin{tikzpicture}[default]
\draw[cred,ultra thick,dashed] (.5,0) rectangle (2.5,1);
\pause
\draw[decorate, decoration={snake}, draw=cblue, line width=2mm] (0,0) -- (3,0);
\mpause[2]{
\node[anchor=west] at (2.5,.5) {$h$};
}
\mpause[3]{
\node[anchor=south] at (1.5,1) {$w$};
}
\mpause[4]{
\node[anchor=south] at (1.5,.4) {$A$};
}
\end{tikzpicture}
\end{center}
\end{minipage}
\begin{minipage}{.64\textwidth}
\mpause[1]{
Introducing notation:
\begin{itemize}
\pause\pause
\item let $h$ be the height of the field
\pause
\item let $w$ be the width (parallel to river)
\pause
\item let $A$ be the area
\end{itemize}
}
\end{minipage}
\pause\smallskip

\mpause[0]{
What do we know?\vspace{-.7ex}
\begin{talign}
&2400 = \mpause[1]{2h + w} \mpause{\;\implies\; \mpause{w = 2400 - 2h}} \quad\mpause{\text{for $h$ in $[\mpause{0},\mpause[5]{1200}]$}} \\[-.5ex]
&\mpause[6]{A = } \mpause{hw} \mpause{ = h(2400 - 2h)} \mpause{ = 2400h - 2h^2} \quad\mpause{\text{for $h$ in $[0,1200]$}}
\end{talign}\vspace{-3ex}
\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause

$A$ is continuous on $[0,1200]$, we use the Closed Interval Method:\hspace{-3ex}\vspace{-.7ex}
\begin{talign}
&\mpause{A'(h) = } \mpause{2400 - 4h} \hspace{1cm} \mpause{A'(h) = 0 \;\iff\; h = 2400/4}\mpause{ = 600}
\end{talign}\vspace{-3ex}
\pause\pause\pause\pause

The value of $A$ at critical number $600$ and the interval ends are:\vspace{-.7ex}
\begin{talign}
A(0) = \mpause[1]{0} && A(600) = \mpause{600\cdot 1200} && A(1200) = \mpause{0}
\end{talign}\vspace{-3ex}
\pause\pause\pause\pause

The dimensions of the field are: $600$ft height, $1200$ft width.
}

\end{exampleblock}
\end{frame}