\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}