\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}