\begin{frame} \frametitle{Optimization} \vspace{-1ex} \begin{exampleblock}{} A man wants wants to get from point $A$ on one side of a $3$km wide river to point $B$, $8$km downstream on the opposite side. He can row $6$km/h and run $8$km/h. Where to land to be fastest? \pause\medskip \begin{minipage}{.3\textwidth} \begin{center} \begin{tikzpicture}[default,yscale=.95] \draw[draw=none,shade,left color=cblue!30,right color=cblue!50] (0,0) rectangle (.75,3); \draw[draw=none,shade,right color=cblue!30,left color=cblue!50] (.74,0) rectangle (1.5,3); \mpause[1]{ \draw (0,2.5) -- node[above,sloped] {row} (1.5,1.5) -- node[above,sloped] {run} (1.5,.5); } \mpause{ \draw[<->] (0,3.1) -- node[above] {$3km$} (1.5,3.1); } \mpause{ \draw[<->] (2.1,2.5) -- node[above,sloped] {$8km$} (2.1,.5); \draw[gray,dashed] (0,2.5) -- (2.1,2.5); \draw[gray,dashed] (1.5,.5) -- (2.1,0.5); } \mpause{} \mpause{ \node[include=cred] (c) at (1.5,1.5) {}; \node[anchor=west,xshift=1mm] at (c) {$C$}; } \mpause{ \draw[ultra thick,cred,<->,shorten <= 0mm, shorten >= 1.2mm] (1.5,2.5) -- node[right] {$x$} (1.5,1.5); } \node[include=cred] (a) at (0,2.5) {}; \node[anchor=east,xshift=-1mm] at (a) {$A$}; \node[include=cred] (b) at (1.5,.5) {}; \node[anchor=west,xshift=1mm] at (b) {$B$}; \end{tikzpicture} \end{center} \end{minipage}~% \begin{minipage}{.69\textwidth} \pause\pause\pause \mpause[1]{ Introducing notation: \begin{itemize} \pause\pause \item let $C$ be the landing point \pause \item let $x = $ downstream distance of $A$ to $C$ \end{itemize} \pause The time for rowing is and running:\vspace{-1ex} \begin{talign} t_{\text{row}}(x) &= \mpause[1]{(\sqrt{3^2 + x^2})/6} \\ \mpause{t_{\text{run}}(x) &=} \mpause{(8-x)/8} \end{talign}\vspace{-4ex} \pause\pause } \end{minipage} \pause\smallskip \mpause[1]{ \pause The total time is \quad $t(x) = t_{\text{row}}(x) + t_{\text{run}}(x)$ \quad for $x$ in $[\mpause[1]{0},\mpause[1]{8}]$ } \pause\pause\vspace{-.7ex} \begin{talign} t'(x) &= \mpause[1]{\frac{x}{6\sqrt{3^2 + x^2}} - \frac{1}{8}} \hspace{1cm} \mpause[8]{t'(x) = 0 \;\iff\; x = 9/\sqrt{7} } \only<-23>{ \\ \mpause[2]{t'(x) &= 0 \;\iff\;} \mpause{3\sqrt{3^2 + x^2} = 4x} \mpause{\;\stackrel{x\ge 0}{\iff}\; 9(3^2 + x^2) = 16x^2}\\[-.5ex] \mpause{&\;\iff\; 7x^2 = 81} \mpause{\;\iff\; x^2 = 81/7} \mpause{\;\stackrel{x\ge 0}{\iff}\; x = 9/\sqrt{7}} } \end{talign} \pause[25]\vspace{-2ex}% Now we apply the \pause Closed Interval Method:\vspace{-.5ex}\pause \begin{talign} t(0) &= \mpause[1]{1.5} & t(9/\sqrt{7}) &= \mpause[2]{1 + \sqrt{7}/8 \approx 1.33} & t(8) &= \mpause[3]{\sqrt{73}/6 \approx 1.42} \end{talign}\vspace{-2.5ex} \pause\pause\pause\pause Thus landing \alert{\mpause[1]{$9/\sqrt{7}$}km downstream} is the fastest. \end{exampleblock} \vspace{10cm} \end{frame}