\begin{frame} \frametitle{Optimization} \vspace{-1ex} \begin{exampleblock}{} Find the point on the parabola $y^2 = 2x$ that is closest to $(1,4)$. \pause\medskip \begin{minipage}{.3\textwidth} \begin{center} \begin{tikzpicture}[default,scale=.5] \diagram{-.5}{4}{-4}{4}{1} \diagramannotatey{1} \diagramannotatex{1,2,3} \diagramannotatez \begin{scope}[ultra thick] \draw[cgreen] plot[smooth,domain=0:4,samples=40] function{sqrt(2*x)}; \draw[cgreen] plot[smooth,domain=0:4,samples=40] function{-sqrt(2*x)}; \end{scope} \node[include=cred] (a) at (1,4) {}; \node[anchor=west,xshift=1mm] at (a) {{\small $(1,4)$}}; \node[include=cgreen] (b) at (2,2) {}; \node[anchor=north west,xshift=1mm] at (b) {{\small $(x,y)$}}; \mpause[2]{ \draw[cred,dashed] (a) -- node[left,yshift=-1mm] {$d$} (b); } \end{tikzpicture} \end{center} \end{minipage} \begin{minipage}{.69\textwidth} \mpause[1]{ Introducing notation: \begin{itemize} \pause\pause \item let $d$ be the distance of $(x,y)$ to $(1,4)$ \end{itemize} \pause Then\vspace{-1ex} \begin{talign} d &= \mpause[1]{\sqrt{(x-1)^2 + (y-4)^2}} & \mpause{x &=}\mpause{ y^2/2 } \end{talign} \pause\pause\pause\pause Square root makes derivative complicated.\\\pause Note that $d$ minimal $\;\iff\;$ \pause $d^2$ minimal.\\\pause Thus, instead of $d$ we minimize $d^2$!\vspace{-.5ex}\pause \begin{talign} f(y) = d^2 = \mpause[1]{(y^2/2-1)^2 + (y-4)^2} \end{talign}\vspace{-3ex} } \end{minipage} \pause\mpause[1]{\pause \begin{talign} &f'(y) = \mpause[1]{2(y^2/2-1)y + 2(y-4)} \mpause{= y^3-8} \\ &\mpause{f'(y) = 0 \;\iff\;} \mpause{y = 2} \end{talign} \pause\pause\pause\pause\pause Moreover $f'(y) < 0$ for all $y < 2$ and $f'(y) > 0$ for all $y > 2$. \pause\smallskip Thus by the First Derivative Test for Absolute Extrema, $f(2)$ is the absolute minimum. \pause Thus the point \alert{$(\mpause[1]{2},\mpause[1]{2})$ is closest to $(1,4)$}. } \end{exampleblock} \vspace{10cm} \end{frame}