94/171
\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}