\begin{frame} \frametitle{Optimization} \vspace{-1ex} \begin{exampleblock}{} Find the area of the largest rectangle that can be inscribed in a semi-circle circle of radius $r$. \pause\medskip \begin{minipage}{.45\textwidth} \begin{center} \begin{tikzpicture}[default,scale=.45,nodes={scale=.9}] \diagram{-3.5}{3.5}{-.5}{3.5}{1} \diagramannotatexx{-3/-r,3/r} \diagramannotatez \begin{scope}[ultra thick] \draw[cgreen] plot[smooth,domain=-3:3,samples=100] function{sqrt(9-x**2)}; \mpause[1]{ \draw[cred] (-2,0) rectangle (2,{pow(9-2*2,.5)}); } \end{scope} \mpause[3]{ \node[include=cred] (a) at (2,{pow(9-2*2,.5)}) {}; \node[anchor=south west] at (a) {$(x,y)$}; } \end{tikzpicture} \end{center} \end{minipage}~% \begin{minipage}{.54\textwidth} \pause \mpause[1]{ Introducing notation: \begin{itemize} \pause\pause \item let $(x,y)$ be the upper right corner of the rectangle \pause \item let $A$ be the area \end{itemize} } \end{minipage} \smallskip \mpause[1]{ \pause The area is \quad $ A(x) = \mpause[1]{2xy} \mpause[2]{= 2x\sqrt{r^2-x^2}} \mpause[3]{\quad\quad\text{for $x$ in $[\mpause[4]{0},\mpause[4]{r}]$}} $ \pause\pause\pause\pause\smallskip $A$ is continuous on $[0,r]$, we use the Closed Interval Method:\vspace{-.7ex} \begin{talign} A'(x) &= \mpause[1]{ 2\sqrt{r^2-x^2} + \frac{2x}{2\sqrt{r^2-x^2}}(-2x) } \mpause{= \frac{2(r^2-2x^2)}{\sqrt{r^2-x^2}} } \\[-.5ex] \mpause{A'(x) &= 0 \;\iff\; }\mpause{x^2 = r^2/2} \mpause{\stackrel{x\ge 0}{\iff} x = r/\sqrt{2}} \end{talign} \pause\pause\pause\pause\pause\pause Note that $A(0) = \mpause[1]{0}$ and $A(r) = \mpause[1]{0}$. \pause\pause Thus the \alert{maximum area} is:\vspace{-.7ex} \begin{talign} \alert{A(r/\sqrt{2}) =} \mpause[1]{ 2\frac{r}{\sqrt{2}}\sqrt{r^2 - \frac{r^2}{\sqrt{2}^2}} } \mpause[2]{ = \sqrt{2}r\sqrt{\frac{r^2}{2}} } \mpause[3]{ = \alert{r^2} } \end{talign}\vspace{-1ex} } \end{exampleblock} \vspace{10cm} \end{frame}