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\begin{frame}
\frametitle{Review - Midterm Exam 3}

\vspace{-1ex}
\begin{exampleblock}{}
A tin can is made to hold $1$L of oil.
Find the dimensions that minimize the cost of the metal to manufacture the can.
\medskip\pause

\begin{minipage}{.3\textwidth}
\begin{center}
\begin{tikzpicture}[default]
\draw[s] (0,0) rectangle (2,1.5);
\draw[s,draw=cblue!35] (1,0) ellipse (1 and .5);
\draw[s,draw=cblue!35] (1,1.5) ellipse (1 and .5);
\end{scope}
\pause
\begin{scope}[cred,dashed]
\draw (1,0) -- node[left] {$h$} (1,1.5);
\draw (1,0) -- node[below] {$r$} (2,0);
\end{scope}
\end{tikzpicture}
\end{center}
\end{minipage}
\begin{minipage}{.69\textwidth}
\mpause[1]{
Introducing notation:
\begin{itemize}
\pause\pause
\item let $h$ be the height
\pause
\item let $r$ be the radius
\pause
\item let $V$ be the volume
\pause
\item let $A$ be the surface area
\end{itemize}
}
\end{minipage}\vspace{-.3ex}
\pause
\mpause[0]{
\begin{talign}
&V = \mpause[1]{\pi r^2 h} \mpause{= 1} \mpause{\;\implies\; h = 1/(\pi r^2)} \\[-.5ex]
&\mpause{A = }\mpause{2\pi r^2 + 2\pi r h} \mpause{= 2\pi r^2 + 2/r} \quad\quad \text{\mpause{for $r$ in }\mpause{$(0,\infty)$}} \\[-.5ex]
&\mpause{A'(r) = }\mpause{4\pi r - 2/r^2} \mpause{= (4\pi r^3 - 2)/r^2} \\[-.5ex]
&\mpause{A'(r) = 0 \;\iff\; r = }\mpause{1/\sqrt[3]{2\pi}}
\mpause{\text{\ \ is the only critical number}}
\end{talign}\vspace{-3ex}
\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause

Cannot use Closed Interval Method since $(0,\infty)$ is not closed.
\pause\smallskip

However, $A(1/\sqrt[3]{2\pi})$ must be the \emph{absolute minimum}\only<-27>{ since:
\begin{itemize}
\pause
\item $A$ is decreasing, $A'(r) < 0$, for all $r < 1/\sqrt[3]{2\pi}$,
\pause
\item $A$ is increasing, $A'(r) > 0$, for all $r > 1/\sqrt[3]{2\pi}$.
\end{itemize}}
\pause\pause\pause\pause\smallskip

Then $h = 1/(\pi r^2) \pause= \sqrt[3]{2\pi}^2/\pi \pause= \sqrt[3]{4\pi^2/\pi^3} \pause= 2 / \sqrt[3]{2\pi} \pause= 2r$
\pause\smallskip

Hence \alert{radius $r = 1/\sqrt[3]{2\pi}$} and \alert{height $h = 2r$} minimizes the cost.
}
\end{exampleblock}
\vspace{10cm}
\end{frame}