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\begin{document}
%MODULE 1: Optimization


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%INPUT THE INSTRUCTIONS PAGE 


%BUILD  TITLE PAGE 



\title{Optimization}


\subtitle{An important application of Calculus is to problems of \emph{\hilite
optimization}. In an optimization problem, someone wants to \emph{minimize}
or \emph{maximize} the value of a variable.}

\makebeamertitle

\lyxframeend{}\lyxframe{{Sample problem}}

\begin{question} 
\begin{itemize}
\item Find the dimensions of a cylindrical can that minimize surface area.
Assume that the thickness of the can is the same everywhere and that
the can will hold 1~L of fluid. 
\item Find the dimensions of a cylindrical can that minimize surface area.
Assume that the top of the can must be twice as thick as the other
sides and that the can will hold 1~L of fluid. 
\end{itemize}
\end{question}


\lyxframeend{}\lyxframe{{SAGE worksheets}}

You will not need a SAGE worksheet for this module.


\lyxframeend{}\lyxframe{\noindent {Strategy}}

\noindent \emph{Optimization} is related to the topic of finding \emph{extrema}.
Thus the procedure is exactly the same, with the exception that in
optimization the function that you want to minimize or maximize is
not usually known beforehand. \begin{method} 
\begin{itemize}
\item Determine the function $\ec f(x)$ that needs to be minimized or maximized. 
\item Find all places where $\ec f'(x)=0$ or $\ec f'(x)$ does not exist.
These $x$ values are called {\hilite critical points}. 
\item Find the $\ec y$ values at each critical points \emph{and} at the
endpoints. 
\item The smallest $\ec y$ value found is the minimum, and the largest
$\ec y$ value found is the maximum. 
\end{itemize}
\noindent \end{method}


\lyxframeend{}\lyxframe{{First example}}
\begin{overlayarea}
{\textwidth}{6in}

The first example is, \begin{question}Find the dimensions of a cylindrical
can that minimize surface area. Assume that the thickness of the can
is the same everywhere and that the can will hold 1~L of fluid.\end{question}


\pause{}


\noindent It asks us to minimize the surface area of a cylinder. The
relevant equation is\vspace{-1ex}
\[
{\ec A=2\pi rh+2\pi r^{2}}.
\]

\begin{onlyenv}%{}
<2>

\noindent \vspace{-4ex}
\begin{explanation} The top and the bottom are circles of radius
$r$, so their area is $\pi r^{2}$ each. Cut and unfold the side
to obtain a rectangle, whose width is $h$ and whose length is the
circumference of the circles on top and bottom: $2\pi r$. Add the
areas and you have \vspace{-2ex}
\begin{align*}
A & =(\mathrm{side})+(\mathrm{top})+(\mathrm{bottom})\\
 & =2\pi rh+\pi r^{2}+\pi r^{2}.
\end{align*}



\noindent Simplifying this gives the equation above.\end{explanation}

\end{onlyenv}%{}

\pause{}


\noindent Here $\ec A$ is a function that depends on the variables
$\ec r$ and $\ec h$.


\pause{}


We encounter a difficulty here: do we differentiate $\ec A$ with
respect to $\ec r$ or to $\ec h$? Before we can take the derivative,
we need to simplify the equation. \emph{How?}

\end{overlayarea}

\lyxframeend{}\lyxframe{{Other information}}

In this problem, you also need to use the assumption that the can
will hold 1~L of fluid. This is information about the \emph{volume}
of the can.


\pause{}

The volume of a cylinder is
\[
{\ec V=\pi r^{2}h}.
\]
 We can substitute 1~L for $\ec V$ and solve for $\ec h$, obtaining
\[
{\ec h=\frac{1}{\pi r^{2}}}.
\]



\lyxframeend{}\lyxframe{Other information: WATCH OUT!}

\noindent {\dims In this problem, you also need to use the assumption
that the can will hold 1~L of fluid. This is information about the
\emph{volume} of the can.

The volume of a cylinder is
\[
{V=\pi r^{2}h}.
\]
 We can substitute 1~L for $V$ and solve for $h$, obtaining
\[
{h=\frac{1}{\pi r^{2}}}.
\]
}

\noindent We have actually made a subtle mistake here. We want to
compute a measure of \emph{distance} ($\ec h$) and $\ec V$ is a
measure of \emph{volume}. Before using it in the problem, we have
to convert 1~L to 1000~cm$^{3}$.


\lyxframeend{}\lyxframe{Other information}

\noindent In this problem, you also need to use the assumption that
the can will hold 1~L of fluid. This is information about the \emph{volume}
of the can.

The volume of a cylinder is
\[
{\ec V=\pi r^{2}h}.
\]
 We can substitute 1000~cm$^{3}$ for $\ec V$ and solve for $\ec h$,
obtaining
\[
{\ec h=\frac{1000}{\pi r^{2}}}.
\]



\lyxframeend{}\lyxframe{{First example}}
\begin{overlayarea}
{\textwidth}{6in}
\begin{onlyenv}%{}
<1>The first example is, \begin{question}Find the dimensions of
a cylindrical can that minimize surface area. Assume that the thickness
of the can is the same everywhere and that the can will hold 1~L
of fluid.\end{question} We want to minimize the surface area of a
cylinder.
\end{onlyenv}%{}

The relevant equation is
\[
\ec{A=2\pi rh+2\pi r^{2}}.
\]
 Here $A$ is the function that depends on the variables $r$ and
$h$.
\begin{onlyenv}%{}
<1-2>

Since $\ec h=1000/(\pi r^{2})$, we can rewrite $\ec A$ as
\[
{\ec A=2\pi r\left(\frac{1000}{\pi r^{2}}\right)+2\pi r^{2}=\frac{2000}{r}+2\pi r^{2}}.
\]


\end{onlyenv}%{}

\pause{}


\noindent We differentiate with respect to $\ec r$, obtaining
\[
{\ec A'=-\frac{2000}{r^{2}}+4\pi r}.
\]



\pause{}


\noindent To optimize, we look for extrema: where is $\ec A'=0$?


\pause{}
\begin{onlyenv}%{}
<4>

\noindent 
\begin{align*}
\ec0 & \ec=-\frac{2000}{r^{2}}+4\pi r
\end{align*}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<5>

\noindent 
\begin{align*}
\dims0 & \dims=-\frac{2000}{r^{2}}+4\pi r\\
\ec0 & \ec=-2000+4\pi r^{3}
\end{align*}
 \begin{explanation}Multiplied both sides by $r^{2}$.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<6>

\noindent 
\begin{align*}
\dims0 & \dims=-2000+4\pi r^{3}\\
\ec r & {\ec=\sqrt[3]{{\frac{500}{\pi}}}}.
\end{align*}
 \begin{explanation}Solved for $r$.\end{explanation}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<7>

\noindent So $\ec A'=0$ when $\ec r=\sqrt[3]{{500/\pi}}$. The problems
asks for the dimensions, so we also need to find $\ec h$:
\[
{\ec h=\frac{1000}{\pi r^{2}}=\frac{1000}{\pi\left(\sqrt[3]{{\frac{500}{\pi}}}\right)^{2}}=\frac{1000}{\sqrt[3]{{250,000\pi}}}}.
\]


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<8>

\noindent So $\ec A'=0$ when $\ec r=\sqrt[3]{{500/\pi}}$ and $\ec h=1000/\sqrt[3]{{250,000\pi}}$.
Rounded to two digits, the dimensions are
\[
{\ec r\approx5.42}\mathrm{\ cm\ and\ }{\ec h\approx10.84}\mathrm{\ cm}.
\]
 \begin{explanation}This value corresponds to real-world measurements
of cans that hold 1~L of oil. Why would someone want to minimize
the surface area of a can?\end{explanation}

\end{onlyenv}%{}
\end{overlayarea}

\lyxframeend{}\lyxframe{{Second example}}
\begin{overlayarea}
{\textwidth}{6in}

The second example is, \begin{question}Find the dimensions of a cylindrical
can that minimize surface area. Assume that the top of the can must
be twice as thick as the other sides and that the can will hold 1~L
of fluid.\end{question} The difference in this problem is that the
top is twice as thick as the other sides, giving\vspace{-2ex}
\[
{\ec A=3\pi r^{2}+2\pi rh}.
\]

\begin{onlyenv}%{}
<1>\begin{explanation}The area of the top circle is doubled, so
the total area is $2\pi r+2\pi r^{2}+\pi r^{2}$.\end{explanation}
\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<2>

\noindent Everything else in the problem works the same. Try to solve
it yourself.

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3>

\vspace{-2ex}
You should find that\vspace{-2ex}
\begin{align*}
\ec r & {\ec=\sqrt[3]{{\frac{1000}{3\pi}}}\approx4.73}\mathrm{\ cm\ and}\\
\ec h & {\ec=\sqrt[3]{{\frac{9000}{\pi}}}\approx14.20}\mathrm{\ cm}.
\end{align*}
 Draw a picture of a can with these dimensions. Can you think of any
cans in the supermarket that look like that picture?

\end{onlyenv}%{}
\end{overlayarea}

\lyxframeend{}\lyxframe{{Conclusion}}
\begin{itemize}
\item We showed how derivatives can be used to solve for maximum and minimum
values of a function. 
\item The method is essentially that of optimization:\setlength{\shadeboxwidth}{4in}
\begin{method} 

\begin{itemize}
\item Determine the function $\ec f(x)$ that needs to be minimized or maximized. 
\item Find all places where $\ec f'(x)=0$ or $\ec f'(x)$ does not exist.
These $x$ values are called {\hilite critical points}. 
\item Find the $\ec y$ values at each critical points \emph{and} at the
endpoints. 
\item The smallest $\ec y$ value found is the minimum, and the largest
$\ec y$ value found is the maximum. 
\end{itemize}

\end{method} 

\item You will often have to use additional information to rewrite $\ec f$
as a function in only \emph{one} variable. 
\end{itemize}

\lyxframeend{}
\end{document}
