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\begin_body

\begin_layout Standard
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%MODULE 1: Optimization
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%USER DEFINED COMMANDS
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%BUILD  TITLE PAGE 
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\begin_layout Title
Optimization
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\begin_layout Subtitle
An important application of Calculus is to problems of 
\emph on

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\backslash
hilite
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\end_inset

 optimization
\emph default
.
 In an optimization problem, someone wants to 
\emph on
minimize
\emph default
 or 
\emph on
maximize
\emph default
 the value of a variable.
\end_layout

\begin_layout BeginFrame
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{
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Sample problem
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}
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\begin_layout Standard
\begin_inset ERT
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\backslash
begin{question}
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\end_inset

 
\end_layout

\begin_layout Itemize
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
\begin_inset space ~
\end_inset

L of fluid.
 
\end_layout

\begin_layout Itemize
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
\begin_inset space ~
\end_inset

L of fluid.
 
\end_layout

\begin_layout Standard
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\backslash
end{question}
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{
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SAGE worksheets
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}
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\end_layout

\begin_layout Standard
You will not need a SAGE worksheet for this module.
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\begin_layout BeginFrame
\noindent
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{
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Strategy
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}
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\begin_layout Standard
\noindent

\emph on
Optimization
\emph default
 is related to the topic of finding 
\emph on
extrema
\emph default
.
 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_inset ERT
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\begin_layout Plain Layout


\backslash
begin{method}
\end_layout

\end_inset

 
\end_layout

\begin_layout Itemize
Determine the function 
\begin_inset Formula $\ec f(x)$
\end_inset

 that needs to be minimized or maximized.
 
\end_layout

\begin_layout Itemize
Find all places where 
\begin_inset Formula $\ec f'(x)=0$
\end_inset

 or 
\begin_inset Formula $\ec f'(x)$
\end_inset

 does not exist.
 These 
\begin_inset Formula $x$
\end_inset

 values are called 
\begin_inset ERT
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\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
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\backslash
hilite
\end_layout

\end_inset

 critical points
\begin_inset ERT
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\begin_layout Plain Layout

}
\end_layout

\end_inset

.
 
\end_layout

\begin_layout Itemize
Find the 
\begin_inset Formula $\ec y$
\end_inset

 values at each critical points 
\emph on
and
\emph default
 at the endpoints.
 
\end_layout

\begin_layout Itemize
The smallest 
\begin_inset Formula $\ec y$
\end_inset

 value found is the minimum, and the largest 
\begin_inset Formula $\ec y$
\end_inset

 value found is the maximum.
 
\end_layout

\begin_layout Standard
\noindent
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\backslash
end{method}
\end_layout

\end_inset


\end_layout

\begin_layout BeginFrame
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{
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\end_inset

First example
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}
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{
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textwidth}{6in}
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\end_layout

\begin_deeper
\begin_layout Standard
The first example is, 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{question}
\end_layout

\end_inset

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
\begin_inset space ~
\end_inset

L of fluid.
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{question}
\end_layout

\end_inset


\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
It asks us to minimize the surface area of a cylinder.
 The relevant equation is
\begin_inset VSpace -1ex
\end_inset


\begin_inset Formula 
\[
{\ec A=2\pi rh+2\pi r^{2}}.
\]

\end_inset


\end_layout

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\begin_layout Plain Layout

<2>
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\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset VSpace -4ex
\end_inset


\begin_inset ERT
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\begin_layout Plain Layout


\backslash
begin{explanation}
\end_layout

\end_inset

 The top and the bottom are circles of radius 
\begin_inset Formula $r$
\end_inset

, so their area is 
\begin_inset Formula $\pi r^{2}$
\end_inset

 each.
 Cut and unfold the side to obtain a rectangle, whose width is 
\begin_inset Formula $h$
\end_inset

 and whose length is the circumference of the circles on top and bottom:
 
\begin_inset Formula $2\pi r$
\end_inset

.
 Add the areas and you have 
\begin_inset VSpace -2ex
\end_inset


\begin_inset Formula 
\begin{align*}
A & =(\mathrm{side})+(\mathrm{top})+(\mathrm{bottom})\\
 & =2\pi rh+\pi r^{2}+\pi r^{2}.
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\noindent
Simplifying this gives the equation above.
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\backslash
end{explanation}
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
Here 
\begin_inset Formula $\ec A$
\end_inset

 is a function that depends on the variables 
\begin_inset Formula $\ec r$
\end_inset

 and 
\begin_inset Formula $\ec h$
\end_inset

.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
We encounter a difficulty here: do we differentiate 
\begin_inset Formula $\ec A$
\end_inset

 with respect to 
\begin_inset Formula $\ec r$
\end_inset

 or to 
\begin_inset Formula $\ec h$
\end_inset

? Before we can take the derivative, we need to simplify the equation.
 
\emph on
How?
\end_layout

\end_deeper
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{
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Other information
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}
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\end_layout

\begin_layout Standard
In this problem, you also need to use the assumption that the can will hold
 1
\begin_inset space ~
\end_inset

L of fluid.
 This is information about the 
\emph on
volume
\emph default
 of the can.
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
The volume of a cylinder is
\begin_inset Formula 
\[
{\ec V=\pi r^{2}h}.
\]

\end_inset

 We can substitute 1
\begin_inset space ~
\end_inset

L for 
\begin_inset Formula $\ec V$
\end_inset

 and solve for 
\begin_inset Formula $\ec h$
\end_inset

, obtaining
\begin_inset Formula 
\[
{\ec h=\frac{1}{\pi r^{2}}}.
\]

\end_inset


\end_layout

\begin_layout BeginFrame
Other information: WATCH OUT!
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\backslash
dims
\end_layout

\end_inset

 In this problem, you also need to use the assumption that the can will
 hold 1
\begin_inset space ~
\end_inset

L of fluid.
 This is information about the 
\emph on
volume
\emph default
 of the can.
\end_layout

\begin_layout Standard
The volume of a cylinder is
\begin_inset Formula 
\[
{V=\pi r^{2}h}.
\]

\end_inset

 We can substitute 1
\begin_inset space ~
\end_inset

L for 
\begin_inset Formula $V$
\end_inset

 and solve for 
\begin_inset Formula $h$
\end_inset

, obtaining
\begin_inset Formula 
\[
{h=\frac{1}{\pi r^{2}}}.
\]

\end_inset


\begin_inset ERT
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}
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\end_layout

\begin_layout Standard
\noindent
We have actually made a subtle mistake here.
 We want to compute a measure of 
\emph on
distance
\emph default
 (
\begin_inset Formula $\ec h$
\end_inset

) and 
\begin_inset Formula $\ec V$
\end_inset

 is a measure of 
\emph on
volume
\emph default
.
 Before using it in the problem, we have to convert 1
\begin_inset space ~
\end_inset

L to 1000
\begin_inset space ~
\end_inset

cm
\begin_inset Formula $^{3}$
\end_inset

.
\end_layout

\begin_layout BeginFrame
Other information
\end_layout

\begin_layout Standard
\noindent
In this problem, you also need to use the assumption that the can will hold
 1
\begin_inset space ~
\end_inset

L of fluid.
 This is information about the 
\emph on
volume
\emph default
 of the can.
\end_layout

\begin_layout Standard
The volume of a cylinder is
\begin_inset Formula 
\[
{\ec V=\pi r^{2}h}.
\]

\end_inset

 We can substitute 1000
\begin_inset space ~
\end_inset

cm
\begin_inset Formula $^{3}$
\end_inset

 for 
\begin_inset Formula $\ec V$
\end_inset

 and solve for 
\begin_inset Formula $\ec h$
\end_inset

, obtaining
\begin_inset Formula 
\[
{\ec h=\frac{1000}{\pi r^{2}}}.
\]

\end_inset


\end_layout

\begin_layout BeginFrame
\begin_inset ERT
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\begin_layout Plain Layout

{
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\end_inset

First example
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}
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\end_layout

\begin_layout OverlayArea
\begin_inset ERT
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{
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textwidth}{6in}
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<1>
\end_layout

\end_inset

The first example is, 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{question}
\end_layout

\end_inset

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
\begin_inset space ~
\end_inset

L of fluid.
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{question}
\end_layout

\end_inset

 We want to minimize the surface area of a cylinder.
\end_layout

\begin_layout Standard
The relevant equation is
\begin_inset Formula 
\[
\ec{A=2\pi rh+2\pi r^{2}}.
\]

\end_inset

 Here 
\begin_inset Formula $A$
\end_inset

 is the function that depends on the variables 
\begin_inset Formula $r$
\end_inset

 and 
\begin_inset Formula $h$
\end_inset

.
\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<1-2>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
Since 
\begin_inset Formula $\ec h=1000/(\pi r^{2})$
\end_inset

, we can rewrite 
\begin_inset Formula $\ec A$
\end_inset

 as
\begin_inset Formula 
\[
{\ec A=2\pi r\left(\frac{1000}{\pi r^{2}}\right)+2\pi r^{2}=\frac{2000}{r}+2\pi r^{2}}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
We differentiate with respect to 
\begin_inset Formula $\ec r$
\end_inset

, obtaining
\begin_inset Formula 
\[
{\ec A'=-\frac{2000}{r^{2}}+4\pi r}.
\]

\end_inset


\end_layout

\begin_layout Pause

\end_layout

\begin_layout Standard
\noindent
To optimize, we look for extrema: where is 
\begin_inset Formula $\ec A'=0$
\end_inset

?
\end_layout

\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<4>
\end_layout

\end_inset


\end_layout

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

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

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<5>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims0 & \dims=-\frac{2000}{r^{2}}+4\pi r\\
\ec0 & \ec=-2000+4\pi r^{3}
\end{align*}

\end_inset

 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{explanation}
\end_layout

\end_inset

Multiplied both sides by 
\begin_inset Formula $r^{2}$
\end_inset

.
\begin_inset ERT
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\backslash
end{explanation}
\end_layout

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

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\begin_inset ERT
status open

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<6>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
\begin_inset Formula 
\begin{align*}
\dims0 & \dims=-2000+4\pi r^{3}\\
\ec r & {\ec=\sqrt[3]{{\frac{500}{\pi}}}}.
\end{align*}

\end_inset

 
\begin_inset ERT
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\begin_layout Plain Layout


\backslash
begin{explanation}
\end_layout

\end_inset

Solved for 
\begin_inset Formula $r$
\end_inset

.
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\backslash
end{explanation}
\end_layout

\end_inset


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\end_deeper
\begin_layout Pause

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\begin_inset ERT
status open

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<7>
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\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
So 
\begin_inset Formula $\ec A'=0$
\end_inset

 when 
\begin_inset Formula $\ec r=\sqrt[3]{{500/\pi}}$
\end_inset

.
 The problems asks for the dimensions, so we also need to find 
\begin_inset Formula $\ec h$
\end_inset

:
\begin_inset Formula 
\[
{\ec h=\frac{1000}{\pi r^{2}}=\frac{1000}{\pi\left(\sqrt[3]{{\frac{500}{\pi}}}\right)^{2}}=\frac{1000}{\sqrt[3]{{250,000\pi}}}}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<8>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
So 
\begin_inset Formula $\ec A'=0$
\end_inset

 when 
\begin_inset Formula $\ec r=\sqrt[3]{{500/\pi}}$
\end_inset

 and 
\begin_inset Formula $\ec h=1000/\sqrt[3]{{250,000\pi}}$
\end_inset

.
 Rounded to two digits, the dimensions are
\begin_inset Formula 
\[
{\ec r\approx5.42}\mathrm{\ cm\ and\ }{\ec h\approx10.84}\mathrm{\ cm}.
\]

\end_inset

 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{explanation}
\end_layout

\end_inset

This value corresponds to real-world measurements of cans that hold 1
\begin_inset space ~
\end_inset

L of oil.
 Why would someone want to minimize the surface area of a can?
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{explanation}
\end_layout

\end_inset


\end_layout

\end_deeper
\end_deeper
\begin_layout BeginFrame
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset

Second example
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset


\end_layout

\begin_layout OverlayArea
\begin_inset ERT
status open

\begin_layout Plain Layout

{
\backslash
textwidth}{6in}
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
The second example is, 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{question}
\end_layout

\end_inset

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
\begin_inset space ~
\end_inset

L of fluid.
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{question}
\end_layout

\end_inset

 The difference in this problem is that the top is twice as thick as the
 other sides, giving
\begin_inset VSpace -2ex
\end_inset


\begin_inset Formula 
\[
{\ec A=3\pi r^{2}+2\pi rh}.
\]

\end_inset


\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<1>
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{explanation}
\end_layout

\end_inset

The area of the top circle is doubled, so the total area is 
\begin_inset Formula $2\pi r+2\pi r^{2}+\pi r^{2}$
\end_inset

.
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{explanation}
\end_layout

\end_inset


\end_layout

\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<2>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\noindent
Everything else in the problem works the same.
 Try to solve it yourself.
\end_layout

\end_deeper
\begin_layout Pause

\end_layout

\begin_layout Only
\begin_inset ERT
status open

\begin_layout Plain Layout

<3>
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
\begin_inset VSpace -2ex
\end_inset

You should find that
\begin_inset VSpace -2ex
\end_inset


\begin_inset Formula 
\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*}

\end_inset

 Draw a picture of a can with these dimensions.
 Can you think of any cans in the supermarket that look like that picture?
\end_layout

\end_deeper
\end_deeper
\begin_layout BeginFrame
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{
\end_layout

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Conclusion
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}
\end_layout

\end_inset


\end_layout

\begin_layout Itemize
We showed how derivatives can be used to solve for maximum and minimum values
 of a function.
 
\end_layout

\begin_layout Itemize
The method is essentially that of optimization:
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
setlength{
\backslash
shadeboxwidth}{4in}
\end_layout

\end_inset

 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
begin{method}
\end_layout

\end_inset

 
\end_layout

\begin_deeper
\begin_layout Itemize
Determine the function 
\begin_inset Formula $\ec f(x)$
\end_inset

 that needs to be minimized or maximized.
 
\end_layout

\begin_layout Itemize
Find all places where 
\begin_inset Formula $\ec f'(x)=0$
\end_inset

 or 
\begin_inset Formula $\ec f'(x)$
\end_inset

 does not exist.
 These 
\begin_inset Formula $x$
\end_inset

 values are called 
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

{
\end_layout

\end_inset


\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
hilite
\end_layout

\end_inset

 critical points
\begin_inset ERT
status collapsed

\begin_layout Plain Layout

}
\end_layout

\end_inset

.
 
\end_layout

\begin_layout Itemize
Find the 
\begin_inset Formula $\ec y$
\end_inset

 values at each critical points 
\emph on
and
\emph default
 at the endpoints.
 
\end_layout

\begin_layout Itemize
The smallest 
\begin_inset Formula $\ec y$
\end_inset

 value found is the minimum, and the largest 
\begin_inset Formula $\ec y$
\end_inset

 value found is the maximum.
 
\end_layout

\begin_layout Standard
\begin_inset ERT
status collapsed

\begin_layout Plain Layout


\backslash
end{method}
\end_layout

\end_inset

 
\end_layout

\end_deeper
\begin_layout Itemize
You will often have to use additional information to rewrite 
\begin_inset Formula $\ec f$
\end_inset

 as a function in only 
\emph on
one
\emph default
 variable.
 
\end_layout

\begin_layout EndFrame

\end_layout

\end_body
\end_document
