\documentclass[12pt]{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage[nomarginpar]{geometry}
\usepackage{color}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{fancyhdr}

\setcounter{MaxMatrixCols}{10}
%TCIDATA{OutputFilter=LATEX.DLL}
%TCIDATA{Version=5.00.0.2570}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{Created=Wednesday, July 12, 2006 00:27:03}
%TCIDATA{LastRevised=Wednesday, March 27, 2019 14:29:11}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="Title" CONTENT="Optimization 2">}
%TCIDATA{<META NAME="DocumentShell" CONTENT="Scientific Notebook\Booklet #1 - with Instructions">}
%TCIDATA{CSTFile=40 LaTeX article.cst}
%TCIDATA{PageSetup=72,72,72,72,1}
%TCIDATA{ComputeGeneralSettings=0,15,15,0,0,0,0}
%TCIDATA{Counters=arabic,1}
%TCIDATA{<META NAME="PrintViewPercent" CONTENT="100">}
%TCIDATA{ComputeDefs=
%$C\left( x\right) =50\sqrt{200^{2}+x^{2}}+30\left( 600-x\right) $
%}

%TCIDATA{AllPages=
%H=36
%F=36,\PARA{038<p type="texpara" tag="Body Text" >\hfill \hfill }
%}


\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}
\geometry{left=0.6in,right=0.7in,top=0.7in,bottom=0.7in}
\pagestyle{fancy}
\lhead{\color{blue} \large Lecture Notes}
\chead{\color{black} \Large Optimization 3}
\rhead{\large page   \ \thepage}
\cfoot{}
\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  Powell,  2011}
\rfoot{\small   Last revised: January 24, 2011}
\textwidth 7.4in 
\textheight 9.3in 
\setlength{\headheight}{28pt}
\setlength{\parindent}{0in}

\begin{document}


\begin{center}
{\Large Sample Problems\bigskip }
\end{center}

\begin{enumerate}
\item Let $a$ and $b$ be positive numbers such that $ab=10$. \ Find the
lowest value of $a^{2}+4b^{2}$.

\item A closed box with a square base is to have a volume of $250$ cubic
meters. The material for the top and bottom of the box costs $\$2$ per
square meter, and the material for the sides costs $\$1$ per square meter.
Can the box be constructed for less than $\$300$?

\item A company determines that the if $n$ is the number of items produced,
they can all be sold at a price of $p\left( n\right) =\sqrt{1200-0.2n}$ \
What is the greatest revenue possible?

\item A company has $\$120\,000$ to spend on the development and promotion
of a new product. \ \ The company estimates that if $x$ is spent on the
development and $y$ is spent on promotion, then approximately $\dfrac{%
x^{1/2}y^{3/2}}{400\,000}$ \ items of new product will be sold. Based on
this estimate, what is the maximum number of products that the company can
sell?

\item A company wants to manufacture cylindrical aluminum cans with a volume
of $1000$ cubic centimeters (one liter). \ What dimensions would guarantee
the minimal amount of aluminum needed to produce a can?

\item An underground telephone cable is to be laid between two boat docks on
opposite banks of a straight river. \ \ One boathouse is $600$ meters
downstream from the other. \ The river is $200$ meters wide. \ If the cost
of laying the cable is $\$50$ per meter under water and $\$30$ per meter on
land, how should the cable to be laid to minimize cost? \ \newline
a) \ Find the lowest cost possible.\newline
b) \ Use the second derivative test to prove that we found a relative
minimum in part a).\newline
c) \ Prove that the relative minimum we found in part a) \ is also an
absolue minimum on the domain $\left[ 0,600\right] .$\newline
d) \ Prove that if we define $C$ on the set of all real numbers, then the
minimum we found in part a) \ is still an absolute minimum.

\item The location function of an object is $s\left( t\right) =\dfrac{120}{%
1+2e^{-t}}$ where $t$ is time, measured in seconds. \ Where is the object
when it is moving with the greatest speed? \ When is that greatest speed
achieved?

\item Consider the function $g\left( x\right) =\dfrac{-2x}{\left(
x^{2}+1\right) ^{2}}$.

a) \ Find all relative extrema of $g$.

b) \ Find all absolute extrema of $g$.

c) \ Find all values of $c$ for which the function $f\left( x\right) =\dfrac{%
1}{x^{2}+1}+cx$ \ is increasing on its entire domain.\bigskip 
\end{enumerate}

\begin{center}
\bigskip

{\Large Sample Problems - Answers\bigskip }
\end{center}

\begin{enumerate}
\item $40$

\item no,the lowest possible cost is\ \ $\$300$ when the box is to be $5%
\unit{m}$ by $5\unit{m}$ by $10\unit{m}$

\item $\$80\,000$

\item $11\,691$

\item $r=\sqrt[3]{\dfrac{500}{\pi }}\simeq \allowbreak 5.\,\allowbreak
419\,26$ \ and \ $h=2\sqrt[3]{\dfrac{500}{\pi }}=2r\simeq \allowbreak
10.\,\allowbreak 838\,521\,402\,\allowbreak 785\,8$

\item a) \ $\,\$26\,000$ when the cable is laid $450$ meters on the ground \
\ \ \ b) \ see solutions \ \ c) \ see solutions \ d) \ see solutions

\item $s=60$ when $t=\ln 2$

\item a) $\ \left( -\dfrac{1}{\sqrt{3}},\dfrac{3\sqrt{3}}{8}\right) \ $%
relative maximum and a relative minimum $\left( -\dfrac{1}{\sqrt{3}},\dfrac{3%
\sqrt{3}}{8}\right) $ \ \ \ \ b) \ see solutions \ \ c) \ $c\geq \dfrac{3%
\sqrt{3}}{8}${\Large \pagebreak }
\end{enumerate}

\begin{center}
{\Large Sample Problems - Solutions\bigskip }
\end{center}

\begin{enumerate}
\item Let $a$ and $b$ be positive numbers such that $ab=10$. \ Find the
lowest value of $a^{2}+4b^{2}$.\newline
Solution: \ We express $a$ in terms of $b:$ \ \ \ $a=\dfrac{10}{b}.$ \ Then
the expression $a^{2}+4b^{2}$ becomes 
\begin{equation*}
P\left( b\right) =\left( \dfrac{10}{b}\right) ^{2}+4b^{2}=4b^{2}+\dfrac{100}{%
b^{2}}=4b^{2}+100b^{-2}
\end{equation*}%
We differentiate this: \ 
\begin{eqnarray*}
P^{\prime }\left( b\right) &=&8b+100\left( -2\right) b^{-3}=8b-\dfrac{200}{%
b^{3}}=\dfrac{8b^{4}-200}{b^{3}} \\
&=&\dfrac{8\left( b^{4}-25\right) }{b^{3}}=\dfrac{8\left( b^{2}+5\right)
\left( b^{2}-5\right) }{b^{3}}=\dfrac{8\left( b^{2}+5\right) \left( b+\sqrt{5%
}\right) \left( b-\sqrt{5}\right) }{b^{3}}
\end{eqnarray*}%
We can figure out when $P^{\prime }$ is positive and negative by sorting out
the signs of each factors in the numerator and denominator.\medskip \medskip

\ \ \ \ \ \ \ \ \ \ \ \ 
\begin{tabular}{|l|l|l|l|l|}
\hline
& $b<-\sqrt{5}$ & $-\sqrt{5}<b<0$ & $0<b<\sqrt{5}$ & $b>\sqrt{5}$ \\ \hline
$\left( b^{2}+5\right) $ & \multicolumn{1}{|c|}{$+$} & \multicolumn{1}{|c|}{$%
+$} & \multicolumn{1}{|c|}{$+$} & \multicolumn{1}{|c|}{$+$} \\ \hline
$\left( b+\sqrt{5}\right) $ & \multicolumn{1}{|c|}{$-$} & 
\multicolumn{1}{|c|}{$+$} & \multicolumn{1}{|c|}{$+$} & \multicolumn{1}{|c|}{%
$+$} \\ \hline
$\left( b-\sqrt{5}\right) $ & \multicolumn{1}{|c|}{$-$} & 
\multicolumn{1}{|c|}{$-$} & \multicolumn{1}{|c|}{$-$} & \multicolumn{1}{|c|}{%
$+$} \\ \hline
$b^{3}$ & \multicolumn{1}{|c|}{$-$} & \multicolumn{1}{|c|}{$-$} & 
\multicolumn{1}{|c|}{$+$} & \multicolumn{1}{|c|}{$+$} \\ \hline
$P^{\prime }$ & \multicolumn{1}{|c|}{$-$} & \multicolumn{1}{|c|}{$+$} & 
\multicolumn{1}{|c|}{$-$} & \multicolumn{1}{|c|}{$+$} \\ \hline
\end{tabular}%
\medskip \medskip

Based on the signs of $P^{\prime }$, \ $P$ has a relative minimum at $b=-%
\sqrt{5}$ and a relative maximum at $\sqrt{5}.$ \ Since $b$ must be
positive, we may consider $P$ on the domain $\left( 0,\infty \right) $. \ On
this domain, $P$ is decreasing on $\left( 0,\sqrt{5}\right) $ and increasing
on $\left( \sqrt{5},\infty \right) $ and so $P$ has an absolute minimum at $%
b=\sqrt{5}$.

If $b=\sqrt{5},$ then 
\begin{equation*}
P\left( \sqrt{5}\right) =\left( \dfrac{10}{\sqrt{5}}\right) ^{2}+4\left( 
\sqrt{5}\right) ^{2}=\dfrac{100}{5}+4\cdot 5=20+20=40
\end{equation*}%
Thus the smallest possible value of $a^{2}+4b^{2}$ is $40$.

\item A closed box with a square base is to have a volume of $250$ cubic
meters. The material for the top and bottom of the box costs $\$2$ per
square meter, and the material for the sides costs $\$1$ per square meter.
Can the box be constructed for less than $\$300$? \newline
Solution: \ Let $x$ denote the side of the square base, and $h$ denote the
height of the box. Then $V=hx^{2}$ gives us%
\begin{eqnarray*}
hx^{2} &=&250 \\
h &=&\dfrac{250}{x^{2}}
\end{eqnarray*}%
We now set up the cost function, $C\left( x\right) $. \ The top and bottom
each cost $\$2$ per square meter, and have area $x^{2}$. \ The four sides
each have area $xh=x\left( \dfrac{250}{x^{2}}\right) =\dfrac{250}{x}$ \ and
cost $\$1$ per square meter. \ Thus 
\begin{equation*}
C\left( x\right) =2\cdot 2\cdot x^{2}+4\cdot 1\cdot \dfrac{250}{x}=4x^{2}+%
\dfrac{1000}{x}=4x^{2}+1000x^{-1}
\end{equation*}%
We are looking for the maximum of $C\left( x\right) $. \ We will
differentiate $C$ first.%
\begin{eqnarray*}
C^{\prime }\left( x\right) &=&8x+1000\left( -1\right) x^{-2}=8x-\dfrac{1000}{%
x^{2}}=\dfrac{8x^{3}-1000}{x^{2}}=\dfrac{8\left( x^{3}-125\right) }{x^{2}} \\
&=&\dfrac{8\left( x-5\right) \left( x^{2}+5x+25\right) }{x^{2}}
\end{eqnarray*}%
The last form shows that $C^{\prime }$ has only one zero, at $x=5$. \ Since
both $x^{2}$ and $x^{2}+5x+25$ \ are positive for all values of $x$, \ $%
C^{\prime }$ will change sign from negative to positive at $x=5$, indicating
a minimum of $C$. \ Thus, the lowest possible cost will be associated with $%
x=5$. \ The actual cost is then%
\begin{equation*}
C\left( 5\right) =4\cdot 5^{2}+\dfrac{1000}{5}=300
\end{equation*}%
Thus, we can not construct this box for less than $\$300.$

\item A company determines that the if $n$ is the number of items produced,
they can all be sold at a price of $p\left( n\right) =\sqrt{1200-0.2n}$ \
What is the greatest revenue possible?\newline
Solution: \ If we sell all $n$ products, the revenue is $R\left( n\right) =n%
\sqrt{1200-0.2n}$ Since the price should be positive, the domain of this
function is determined by $n>0$ and $1200-0.2n>0.$ \ We solve these
inequalities and obtain the domain. $\left( 0,6000\right) .$ \ \ We will
find the maximum of this function by finding the zeroes of the derivative.%
\begin{eqnarray*}
R^{\prime }\left( n\right) &=&\sqrt{1200-0.2n}+n\dfrac{1}{2\sqrt{1200-0.2n}}%
\left( -0.2\right) =\sqrt{1200-0.2n}-\dfrac{0.1n}{\sqrt{1200-0.2n}} \\
&=&\dfrac{1200-0.2n}{\sqrt{1200-0.2n}}-\dfrac{0.1n}{\sqrt{1200-0.2n}}=\dfrac{%
1200-0.2n-0.1n}{\sqrt{1200-0.2n}}=\dfrac{-0.3n+1200}{\sqrt{1200-0.2n}} \\
&=&\dfrac{-0.3\left( n-4000\right) }{\sqrt{1200-0.2n}}
\end{eqnarray*}%
$\allowbreak $The derivative has only one zero, at $n=4000$. \ The
denominator is positive for all $n$ with $0<n<6000$ and the numerator is
positive before $4000$ and negative after $4000.$ \ Consequently, $R\left(
n\right) $ \ has a relative and absolute maximum at $n=4000$. \ The maximal
possible revenue is then $R\left( 4000\right) $.%
\begin{equation*}
R\left( 4000\right) =4000\sqrt{1200-0.2\left( 4000\right) }=4000\sqrt{%
1200-800}=4000\sqrt{400}=4000\left( 20\right) =80\,000
\end{equation*}

\item A company has $\$120\,000$ to spend on the development and promotion
of a new product. \ \ The company estimates that if $x$ is spent on the
development and $y$ is spent on promotion, then approximately $\dfrac{%
x^{1/2}y^{3/2}}{400\,000}$ \ items of new product will be sold. Based on
this estimate, what is the maximum number of products that the company can
sell? \newline
Solution: \ Since $y=120\,000-x$, we can write the number sold as a function
of $x$.%
\begin{equation*}
N\left( x\right) =\dfrac{1}{400\,000}x^{1/2}\left( 120\,000-x\right) ^{3/2}
\end{equation*}%
The domain is clearly $\left[ 0,120\,000\right] $. \ We differentiate $N$.%
\begin{eqnarray*}
N^{\prime }\left( x\right) &=&\dfrac{1}{400\,000}\left[ \dfrac{1}{2}%
x^{-1/2}\left( 120\,000-x\right) ^{3/2}+x^{1/2}\left( \dfrac{3}{2}\right)
\left( 120\,000-x\right) ^{1/2}\left( -1\right) \right] \\
&=&\dfrac{1}{400\,000}\left( \dfrac{1}{2}\right) \left[ \dfrac{\left(
120\,000-x\right) ^{3/2}}{\sqrt{x}}-3\sqrt{x}\left( 120\,000-x\right) ^{1/2}%
\right] \\
&=&\dfrac{1}{800\,000}\left( \dfrac{\sqrt{120\,000-x}\left(
120\,000-x\right) }{\sqrt{x}}-3\sqrt{x}\sqrt{120\,000-x}\right) \text{ \ \ \
\ \ \ \ factor out }\sqrt{120\,000-x} \\
&=&\dfrac{\sqrt{120\,000-x}}{800\,000}\left( \dfrac{\left( 120\,000-x\right) 
}{\sqrt{x}}-3\sqrt{x}\right) \text{ \ \ \ \ \ \ \ \ bring difference to
common denominator} \\
&=&\dfrac{\sqrt{120\,000-x}}{800\,000}\left( \dfrac{\left( 120\,000-x\right) 
}{\sqrt{x}}-3\dfrac{x}{\sqrt{x}}\right) =\dfrac{\sqrt{120\,000-x}}{800\,000}%
\left( \dfrac{120\,000-x-3x}{\sqrt{x}}\right) \\
&=&\dfrac{\sqrt{120\,000-x}}{800\,000}\left( \dfrac{120\,000-4x}{\sqrt{x}}%
\right) =\dfrac{-4\sqrt{120\,000-x}\left( x-30\,000\right) }{800\,000\sqrt{x}%
} \\
&=&\dfrac{\sqrt{120\,000-x}}{800\,000\sqrt{x}}\left[ -4\left(
x-30\,000\right) \right]
\end{eqnarray*}%
\ \ \ \ 
\begin{equation*}
N^{\prime }\left( x\right) =\dfrac{\sqrt{120\,000-x}}{800\,000\sqrt{x}}%
~~\left( -4\right) \left( x-30\,000\right)
\end{equation*}%
The first factor, $\dfrac{\sqrt{120\,000-x}}{800\,000\sqrt{x}}$ is positive
for all $x$ in the domain, except for the endpoints. \ The second factor, \ $%
-4\left( x-30\,000\right) $ \ is positive before $30\,000$, zero at $%
x=30\,000,$ and negative after $30\,000$, indicating an absolute maximum at $%
x=30\,000$. \ Then $y=120\,000-30\,000=90\,000$ and the maximal number sold
is then 
\begin{equation*}
N\left( 30\,000\right) =\dfrac{x^{1/2}y^{3/2}}{400\,000}=\dfrac{%
30\,000^{1/2}90\,000^{3/2}}{400\,000}\approx 11691.\,\allowbreak 342\,95
\end{equation*}%
The integer nearest to this number is $11\,691$.

\item Consider the function $g\left( x\right) =\dfrac{-2x}{\left(
x^{2}+1\right) ^{2}}$.

a) \ Find all relative extrema of $g$.\newline
Solution: \ We differentiate $g$ using the quotient rule.%
\begin{eqnarray*}
g^{\prime }\left( x\right) &=&\dfrac{-2\left( x^{2}+1\right) ^{2}-\left(
-2x\right) 2\left( x^{2}+1\right) \left( 2x\right) }{\left( x^{2}+1\right)
^{4}}\text{ \ \ \ \ \ \ simplify by }x^{2}+1 \\
&=&\dfrac{-2\left( x^{2}+1\right) -\left( -2x\right) 2\left( 2x\right) }{%
\left( x^{2}+1\right) ^{3}}=\dfrac{-2x^{2}-2+8x^{2}}{\left( x^{2}+1\right)
^{3}}=\dfrac{6x^{2}-2}{\left( x^{2}+1\right) ^{3}}
\end{eqnarray*}%
We now factor the numerator to see when $g^{\prime }$ is positive and
negative.%
\begin{equation*}
g^{\prime }\left( x\right) =\dfrac{6x^{2}-2}{\left( x^{2}+1\right) ^{3}}=%
\dfrac{6\left( x^{2}-\dfrac{1}{3}\right) }{\left( x^{2}+1\right) ^{3}}=%
\dfrac{6\left( x+\dfrac{1}{\sqrt{3}}\right) \left( x-\dfrac{1}{\sqrt{3}}%
\right) }{\left( x^{2}+1\right) ^{3}}
\end{equation*}%
The denominator is always positive, the numerator is a quadratic expression
with a positive leading coefficient, and so $g^{\prime }$ is positive on $%
\left( -\infty ,-\dfrac{1}{\sqrt{3}}\right) \cup \left( \dfrac{1}{\sqrt{3}}%
,\infty \right) $ \ \ and negative on $\left( -\dfrac{1}{\sqrt{3}},\dfrac{1}{%
\sqrt{3}}\right) .$ \ Consequently, $g$ is increasing on $\left( -\infty ,-%
\dfrac{1}{\sqrt{3}}\right) $, decreasing on $\left( -\dfrac{1}{\sqrt{3}},%
\dfrac{1}{\sqrt{3}}\right) ,$ and increasing on $\left( \dfrac{1}{\sqrt{3}}%
,\infty \right) $. \ That means that $g$ has a relative maximum at $x=-%
\dfrac{1}{\sqrt{3}}$ and a relative minimum at $x=\dfrac{1}{\sqrt{3}}$. \ We
compute the function values at $-\dfrac{1}{\sqrt{3}}$ and $\dfrac{1}{\sqrt{3}%
}.$ \ 
\begin{equation*}
g\left( -\dfrac{1}{\sqrt{3}}\right) =\dfrac{-2\left( -\dfrac{1}{\sqrt{3}}%
\right) }{\left( \left( -\dfrac{1}{\sqrt{3}}\right) ^{2}+1\right) ^{2}}=%
\dfrac{\dfrac{2}{\sqrt{3}}}{\left( \dfrac{1}{3}+1\right) ^{2}}=\dfrac{\dfrac{%
2}{\sqrt{3}}}{\left( \dfrac{4}{3}\right) ^{2}}=\dfrac{2}{\sqrt{3}}\cdot 
\dfrac{9}{16}=\dfrac{3\sqrt{3}}{8}
\end{equation*}%
This is a good time to notice that $g$ is an odd function%
\begin{equation*}
g\left( -x\right) =\dfrac{-2\left( -x\right) }{\left( \left( -x\right)
^{2}+1\right) ^{2}}=\dfrac{2x}{\left( x^{2}+1\right) ^{2}}=-g\left( x\right)
\end{equation*}%
and so $g$ has a relative maximum: $\left( -\dfrac{1}{\sqrt{3}},\dfrac{3%
\sqrt{3}}{8}\right) $ and a relative minimum $\left( -\dfrac{1}{\sqrt{3}},%
\dfrac{3\sqrt{3}}{8}\right) $.

b) \ Find all absolute extrema of $g$.\newline
Solution: \ We will show that the relateive extrema we found in part a) are
in fact absolute extrema. \ This is not a fact that simply follows from the
signs of the derivative. \ For all we know, a decreasing, then increasing,
then decreasing function may look like an upside down cubic function that
has neither absolute minumum nor absolute maximum.\newline
Claim: \ $\left( -\dfrac{1}{\sqrt{3}},\dfrac{3\sqrt{3}}{8}\right) $ is an
absolute maximum, i.e. for all real numbers $x$, $g\left( x\right) \leq 
\dfrac{3\sqrt{3}}{8}$.\newline
In part a) \ we determined that $g$ is increasing on $\left( -\infty ,-%
\dfrac{1}{\sqrt{3}}\right) $, decreasing on $\left( -\dfrac{1}{\sqrt{3}},%
\dfrac{1}{\sqrt{3}}\right) ,$ and increasing on $\left( \dfrac{1}{\sqrt{3}}%
,\infty \right) $.\newline
Because $g$ is increasing on $\left( -\infty ,-\dfrac{1}{\sqrt{3}}\right) $, 
\begin{equation*}
\text{for all }x\leq -\dfrac{1}{\sqrt{3}}\text{, \ \ \ \ \ }g\left( x\right)
\leq g\left( -\dfrac{1}{\sqrt{3}}\right)
\end{equation*}%
Because $g$ is decreasing on $\left( -\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}%
\right) $,%
\begin{equation*}
\text{for all }-\dfrac{1}{\sqrt{3}}\leq x\leq \dfrac{1}{\sqrt{3}}\text{, \ \
\ \ \ }g\left( x\right) \leq g\left( -\dfrac{1}{\sqrt{3}}\right)
\end{equation*}%
We only have to prove that for all $x\geq \dfrac{1}{\sqrt{3}}$, \ \ \ \ \ $%
g\left( x\right) \leq g\left( -\dfrac{1}{\sqrt{3}}\right) $ is also true,
but the increasing/decreasing behavior does not help here, since $g$ is
increasing on $\left( \dfrac{1}{\sqrt{3}},\infty \right) $. \ Luckily, there
is a simple, elementary way to finish the proof. \ Recall that $g\left(
x\right) =\dfrac{-2x}{\left( x^{2}+1\right) ^{2}}$. \ It is easy to see that
if $x$ is positive, then $g\left( x\right) $ is negative. \ Thus $g$ is
negative on $\left( \dfrac{1}{\sqrt{3}},\infty \right) $ and so 
\begin{equation*}
\text{for all }x\geq \dfrac{1}{\sqrt{3}}\text{, \ \ \ \ \ }g\left( x\right)
\leq g\left( -\dfrac{1}{\sqrt{3}}\right) =\dfrac{3\sqrt{3}}{8}
\end{equation*}%
Thus $g$ has an absolute maximum at $x=-\dfrac{1}{\sqrt{3}}$.\newline
Claim: \ $\left( \dfrac{1}{\sqrt{3}},-\dfrac{3\sqrt{3}}{8}\right) $ is an
absolute minimum, i.e. for all real numbers $x$, $g\left( x\right) \leq -%
\dfrac{3\sqrt{3}}{8}$.\newline
proof: \ a very similar argument could work. $\ g$ is decreasing on $\left( -%
\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right) ,$ and increasing on $\left( 
\dfrac{1}{\sqrt{3}},\infty \right) $ and positive on $\left( -\infty ,-%
\dfrac{1}{\sqrt{3}}\right) $ \ Another way to prove this is to use the fact
that $g$ is an odd function.%
\begin{eqnarray*}
\text{for all }x\text{, }g\left( x\right) &\leq &\dfrac{3\sqrt{3}}{8}\text{
\ \ \ \ \ \ multiply by }-1 \\
\text{for all }x\text{, }-g\left( x\right) &\geq &-\dfrac{3\sqrt{3}}{8}\text{
\ \ \ \ since }g\text{ is odd, }-g\left( x\right) =g\left( -x\right) \\
\text{for all }x\text{, }g\left( -x\right) &\geq &-\dfrac{3\sqrt{3}}{8}
\end{eqnarray*}%
and so $g$ has an absolute minimum at $x=\dfrac{1}{\sqrt{3}}$.\FRAME{dtbpFX}{%
3.5129in}{1.5056in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type
"MAPLEPLOT";width 3.5129in;height 1.5056in;depth 0pt;display
"USEDEF";plot_snapshots TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin
"-5";xmax "5";xviewmin "-5.01";xviewmax "5.01";yviewmin
"-0.650105019579922";yviewmax "0.650105019579922";plottype 4;plotticks
1;num-x-ticks 11;numpoints 100;plotstyle "patch";axesstyle "normal";xis
\TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function
\TEXUX{$\dfrac{-2x}{\left( x^{2}+1\right) ^{2}}$};linecolor
"black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes
"Solid";var1range "-5,5";num-x-gridlines 100;curveColor
"[flat::RGB:0000000000]";curveStyle "Line";valid_file "T";tempfilename
'LIFPQS00.wmf';tempfile-properties "PR";}}

c) \ Find all values of $c$ for which the function $f\left( x\right) =\dfrac{%
1}{x^{2}+1}+cx$ \ is increasing on its entire domain.\newline
Solution: \ $f\left( x\right) $ is always increasing if $f^{\prime }\left(
x\right) $ \ is always positive. \ $f^{\prime }\left( x\right) =\dfrac{-2x}{%
\left( x^{2}+1\right) ^{2}}+c$. \ We proved that the absolute minimum of $%
\dfrac{-2x}{\left( x^{2}+1\right) ^{2}}$ is $-\dfrac{3\sqrt{3}}{8}$ and so
if we set $c\geq \dfrac{3\sqrt{3}}{8},$ then \ $f^{\prime }\left( x\right) =%
\dfrac{-2x}{\left( x^{2}+1\right) ^{2}}+c$ will be non-negative for all $x$.

\item A company wants to manufacture cylindrical aluminum cans with a volume
of $1000$ cubic centimeters (one liter). \ What dimensions would guarantee
the minimal amount of aluminum needed to produce a can? \ \newline
Solution: Let h denote the height of the can, and $r$ denote the radius of
the base circle.\ 
\begin{equation*}
\pi r^{2}h=1000~~~~h=\dfrac{1000}{\pi r^{2}}
\end{equation*}%
The domain is $\left( 0,\infty \right) $%
\begin{eqnarray*}
S\left( r\right) &=&2\pi rh+2\pi r^{2}=2\pi r\left( \dfrac{1000}{\pi r^{2}}%
\right) +2\pi r^{2}=2\pi r^{2}+\dfrac{2000}{r} \\
S^{\prime }\left( r\right) &=&4\pi r-\dfrac{2000}{r^{2}}=\dfrac{4\pi
r^{3}-2000}{r^{2}}
\end{eqnarray*}%
\begin{eqnarray*}
4\pi r-\dfrac{2000}{r^{2}} &=&0 \\
4\pi r &=&\dfrac{2000}{r^{2}} \\
\pi r^{3} &=&500~~~\Longrightarrow ~~r=\sqrt[3]{\dfrac{500}{\pi }}\simeq
\allowbreak 5.\,\allowbreak 419\,26
\end{eqnarray*}%
and 
\begin{eqnarray*}
h &=&\dfrac{1000}{\pi \left( \sqrt[3]{\dfrac{500}{\pi }}\right) ^{2}}=\dfrac{%
1000}{\pi \left( 500^{2/3}\right) \left( \pi ^{-2/3}\right) }=\dfrac{2\cdot
500}{\left( 500^{2/3}\right) \left( \pi ^{1/3}\right) }=\dfrac{2\cdot
500^{1/3}}{\left( \pi ^{1/3}\right) } \\
&=&2\sqrt[3]{\dfrac{500}{\pi }}=2r\simeq \allowbreak 10.\,\allowbreak
838\,521\,402\,\allowbreak 785\,8
\end{eqnarray*}%
But is this an absolute minimum we found?%
\begin{equation*}
S^{\prime \prime }\left( r\right) =4\pi +\dfrac{4000}{r^{3}}
\end{equation*}%
Since $S^{\prime \prime }$ is positive on the entire domain (recall $r>0$), $%
S^{\prime }$ is strictly increasing on its entire domain. \ This means that $%
S^{\prime }$ is negative before its only zero and positive after. \ This
implies that $S$ is decreasing before and increasing after, and so we indeed
found the absolute minimum.

\item An underground telephone cable is to be laid between two boat docks on
opposite banks of a straight river. \ \ One boathouse is $600$ meters
downstream from the other. \ The river is $200$ meters wide. \ If the cost
of laying the cable is $\$50$ per meter under water and $\$30$ per meter on
land, how should the cable to be laid to minimize cost? \ \newline
a) \ Find the lowest cost possible.\newline
Solution: \ Let us denote by $x$ - as shown on the picture below - the
distance alongside the river of the part of the cable to be laid under the
water.\FRAME{dtbpF}{3.8778in}{1.6397in}{0pt}{}{}{image1.bmp}{\special%
{language "Scientific Word";type "GRAPHIC";display "USEDEF";valid_file
"F";width 3.8778in;height 1.6397in;depth 0pt;original-width
0.141in;original-height 0.0778in;cropleft "0";croptop "0.9063";cropright
"1";cropbottom "0.0937";filename 'image1.bmp';file-properties "XNPEU";}}\
Then the length of the cable laid under water is $\sqrt{200^{2}+x^{2}}$ \
and $600-x$ on the ground. \ We can now express the cost as a function of $x$%
.%
\begin{equation*}
C\left( x\right) =50\sqrt{200^{2}+x^{2}}+30\left( 600-x\right) \text{ \ \ \
on domain }\left[ 0,600\right]
\end{equation*}%
We find the minimum of $C$ by differentiating $C$.%
\begin{equation*}
C^{\prime }\left( x\right) =50\dfrac{1}{2\sqrt{200^{2}+x^{2}}}\left(
2x\right) +30\left( -1\right) =\dfrac{50x}{\sqrt{200^{2}+x^{2}}}-30
\end{equation*}%
To find the extrema, we solve for the zeroes of the derivative.%
\begin{eqnarray*}
C^{\prime }\left( x\right) &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }25x^{2}=9\left( x^{2}+40\,000\right) \\
\dfrac{50x}{\sqrt{200^{2}+x^{2}}}-30 &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }25x^{2}=9x^{2}+360\,000 \\
\dfrac{50x}{\sqrt{200^{2}+x^{2}}} &=&30\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }16x^{2}=360\,000 \\
50x &=&30\sqrt{200^{2}+x^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }%
x^{2}=22\,500 \\
5x &=&3\sqrt{200^{2}+x^{2}}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ }x=\pm 150 \\
25x^{2} &=&9\left( 200^{2}+x^{2}\right)
\end{eqnarray*}%
Since the domain is $\left[ 0,600\right] ,$ the only possibility is $x=150.$
\ Thus the minimum cost is $C\left( 150\right) $.\FRAME{dtbpF}{1.7997in}{%
3.4774in}{0pt}{}{}{first2.bmp}{\special{language "Scientific Word";type
"GRAPHIC";display "USEDEF";valid_file "F";width 1.7997in;height
3.4774in;depth 0pt;original-width 0.0839in;original-height 0.1496in;cropleft
"0.0625";croptop "1";cropright "0.9375";cropbottom "0";filename
'first2.bmp';file-properties "XNPEU";}}%
\begin{eqnarray*}
C\left( 150\right) &=&\$50\sqrt{200^{2}+150^{2}}+\$30\left( 600-150\right)
=\$50\cdot 250+\$30\cdot 450 \\
&=&\$26\,000
\end{eqnarray*}%
b) \ Use the second derivative test to prove that we found a relative
minimum in part a).\newline
Solution: \ We already know that $C^{\prime }\left( 150\right) =0$. \ If $%
C^{\prime \prime }\left( 150\right) $ is positive, then $C$ has a relative
minimum at $x=150$. \ If $C^{\prime \prime }\left( 150\right) $ is negative,
then $C$ has a relative maximum at $x=150$. \ If $C^{\prime \prime }=0,$ the
second derivative test is inconclusive.%
\begin{eqnarray*}
C^{\prime }\left( x\right) &=&\dfrac{50x}{\sqrt{200^{2}+x^{2}}}-30 \\
C^{\prime \prime }\left( x\right) &=&\dfrac{50\sqrt{200^{2}+x^{2}}-50x\dfrac{%
1}{2\sqrt{200^{2}+x^{2}}}\left( 2x\right) }{200^{2}+x^{2}}=\dfrac{50\sqrt{%
200^{2}+x^{2}}-\dfrac{50x^{2}}{\sqrt{200^{2}+x^{2}}}}{200^{2}+x^{2}} \\
C^{\prime \prime }\left( x\right) &=&\dfrac{50}{200^{2}+x^{2}}\left( \sqrt{%
200^{2}+x^{2}}-\dfrac{x^{2}}{\sqrt{200^{2}+x^{2}}}\right) \\
C^{\prime \prime }\left( 150\right) &=&\dfrac{50}{200^{2}+150^{2}}\left( 
\sqrt{200^{2}+150^{2}}-\dfrac{150^{2}}{\sqrt{200^{2}+150^{2}}}\right) =%
\dfrac{16}{125}>0
\end{eqnarray*}%
Since its second derivative is positive at $x=150$, the function $C$ has a
relative minimum at $x=150$.\newline
c) \ Prove that the relative minimum we found in part a) \ is also an
absolue minimum on the domain $\left[ 0,600\right] .$\newline
Solution: \ Since the function $C\left( x\right) $ is continuous on the
closed interval $\left[ 0,600\right] ,$ it achieves the ansolute minimum and
absolute maximum. \ To find these, we only need to evaluate the function at
the relative extrema and the endpoints of the interval. \ We need to compute 
$C\left( 0\right) $, $C\left( 150\right) ,$ and $C\left( 600\right) $. \
These values are $C\left( 0\right) =\$28\,000$, \ $C\left( 150\right)
=\$26\,000$, and $C\left( 600\right) =\$10\,000\sqrt{10}\approx
\$31622.\,\allowbreak 776\,602$. \ Thus $C$ has an absolute minimum at $%
x=150 $.\newline
d) \ Prove that if we define $C$ on the set of all real numbers, then the
minimum we found in part a) \ is still an absolute minimum.\newline
Solution: \ Finding absolute extrema on domains others than a closed
interval is often tricky. \ In this case, we can prove the statement by
looking at $C^{\prime \prime }$ more carefully.%
\begin{eqnarray*}
C^{\prime \prime }\left( x\right) &=&\dfrac{50}{200^{2}+x^{2}}\left( \sqrt{%
200^{2}+x^{2}}-\dfrac{x^{2}}{\sqrt{200^{2}+x^{2}}}\right) =\dfrac{50}{%
200^{2}+x^{2}}\left( \dfrac{\left( \sqrt{200^{2}+x^{2}}\right) ^{2}}{\sqrt{%
200^{2}+x^{2}}}-\dfrac{x^{2}}{\sqrt{200^{2}+x^{2}}}\right) \\
&=&\dfrac{50}{200^{2}+x^{2}}\left( \dfrac{200^{2}+x^{2}-x^{2}}{\sqrt{%
200^{2}+x^{2}}}\right) =\dfrac{50}{200^{2}+x^{2}}\left( \dfrac{200^{2}}{%
\sqrt{200^{2}+x^{2}}}\right)
\end{eqnarray*}%
Since $C^{\prime \prime }$ is positive for all $x$, $C^{\prime }$ is
increasing for all $x$. \ So the zero at $x=150$ is the only zero of $%
C^{\prime }$; \ $C^{\prime }$ is negative on $\left( -\infty ,150\right) $
and positive on $\left( 150,\infty \right) $. \ Consequently, $C$ is
decreasing on on $\left( -\infty ,150\right) $ and increasing on $\left(
150,\infty \right) $, and so $C$ has an absolute minimum at $x=150$.

\item The location function of an object is $s\left( t\right) =\dfrac{120}{%
1+2e^{-t}}$ where $t$ is time, measured in seconds. \ Where is the object
when it is moving with the greatest speed? \ When is that greatest speed
achieved?\newline
Solution: 
\begin{eqnarray*}
s\left( t\right) &=&\dfrac{120}{1+2e^{-t}}=120\left( 1+2e^{-t}\right) ^{-1}
\\
v\left( t\right) &=&s^{\prime }\left( t\right) =120\left( -1\right) \left(
1+2e^{-t}\right) ^{-2}\left( 2e^{-t}\right) \left( -1\right) =\dfrac{%
240e^{-t}}{\left( 1+2e^{-t}\right) ^{2}}=240\dfrac{e^{-t}}{\left(
1+2e^{-t}\right) ^{2}}
\end{eqnarray*}%
We need to find the maximum of $v$ which means we need to differentiate
again. \ This time we will need to use the quotient rule.%
\begin{eqnarray*}
a\left( t\right) &=&v^{\prime }\left( t\right) =240\dfrac{e^{-t}\left(
-1\right) \left( 1+2e^{-t}\right) ^{2}-e^{-t}\left( 2\right) \left(
1+2e^{-t}\right) \left( 2e^{-t}\right) \left( -1\right) }{\left(
1+2e^{-t}\right) ^{4}}=\text{ \ \ \ simplify by }\left( 1+2e^{-t}\right) \\
&=&240\dfrac{e^{-t}\left( -1\right) \left( 1+2e^{-t}\right) -e^{-t}\left(
2\right) \left( 2e^{-t}\right) \left( -1\right) }{\left( 1+2e^{-t}\right)
^{3}}\text{ \ \ \ \ \ \ factor out }e^{-t} \\
&=&240e^{-t}\dfrac{-\left( 1+2e^{-t}\right) +4e^{-t}}{\left(
1+2e^{-t}\right) ^{3}}=240e^{-t}\dfrac{-1-2e^{-t}+4e^{-t}}{\left(
1+2e^{-t}\right) ^{3}}=240e^{-t}\dfrac{2e^{-t}-1}{\left( 1+2e^{-t}\right)
^{3}}
\end{eqnarray*}%
$240e^{-t}$ and the denominator are always positive. \ The only way $%
v^{\prime }\left( t\right) =0$ is when $2e^{-t}-1=0$. \ We solve this
equation for $t$.%
\begin{eqnarray*}
2e^{-t}-1 &=&0 \\
2e^{-t} &=&1 \\
e^{-t} &=&\dfrac{1}{2} \\
-t &=&\ln \left( \dfrac{1}{2}\right) \\
t &=&-\ln \left( \dfrac{1}{2}\right) =\ln 2
\end{eqnarray*}%
Thus $t=\ln 2$ is when the object is fastest. \ The location of the object
is then 
\begin{equation*}
s\left( \ln 2\right) =\dfrac{120}{1+2e^{-\ln 2}}=\dfrac{120}{1+2\left( 
\dfrac{1}{2}\right) }=\dfrac{120}{2}=60
\end{equation*}
\end{enumerate}

\bigskip

\bigskip

\bigskip

\bigskip

Save for quizzes:

\bigskip

\begin{enumerate}
\item A manufacturer estimates that when $q$ units of a particular product
are produced each month, the total cost will be $C(q)=5q+17000$ dollars, and
all $q$ units can be sold at a price of $p(q)=65-\dfrac{q}{100}$ \ dollars
per unit.

a) \ Find the fixed cost. $\ \ \ \ \ \ \ \ \ \ \ \$17000$

b) \ Determine the level of production that results in a maximum profit. $~~~%
%TCIMACRO{\TeXButton{red}{\color{red}}}%
%BeginExpansion
\color{red}%
%EndExpansion
3000$ \ \ 
%TCIMACRO{\TeXButton{red}{\color{red}}}%
%BeginExpansion
\color{red}%
%EndExpansion
unit per month%
%TCIMACRO{\TeXButton{black}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion
\newline
Solution:%
\begin{eqnarray*}
P\left( q\right) &=&R\left( q\right) -C\left( q\right) =q\cdot p\left(
q\right) -C\left( q\right) \\
&=&q\left( 65-\dfrac{q}{100}\right) -\left( 5q+17000\right) \\
&=&65q-\dfrac{1}{100}q^{2}-5q-17000 \\
&=&-\dfrac{1}{100}q^{2}+60q-17000
\end{eqnarray*}%
Since we have an upside down parabola, the extrema is indeed a maximum.%
\begin{equation*}
P^{\prime }\left( q\right) =-\dfrac{1}{100}\left( 2q\right) +60=-\dfrac{1}{50%
}q+60
\end{equation*}%
The extrema is where the derivative is zero.%
\begin{eqnarray*}
-\dfrac{1}{50}q+60 &=&0 \\
60 &=&\dfrac{q}{50} \\
3000 &=&q
\end{eqnarray*}%
Thus $3000$ \ unit per month production level will yield for maximal profit.
\ How can be sure that this is not a minimum? \ The derivative \ $-\dfrac{1}{%
50}q+60=-\dfrac{1}{50}\left( q-3000\right) $ \ changes sign from positive to
negative, indicating a relative maximum.

\begin{enumerate}
\item What is the maximum profit?$~~~%
%TCIMACRO{\TeXButton{red}{\color{red}}}%
%BeginExpansion
\color{red}%
%EndExpansion
\$~73\,000$%
%TCIMACRO{\TeXButton{black}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion
\end{enumerate}

\item A box manufacturer desires to create a box with a surface area of $24%
\unit{ft}^{2}$. \ What is the maximum size volume that can be formed by
bending this material into a box? The box is to be closed. The box is to
have a square base, square top, and rectangular sides. \ \ $8\unit{ft}^{3}$%
\newline
Solution: \ Let $x$ denote the dimensions of the base and $h$ denote the
height of the box. \ We start by expressing the surface of the box.%
\begin{equation*}
24=2x^{2}+4xh
\end{equation*}%
We solve this for $h:$%
\begin{eqnarray*}
24 &=&2x^{2}+4xh \\
24-2x^{2} &=&4xh \\
\dfrac{24-2x^{2}}{4x} &=&h \\
h &=&\dfrac{24-2x^{2}}{4x}=\dfrac{24}{4x}-\dfrac{2x^{2}}{4x}=\dfrac{6}{x}-%
\dfrac{x}{2}=-\dfrac{1}{2}x+\dfrac{6}{x}
\end{eqnarray*}%
We now express the volume of the box: $V=x^{2}h.$ \ Thus $V\left( x\right)
=x^{2}\left( -\dfrac{1}{2}x+\dfrac{6}{x}\right) =-\dfrac{1}{2}x^{3}+6x$.%
\begin{equation*}
V\left( x\right) =-\dfrac{1}{2}x^{3}+6x\text{ \ \ \ \ \ \ \ \ }V^{\prime
}\left( x\right) =-\dfrac{3}{2}x^{2}+6
\end{equation*}%
We solve for the zeroes of the derivative:%
\begin{eqnarray*}
-\dfrac{3}{2}x^{2}+6 &=&0\text{ \ \ \ \ \ \ \ \ \ \ add }\dfrac{3}{2}x^{2} \\
6 &=&\dfrac{3}{2}x^{2}\text{ \ \ \ \ \ divide by }\dfrac{3}{2} \\
4 &=&x^{2} \\
\pm 2 &=&x
\end{eqnarray*}%
Since $x$ denotes a distance, it cannot be negative, and so $x=2.$ \ $%
V\left( 2\right) =-\dfrac{1}{2}\cdot 2^{3}+6\cdot 2=8$ \ The greatest
possible volume is $8\unit{ft}^{3}.$
\end{enumerate}

\end{document}