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\lhead{\color{blue} \large Lecture Notes}
\chead{\color{black} \Large Optimization 2B}
\rhead{\small page   \ \thepage}
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\lfoot{\footnotesize   \copyright $\;$  Hidegkuti,  Powell,  2011}
\rfoot{\footnotesize  Last revised: March 27, 2019}
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\begin{document}


\begin{center}
{\Large Sample Problems}
\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 An open box (no top) with a square base is to have a volume of $60%
\unit{in}^{3}$. \ What dimensions would guarantee the least amount of
material needed?

\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 We are designing a poster to contain $60\unit{in}^{2}$ of printing
with a $2-$ inch wide margin at the top and bottom and a $1-$ inch wide
margin at each side. \ What overall dimensions will minimize the amount of
paper used?

\item Consider all lines with negative slopes that pass through the point $%
P\left( 8,2\right) $. \ Let us denote the origin by $O,$ the $x-$intercept
of the line by $A$ and its $y-$intercept by $B$. \ What is the smallest
possible area of triangle $OAB$?
\end{enumerate}

\bigskip

\begin{center}
{\Large Practice Problems}
\end{center}

\begin{enumerate}
\item We are designing a poster to contain $60\unit{in}^{2}$ of printing
with a $3-$ inch wide margin at the top and bottom and a $2-$ inch wide
margin at each side. \ What overall dimensions will minimize the amount of
paper used?

\item Let $x$ and $y$\ be positive numbers such that $xy=1$. \ Find the
lowest possible value of $x^{3}+2y^{3}$.

\item We would like to constract an open box with a square base. \ The box
to have a volume of $200\unit{in}^{3}$. \ What dimensions would guarantee
that the box can be made using the least amount of material?

\item A company wants to manufacture cylindrical aluminum cans with a volume
of $200\pi $ cubic centimeters. \ What dimensions would guarantee the
minimal amount of aluminum needed to produce a can?

\item Consider all lines with negative slopes that pass through the point $%
P\left( 3,12\right) $. \ Let us denote the origin by $O,$ the $x-$intercept
of the line by $A$ and its $y-$intercept by $B$. \ What is the smallest
possible area of triangle $OAB$? \ \ \ \ 

\item We would like to design a flowerbed in the shape of a circular sector.
\ If the area of the sector needs to be $20\unit{m}^{2}$, then what is the
smallest possible perimeter?

\pagebreak
\end{enumerate}

\begin{center}
{\Large Sample Problems - Answers}
\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 $x=\sqrt[3]{120}$ and $y=\dfrac{1}{2}\sqrt[3]{120}$

\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 horizontal side $\sqrt{30}+2$ inches long, and the vertical side $2%
\sqrt{30}+4$ inches long

\item $m=-\dfrac{1}{4}$
\end{enumerate}

\begin{center}
{\Large Practice Problems - Answers}
\end{center}

\begin{enumerate}
\item the print area is $2\sqrt{10}\unit{in}$ (horizontal) \ and $3\sqrt{10}%
\unit{in}$ (vertical),

the paper is $2\sqrt{10}+4\unit{in}$ (horizontal) \ and $3\sqrt{10}+6\unit{in%
}$ (vertical)

\item $2\sqrt{2}$ \ \ (when $x=\sqrt[6]{2}$)

\item base: $\sqrt[3]{400}\unit{in}$ by $\sqrt[3]{400}\unit{in}$ \ \ height: 
$\dfrac{1}{2}\sqrt[3]{400}\unit{in}$

\item $r=\sqrt[3]{100}\unit{cm}$ \ and $h=2\sqrt[3]{100}\unit{cm}$

\item $m=-4$

\item $r=2\sqrt{5}$ \ \ \ \ \ $\alpha =2$rad
\end{enumerate}

{\Large \pagebreak }

\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 $a^{2}+4b^{2}$.

Solution: \ We solve for $a$ 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*}%
The critical numbers for $P$ are $-\sqrt{5}$, $0$, and $\sqrt{5}$. \ All
relative maximums or minimums will be here. \ We can figure out when $%
P^{\prime }$ is positive and negative by sorting out the signs of each
factor 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\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 $P^{\prime }$ only, \ $P$ has a relative minimum at $%
b=-\sqrt{5}$ and \ $\sqrt{5}$ and a relative maximum at $0$. \ However, the
function does not have a relative maximum at zero. \ Looking at the formula
for the original function, $P\left( b\right) =4b^{2}+\dfrac{100}{b^{2}}$, we
see that there is a vertical asymptote and the graph shoots up toward plus
infinity on both sides of the asymptote. \ \ Not to mention tha fact that $a$
and $b$ must both be positive. \ Since $b$ must be positive, we may consider 
$P$ on the domain $\left( 0,\infty \right) $. \ On this domain, $P$ is
continuous and differentiable everywhere, 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 \fbox{$40$}.

\pagebreak

\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 An open box (no top) with a square base is to have a volume of $60%
\unit{in}^{3}$. \ What dimensions would guarantee the least amount of
material needed?

Solution 1: \ Let us denote the sides of the square base by $x$ and the
vertical side by $y$. \ 

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The volume of the box is $V=x^{2}y,$ so we have that 
\begin{equation*}
60=x^{2}y\text{ \ \ \ \ \ solve for }y\text{: \ \ \ \ }y=\dfrac{60}{x^{2}}
\end{equation*}%
The amount of material needed: $x^{2}$ for the bottom and $4xy$ for the
vertical sides. \ So the function, whose minimum we are to find, is%
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\begin{equation*}
A=x^{2}+4xy=x^{2}+4x\left( \dfrac{60}{x^{2}}\right) =x^{2}+\dfrac{240}{x}%
=x^{2}+240x^{-1}
\end{equation*}

We differentiate: $\dfrac{d}{dx}\left( x^{2}+\dfrac{240}{x}\right) =2x-%
\dfrac{240}{x^{2}}=\dfrac{2x^{3}-240}{x^{2}}$

Recall the difference of cubes theorem:%
\begin{equation*}
A^{3}-B^{3}=\left( A-B\right) \left( A^{2}+AB+B^{2}\right)
\end{equation*}%
where the second factor is a sum of two squares, always positive. \ (same as 
$\left( A+\dfrac{B}{2}\right) ^{2}+\dfrac{3}{4}B^{2}$). \ Using this, we can
factor the denominator:%
\begin{equation*}
f^{\prime }\left( x\right) =\dfrac{2x^{3}-240}{x^{2}}=\dfrac{2\left(
x^{3}-120\right) }{x^{2}}=\dfrac{2\left( x-\sqrt[3]{120}\right) \left( x^{2}+%
\sqrt[3]{120}x+\left( \sqrt[3]{120}\right) ^{2}\right) }{x^{2}}
\end{equation*}%
The denominator is always positive, and so is the second, longer factor. \
The only factor that changes sign is the line $y=x-\sqrt[3]{120}$. \ At $x=%
\sqrt[3]{120}$, this line changes sign from negative to positive, and so
does $f^{\prime }$. \ Therefore, $f$ has a relative minimum at $x=\sqrt[3]{%
120}$.

Recall that $y=\dfrac{60}{x^{2}}$. \ So $y=\dfrac{60}{\left( \sqrt[3]{120}%
\right) ^{2}}=\dfrac{60}{\left( \sqrt[3]{120}\right) ^{2}}\cdot \dfrac{\sqrt[%
3]{120}}{\sqrt[3]{120}}=\dfrac{60\sqrt[3]{120}}{\left( \sqrt[3]{120}\right)
^{3}}=\dfrac{60\sqrt[3]{120}}{120}=\dfrac{\sqrt[3]{120}}{2}=\dfrac{x}{2}$. \
So the dimensions that need the least amount of material are: \fbox{$x=\sqrt[%
3]{120}$ and $y=\dfrac{1}{2}\sqrt[3]{120}$}.

Solution 2. \ Let us denote the sides of the square base by $x$ and the
vertical side by $y$. \ 

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The volume of the box is $V=x^{2}y,$ so we have that 
\begin{equation*}
60=x^{2}y\text{ \ \ \ \ \ solve for }y\text{: \ \ \ \ }y=\dfrac{60}{x^{2}}
\end{equation*}%
The amount of material needed: $x^{2}$ for the bottom and $4xy$ for the
vertical sides. \ So the function, whose minimum we are to find, is%
%TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}%
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\begin{equation*}
A=x^{2}+4x=x^{2}+4x\left( \dfrac{60}{x^{2}}\right) =x^{2}+\dfrac{240}{x}%
~~~~~~f\left( x\right) =x^{2}+\dfrac{240}{x}
\end{equation*}

We differentiate: $f^{\prime }\left( x\right) =2x-\dfrac{240}{x^{2}}$ \ and
again: \ $f^{\prime \prime }\left( x\right) =2+\dfrac{480}{x^{3}}$

Find the critical numbers: \ Solve $f^{\prime }\left( x\right) =0$ for $x.$%
\begin{eqnarray*}
f^{\prime }\left( x\right) &=&0 \\
2x-\dfrac{240}{x^{2}} &=&0 \\
2x &=&\dfrac{240}{x^{2}} \\
2x^{3} &=&240 \\
x^{3} &=&120 \\
x &=&\sqrt[3]{120}
\end{eqnarray*}%
Since $x$ is a side of the box, it must be a positive number. \ If so, the
second derivative%
\begin{equation*}
f^{\prime \prime }\left( x\right) =2+\dfrac{480}{x^{3}}
\end{equation*}%
is positive for all values of $x$ in the domain, including \ at $x=\sqrt[3]{%
120}$. \ By the second derivative test, $f$ has a minimum at $x=\sqrt[3]{120}
$. \ Recall that $y=\dfrac{60}{x^{2}}$. \ So $y=\dfrac{60}{\left( \sqrt[3]{%
120}\right) ^{2}}=\dfrac{60}{\left( \sqrt[3]{120}\right) ^{2}}\cdot \dfrac{%
\sqrt[3]{120}}{\sqrt[3]{120}}=\dfrac{60\sqrt[3]{120}}{\left( \sqrt[3]{120}%
\right) ^{3}}=\dfrac{60\sqrt[3]{120}}{120}=\dfrac{\sqrt[3]{120}}{2}=\dfrac{x%
}{2}$. \ So the dimensions that need the least amount of material are: \fbox{%
$x=\sqrt[3]{120}$ and $y=\dfrac{1}{2}\sqrt[3]{120}$}.

\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 }}\approx
5.\,\allowbreak 419\,26
\end{eqnarray*}%
and 
\begin{equation*}
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\approx
10.\,\allowbreak 838\,52
\end{equation*}%
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.

\pagebreak

\item We are designing a poster to contain $60\unit{in}^{2}$ of printing
with a $2-$ inch wide margin at the top and bottom and a $1-$ inch wide
margin at each side. \ What overall dimensions will minimize the amount of
paper used?

Solution: \ Let $x$ be the horizontal side of the printing and $y$ the
vertical side of the printing. \ 

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Then%
\begin{equation*}
xy=60\text{ \ \ \ }\Longrightarrow \text{ \ \ \ }y=\dfrac{60}{x}
\end{equation*}%
The entire page then has sides $x+2$ and $y+4$ and therefore area $A=\left(
x+2\right) \left( y+4\right) $. \ We are looking for the minimum value of
this expression.%
\begin{eqnarray*}
A &=&\left( x+2\right) \left( y+4\right) =xy+4x+2y+8\text{ \ \ \ \ \ recall
that }xy=60 \\
&=&60+4x+2y+8=68+4x+2y\text{ \ \ \ \ \ \ \ recall that \ }y=\dfrac{60}{x} \\
A\left( x\right) &=&68+4x+2\left( \dfrac{60}{x}\right) =68+4x+\dfrac{120}{x}
\end{eqnarray*}%
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We differentiate $A\left( x\right) $%
\begin{equation*}
A^{\prime }\left( x\right) =4-\dfrac{120}{x^{2}}=\dfrac{4x^{2}-120}{x^{2}}=%
\dfrac{4\left( x^{2}-30\right) }{x^{2}}=\dfrac{4\left( x+\sqrt{30}\right)
\left( x-\sqrt{30}\right) }{x^{2}}
\end{equation*}

The denominator is always positive. \ The numerator is a quadratic
expression, positive on $\left( -\infty ,-\sqrt{30}\right) $ and $\left( 
\sqrt{30},\infty \right) $ \ and negative on $\left( -\sqrt{30},\sqrt{30}%
\right) $. \ At $x=\sqrt{30}$, the derivative $A^{\prime }$ changes sign
from negative to positive, indicating a minimum. \ \ Therefore, $x=\sqrt{30}%
, $ and $y=\dfrac{60}{x}=\dfrac{60}{\sqrt{30}}=\dfrac{60\sqrt{30}}{30}=2%
\sqrt{30}$. \ The sheet of paper than needs to have sides $x+2=\sqrt{30}+2$
and $y+4=2\sqrt{30}+4$. \ So the sides must be \fbox{$\sqrt{30}+2$ inches
and $2\sqrt{30}+4$ inches} long.

\item Consider all lines with negative slopes that pass through the point $%
P\left( 8,2\right) $. \ Let us denote the origin by $O,$ the $x-$intercept
of the line by $A$ and its $y-$intercept by $B$. \ What is the smallest
possible area of triangle $OAB$?

Solution: \ Let $m<0$ be the slope of the line. \ Using the point-slope
form, the equation of the line $AB$ is $y-2=m\left( x-8\right) $ \ or $%
y=m\left( x-8\right) +2$.

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Let us find the intercepts in terms of $m$. \ If $x=0$, then \vspace{0.03in}

$y=m\left( x-8\right) +2$ becomes $y=m\left( -8\right) +2=-8m+2$.\vspace{%
0.03in} \ 

Thus the $y-$intercept is $\left( 0,-8m+2\right) $.\vspace{0.03in}\vspace{%
0.03in}\vspace{0.03in}

If $y=0$, then\vspace{0.03in} $y=m\left( x-8\right) +2$ becomes $0=m\left(
x-8\right) +2$ \ \ We solve for $x.$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $%
0=m\left( x-8\right) +2$\vspace{0.03in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $-2=m\left(
x-8\right) $\vspace{0.03in}\ 

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $-\dfrac{2}{m}%
=x-8$\vspace{0.03in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $8-\dfrac{2}{m}=x$%
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Thus the $x-$intercept is $\left( 8-\dfrac{2}{m},0\right) $.

The area of triangle $OAB$ is $A\left( m\right) =\dfrac{1}{2}\left(
-8m+2\right) \left( 8-\dfrac{2}{m}\right) =-32m+16-\dfrac{2}{m}$. \ We
differentiate $A\left( m\right) $%
\begin{equation*}
A^{\prime }\left( m\right) =-32+\dfrac{2}{m^{2}}\text{ \ \ we solve for the
zeroes of }A^{\prime }\left( m\right)
\end{equation*}%
\begin{eqnarray*}
-32+\dfrac{2}{m^{2}} &=&0 \\
\dfrac{2}{m^{2}} &=&32 \\
2 &=&32m^{2} \\
\dfrac{1}{16} &=&m^{2} \\
m &=&\pm \dfrac{1}{4}
\end{eqnarray*}%
Because $m<0$, $m=-\dfrac{1}{4}$. \ We differentiate again: \ 
\begin{equation*}
A^{\prime }\left( m\right) =-32+\dfrac{2}{m^{2}}~~~~\Longrightarrow
~~~~A^{\prime \prime }\left( m\right) =-\dfrac{4}{m^{3}}
\end{equation*}

Since $m$ is negative, so is $m^{3}$. \ Therefore, $-\dfrac{4}{m^{3}}$ is
positive for all $m<0$, indicating a minimum at $m=-\dfrac{1}{4}$ by the
second derivative test.

Another way to veify that $A$ has a minimum at $x=-\dfrac{1}{4}$ is to look
at $A^{\prime }\left( m\right) $.

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $A^{\prime }\left(
m\right) =-32+\dfrac{2}{m^{2}}=\dfrac{-32m^{2}+2}{m^{2}}=\dfrac{-32\left(
m^{2}-\dfrac{1}{16}\right) }{m^{2}}=\dfrac{-32\left( m+\dfrac{1}{4}\right)
\left( m-\dfrac{1}{4}\right) }{m^{2}}$

The denominator is always positive. \ The numerator is a downward opening
parabola, negative on $\left( -\infty ,-\dfrac{1}{4}\right) $ and on $\left( 
\dfrac{1}{4},\infty \right) $, and positive on $\left( -\dfrac{1}{4},\dfrac{1%
}{4}\right) $. \ Therefore, $A^{\prime }\left( m\right) $ changes sign from
negative to positive, indicating a minimum at \fbox{$m=-\dfrac{1}{4}$}.
\end{enumerate}

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\href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{%
For more documents like this, visit our page at\
https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail
questions or comments to mhidegkuti@ccc.edu.}

\end{document}