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\begin{document}

\newcommand{\limp}{\rightarrow}
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\noindent
\noindent
\parbox{2in}{\bf Math 1030} 
\hfill {\Large \bf Workshop \#3} \hfill
\parbox{2in}{\bf \hfill Due: July $17^{\mathrm{th}}$, 2015}

\vspace{0.15in}

\noindent {\large\bf Name:} \underline{\hskip 3.0 truein} 


\vspace{0.2in}

\noindent {\bf 1.} Stewart sells specialty bicycles. He has collected the following data about a particular bike:

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
Bikes Sold                   &               50 &        100 &         200 &           300  \\ \hline
Demand Price             &     \$1,925 &     \$935 &     \$960 &     \$1,155  \\ \hline
   Supply Price             &        \$480 &     \$520 &     \$750  &    \$1,100  \\ \hline
\end{tabular}
\end{center}

For example, if Stewart charges \$935 per bike, he can expect to sell 100 bikes. He can obtain 100 of these bikes for \$52,000 (i.e. \$520 a piece).

\begin{enumerate}[(a)]
   \item Use the table of data to find supply and demand price functions. 
            Use a {\bf cubic} model for the demand function and a {\bf quadratic} model for the supply function.

\vspace{0.1in}

Demand function: $p_d=$ \underline{\hspace{3in}} 

\vspace{0.2in}

Supply function: $p_s=$ \underline{\hspace{3in}}

\vspace{0.1in}

The market equilibrium is $\displaystyle{(q_E, p_E) = \Big(}$ \underline{\hspace{1.5in}} , \underline{\hspace{1.5in}} $\displaystyle{\Big)}$.\\
{}[Round the quantity to 3 decimal places and the price to dollars and change.]

\vspace{0.1in}

If Stewart charges \$1,500 per bike, how many can he expect to sell? \hfill \underline{\hspace{1.25in}}\\
{}[Round to 3 decimal places.]

\vspace{0.1in}

If Stewart wants to sell 250 bikes, what will his (average supply) cost per bike be? \hfill \$\underline{\hspace{1.25in}}

\vspace{0.1in}

\item Use your model for the demand function to find a revenue function. Use your supply function to model the variable cost per bike and the fact that Stewart has fixed costs of \$10,000 to find his cost function. Finally, use your revenue and cost functions to find profit, marginal revenue, marginal cost, and marginal profit functions.

\vspace{0.1in}

Stewart has \underline{\hspace{0.5in}} break even points. These occur at $q=$ \underline{\hspace{2in}} bikes.\\
{}[Round all break even quantities to 3 decimal places.]

\vspace{0.1in}

Stewart's $75^{\mathrm{th}}$ bike brings in \$\underline{\hspace{1.5in}} revenue. 

\vspace{0.1in}

The $75^{\mathrm{th}}$ bike cost him \$\underline{\hspace{1.5in}}.

\vspace{0.1in}

Stewart's marginal profit is \$0 when $q=$ \underline{\hspace{2in}} bikes.\\
{}[List \underline{\bf all} quantities where $MP(q)=0$. Round quantities to 3 decimal places.] 

\vspace{0.1in}

Stewart will maximize his profit if he sells \underline{\hspace{1in}} bikes at a price of \$\underline{\hspace{1.5in}} each.

\vspace{0.1in}

His maximum possible profit is \$\underline{\hspace{2.25in}}.

\item Does Stewart's cost function have a minimum? In a few sentences explain why or why not.

\end{enumerate}

\newpage

\noindent {\bf 2.} Use Excel to compute the following limits. If the limit does not exist write ``DNE''.

\begin{enumerate}[(a)]

\item Let $\displaystyle f(x) = \left\{ \begin{array}{ccc} \ln(x^2+1) & & \mbox{if } x \leq 2 {\large\strut} \\  e^{-2x} & & \mbox{if } x > 2 {\large\strut} \end{array}\right.$

\vspace{0.5in}

\hspace{0.25in} $\displaystyle{ \lim_{x \rightarrow 2} \  f(x)  = }$ \underline{\hspace{1.5in}}

\vspace{0.5in}

\hspace{0.25in} $\displaystyle{ \lim_{x \rightarrow 1} \  f(x)  = }$ \underline{\hspace{1.5in}}


\vspace{0.2in}

\item \hspace{0.25in} $\displaystyle{ \lim_{x \rightarrow -3} \; \frac{(x+3)(x+5)}{\ln \,|x+4|} = }$ \underline{\hspace{1.5in}}

{}[{\it Note:} $|x+4|$ is the absolute value of $x+4$. In Excel, ``ABS'' is the absolute value function.]


\vspace{0.1in}

\end{enumerate}

\noindent {\bf 3.} Recall that the {\bf derivative} of $f(x)$ is defined to be 

\[ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \]

        where $\displaystyle{\frac{f(x+h)-f(x)}{h}}$ is called the {\bf difference quotient} of $f(x)$. 

\vspace{0.25in}      

 Let $f(x)=(x-4)^{2/3}$.
 
 \vspace{0.1in}
 
 {\large\bf\color{red}Warning:} Excel doesn't handle negative cube roots well.  If you get ``NUM!'' errors, try this fix: You will need to enter $f(x)$ as $((x-4)^2)^{1/3}$. If $x$ is in A1, this is ``\verb|=((A1-4)^2)^(1/3)|''.

\vspace{0.1in}      

\begin{enumerate}[(a)]

\item Compute the difference quotient of $f(x)$ when $x=2$ and $h=0.1$. 

\vspace{0.1in}

        $\displaystyle{\frac{f(2+0.1) - f(2)}{0.1} = }$ \underline{\hspace{1.5in}}

\vspace{0.1in}

         Now use Excel to compute the limit as $h \rightarrow 0$. This shows that $f'(2) \approx$ \underline{\hspace{1.25in}}.

        \vspace{0.1in}

\item Use Excel to repeat the previous part for $x=4$ (try to compute $f'(4)$). Does the limit ($h \to 0$) of the difference quotient at $x=4$ exist? If it does exist, what is it? If it does not exist, why not? Explain your answer in a sentence or two.

\end{enumerate}

\end{document}

