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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 $19^{\mathrm{th}}$, 2016}

\vspace{0.15in}

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


\vspace{0.2in}

\noindent {\bf 1.} Hank is a ticket scalping middleman. He has collected the following data about tickets to a local themepark: 
{}[{\it Note:} For this whole problem, restrict your attention to the range \fbox{$\bf 0 \leq q \leq 300$}. Our models will only be useful for these quantities.]

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
Tickets Sold                   &       50 &     100 &   200 &  300  \\ \hline
Demand Price             & \$120 & \$115 &  \$80 & \$65  \\ \hline
   Supply Price             &   \$45 &   \$65 &   \$70 & \$80  \\ \hline
\end{tabular}
\end{center}

For example, if Hank charges \$120 per ticket, he can expect to sell 50 tickets. On the other hand, he can obtain 50 of these tickets for \$2,250 (i.e. \$45 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.
            {}[Use ``Format trendline label" to \underline{\bf show 6 decimal places} of accuracy for the {\bf cubic} model.]

\vspace{0.1in}

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

\vspace{0.1in}

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. Stay inside $0 \leq q \leq 300$.]

\vspace{0.1in}

If Hank charges \$90 per ticket, how many can he expect to sell? \hfill \underline{\hspace{1.25in}}\\
{}[Round to 3 decimal places.]

\vspace{0.1in}

If Hank wants to sell 125 tickets, what will his (average supply) cost per ticket 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 Hank has fixed costs of \$2,000 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}

Hank has \underline{\hspace{0.5in}} break even points. These occur at $q=$ \underline{\hspace{2in}} tickets.\\
{}[Round all quantities to 3 decimal places and remember to ignore answers outside $0 \leq q \leq 300$.]

\vspace{0.1in}

Hank's $220^{\mathrm{th}}$ ticket brings in \$\underline{\hspace{1.5in}} revenue. 
This ticket cost him \$\underline{\hspace{1.5in}}.

\vspace{0.1in}

Hank's marginal profit is \$0 when $q=$ \underline{\hspace{2in}} tickets.\\
{}[List \underline{\bf all} quantities where $MP(q)=0$ and $0 \leq q \leq 300$. Round quantities to 3 decimal places.] 

\vspace{0.1in}

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

\vspace{0.1in}

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

\item Does Hank's profit function have a minimum? Either explain why it does not have one or if it does find where it is.

\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 $\displaystyle f(x) = (x-3)\ln\left((x-3)^2\right)+\dfrac{1}{x+1}$

\vspace{0.5in}

\hspace{0.25in} $\displaystyle{ \lim_{x \rightarrow 3} \  f(x)  = }$ \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} $\displaystyle{ \lim_{x \rightarrow 0} \; e^{-1/x^2} = }$ \underline{\hspace{1.5in}}

\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 $f(x)=\left\{ \begin{array}{ccc} x\ln(x^2) & & x \not= 0 \\ 0 & & x=0 \end{array}\right.$.
 
 \vspace{0.1in}
  
\begin{enumerate}[(a)]

\item Compute the difference quotient of $f(x)$ when $x=3$ and $h=0.1$. 

\vspace{0.1in}

        $\displaystyle{\frac{f(3+0.1) - f(3)}{0.1} = }$ \underline{\hspace{1.5in}}

\vspace{0.1in}

         Now use Excel to compute the limit as $h \rightarrow 0$. This shows that $f'(3) \approx$ \underline{\hspace{1.25in}}.

        \vspace{0.1in}

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

{}[Be careful that $f(0)$ is defined to be $0$. The formula $x\ln(x^2)$ is undefined there: $0\cdot \ln(0^2)$ is undefined because $\ln(0)$ is undefined.]

\end{enumerate}

\end{document}

