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\noindent
{\bf Math 1030 Workshop \#6} \hfill
{\bf Name: \underline{\Large\color{red}\sc\quad Answer Key \quad}}

\vspace{0.05in}

\noindent
{\Large \bf DUE: Thursday, July 30${}^{\mbox{st}}$} Please turn in a paper copy and
{\bf SHOW YOUR WORK!}

\vspace{-0.05in}

\begin{enumerate}
   \item  Apple has rolled out its latest product\dots{}the iPen. The local iStore has collected the following iData about their iProduct's demand:

\hfill
\begin{tabular}{|c||c|c|c|c|} \hline
iPens Sold \varspace{0.25in}{0.01in} & 15   & 100  & 200 &  350 \\ \hline
            Price \varspace{0.25in}{0.01in} & \$1,000 & \$225 & \$100 & \$50 \\ \hline
\end{tabular}  
\hfill \hfill

\begin{enumerate}[(a)]
   \item Compute elasticity (round to 3 decimals places):

\begin{tabular}{|c||c|c|c|c|} \hline
\hspace{0.05in} iPens Sold \varspace{0.25in}{0.01in} & 15   & 100  & 200 &  350 \\ \hline
            Price \varspace{0.25in}{0.01in} & \quad \$1,000 \quad{} & \quad{} \$225 \quad{} & \quad \$100 \quad{} & \quad \$50 \quad{} \\ \hline
\end{tabular}  

\vspace{-0.05in}
    
\begin{tabular}{|c|c|c|c|c|} \hline
 \mbox{\parbox[c][0.4in]{1.3in}{Elasticity}} & \ \  1.1683  \ \  \;{} &  \ \   0.8667  \ \  {} &  \ \  0.8182  \  {} & \mbox{\parbox[c][0.2in]{0.155in}{ $\times$\\$\times$ }} \\ \hline
\end{tabular}

\vspace{0.1in}

   \item Model this demand price (in Excel) using an {\bf power} trendline: \hspace{0.25in}    $p_d(q) =$ \underline{\quad $13564 \; q^{-0.932}$ \quad}

\vspace{0.05in}

At this point, I'll switch from Excel to using Wolfram Alpha. Alpha does a better job of solving equations. 

To find what quantity goes with $p_d = \$75$ we use\dots \hspace{0.5in}
\verb|13564 q^-0.932 = 75| \href{https://www.wolframalpha.com/input/?i=13564+q%5E-0.932+%3D+75}{\ALPHA}

Alpha solves this equation and tell us that $q \approx 264.256$. 

\vspace{0.1in}

           According to this model, the store will sell \underline{\quad 264 \quad} iPens if they set their price at \$75. 

\vspace{0.05in}

Recall that point elasticity is $\varepsilon = - \dfrac{p}{q} \cdot \dfrac{dq}{dp} = -\dfrac{p}{d} \Big/ \dfrac{dp}{dq}$. Since we have our demand function: $p_d(q) = 13564 q^{-0.932}$ (price in terms of quantity), we need to use the second expression (we will be computing $\dfrac{dp}{dq}$ not $\dfrac{dq}{dp}$).
Let's use Alpha to find this function, then we can plug in $q=264.256$ and $p=75$.

\verb|derivative 13564 q^-0.932| \href{https://www.wolframalpha.com/input/?i=derivative+13564+q%5E-0.932}{\ALPHA}
\quad Alpha finds that the derivative is $-\dfrac{12641.6}{q^{1.932}}$. Putting this together, we find that  $\varepsilon = -\dfrac{p}{q} \cdot \dfrac{dq}{dp} = -\dfrac{13564 \; q^{-0.932}}{q} \Big/ \dfrac{-12641.6}{q^{1.932}}$. Now plug in $q=264.256$\dots

\vspace{0.05in}

\verb|-(13564q^-0.932 / q)/(-12641.6/q^1.932) at q=264.256| \href{https://www.wolframalpha.com/input/?i=-(13564q%5E-0.932%20%2F%20q)%2F(-12641.6%2Fq%5E1.932)%20at%20q%3D264.256&t=crmtb01}{\ALPHA}

\vspace{0.05in}

           When $p_d(q) = \$75$, our point elasticity (using this model) is $\varepsilon =$ \underline{\quad $1.073$ \quad} (round to 3 decimal places).

\vspace{0.05in}

Since $\varepsilon > 1$, quantity is dominant and so revenue moves against price. To increase revenue they should drop the price.

\vspace{0.05in}

          \underline{Circle the correct answer:} \qquad The store is currently charging \$75 for an iPen and wants to {\bf increase} its revenue, they should \varspace{0.5in}{0.01in}
         \quad {\Large raise \quad / \quad \fbox{lower}} \quad their price.
        
\vspace{0.05in}

To find where revenue is maximized we must solve the equation: $\varepsilon = 1$.

\verb|-(13564q^-0.932 / q)/(-12641.6/q^1.932) = 1| \href{https://www.wolframalpha.com/input/?i=-%2813564q%5E-0.932+%2F+q%29%2F%28-12641.6%2Fq%5E1.932%29+%3D+1}{\ALPHA}

Alpha correctly returns ``False''. This happens because our revenue function has \underline{\color{red}\bf no maximum}. 

\vspace{0.05in}

         \sout{What quantity and price will {\bf maximize} revenue? \quad $q =$ \underline{\qquad\qquad}  \quad $p=$ \underline{\qquad\qquad}}
        
\end{enumerate}

\vspace{-0.05in}

   \item Suppose that some items point elasticity is $\varepsilon = 1.5$ at some price. If the price is lowered by 5\%, \dots
   
%   \vspace{0.05in}
   
   We have that $1.5 = \varepsilon = - \dfrac{\mbox{percent change in quantity}}{\mbox{percent change in price}} = -\dfrac{\mbox{percent change in quantity}}{-0.05}$. Thus the percent change in quantity is $1.5 \times 0.05 = 0.075$.
   
   Suppose that revenue used to be $R=pq$. Then if the price is lowered by 5\%, it becomes $0.95p$. If the quantity is raised by 7.5\%, it becomes $1.075q$. Putting this together, our new revenue is $R=(0.95p)(1.075q) = 1.02125pq$. Thus the new revenue is 102.125\% of the old revenue.
   
   The quantity is {\Large \fbox{raised} } by \underline{\quad 7.5\% \quad} 
   and revenue is {\Large \fbox{raised} } by \underline{\quad 2.125\% \quad}. 

\vspace{0.5in}
             
   \item Kyle is painting a very odd wall. Its height varies quite a lot. He needs to estimate the square footage of the wall so he can figure out how much paint to buy. Suppose that the wall is 30 feet long. Let $x$ be the number of feet from the start of the wall and $y$ be the height (in feet) of the wall as measured by Kyle\dots
   
   \begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|c|} \hline
   $x=$ & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 & 30 \\ \hline
   $y=$ & 5 & 7 & 8 & 7 & 10 & 9 & 6 & 5 & 4 & 6 & 7 \\ \hline
   \end{tabular}

\vspace{0.05in}

Approximate the area of this wall in 2 different ways: (1) Using a right hand rule approximation with $n=10$ rectangles and (2) Using Simpson's rule with $n=10$. [Round each answer to 4 decimal places.]

\vspace{0.05in}

For both approximations we can use our textbook's Excel sheets. In both cases $a=0$, $b=30$, $n=10$ (so $\Delta x = 3$). Once $a$, $b$, and $n$ are typed in, the sheets list a bunch of $x$ values (partition points). Notice that Simpson's rule lists $0,3,\dots, 30$ while the Righthand rule sheet leaves off $0$ (it's the leftmost point, so righthand rule doesn't use it). 

We don't have a formula for our $f(x)$, but Kyle provided us with the values of $f(x)$ that we need. Just type in each of the values. So for the righthand rule sheet type in $7,8,7,10,$ etc. For Simpson's type in $5,7,8,7,$ etc.  The sheets then compute our desired answers.

\vspace{0.1in}

Right hand rule: \underline{\quad 207 square feet \quad}  \hfill
Simpson's rule: \underline{\quad 204 square feet \quad}  \hfill

\vspace{0.1in}

   \item Suppose that $R(t)=(t+0.5)^3e^{-t}$ models the construction rate of micro-houses (in millions of houses built per year). $t$ is the number of years since January 1, 2015. 
   
   \vspace{0.05in}
   
   Our first question asks about the rate function itself. We need to solve $R(t)=0.01$ (remember the rate is in millions of houses so 1 = 1 million houses, 0.1 = 100,000 houses, and 0.01 = 10,000 houses). 
   
   \verb|(t+0.5)^3e^(-t) = 0.01| \href{https://www.wolframalpha.com/input/?i=%28t%2B0.5%29%5E3e%5E%28-t%29+%3D+0.01}{\ALPHA}
   
   When (after the initial peak) will the construction rate drop to 10,000 houses per year? $t=$ \underline{\quad 12.239 years \quad}
   
   
   \vspace{0.05in}
   
   January 1, 2025 corresponds with $t=10$. We need to ``add up'' all of the houses built after $t=10$. In other words, we need to compute $\int_{10}^\infty R(t)\,dt$. 
   
   \verb|int (t+0.5)^3e^(-t) from 10 to infinity| \href{https://www.wolframalpha.com/input/?i=int+%28t%2B0.5%29%5E3e%5E%28-t%29+from+10+to+infinity}{\ALPHA}
   
   Alpha finds that $\int_{10}^\infty R(t)\,dt \approx 0.070705$ (millions of houses).
   
   
   How many micro-houses will be built after January 1, 2025? \underline{\quad 70,705 houses \quad}.\\
      
   \vspace{0.05in}
   
   We need to find $X$ such that the number of houses built from $t=0$ to $t=X$ is 5,000,000. In other words, we need to solve $\int_0^X R(t)\,dt = 5$.
   
   \verb|(int (t+0.5)^3e^(-t) from 0 to X) = 5| \href{https://www.wolframalpha.com/input/?i=%28int+%28t%2B0.5%29%5E3e%5E%28-t%29+from+0+to+X%29+%3D+5}{\ALPHA}
   
   Alpha finds that $X \approx 3.20646$ years. Notice that $0.20646$ years = $0.20646 \times 12 = 2.47752$ months. 3 years and 2.468 months from January 1, 2015 is somewhere in the middle of March 2018.
      
   When will the 5,000,000${}^{\mathrm{th}}$ micro-house be completed? \underline{\quad March 2018 \quad}. 
   
      
   \vspace{0.1in}
   
      \item After surveying the residents of Raccoon City, we have found that the lifespan of an average citizen is 35 years with a standard deviation of 3. Assume that these lifespans are normally distributed.
   
   \vspace{0.05in}
   
   Type in ``normal distribution'' to get the required formula. Then set $\mu = 35$ and $\sigma =3$. Our first question asks us to compute $\int_{20}^{40} n(x)\,dx$. 
   
   \verb|int e^(-(x-35)^2/(2 3^2))/(sqrt(2 pi) 3) from 20 to 40| \href{https://www.wolframalpha.com/input/?i=int+e%5E%28-%28x-35%29%5E2%2F%282+3%5E2%29%29%2F%28sqrt%282+pi%29+3%29+from+20+to+40}{\ALPHA}

    What percentage of residents live between 20 and 40 years? \underline{\quad 95.221\% \quad}
  
   \vspace{0.05in}
  
  For the next question we need to know the percentage of residents that live more than 50 years and then multiply by the number of residents.
  
  \verb|10000000 int e^(-(x-35)^2/(2 3^2))/(sqrt(2 pi) 3) from 50 to infinity| \href{https://www.wolframalpha.com/input/?i=10000000+int+e%5E%28-%28x-35%29%5E2%2F%282+3%5E2%29%29%2F%28sqrt%282+pi%29+3%29+from+50+to+infinity}{\ALPHA}
  
   If Raccoon City has 10,000,000 residents, how many live more than 50 years? \underline{\quad 2.867 = about 3 people \quad}
  
   \vspace{0.05in}
   
For the last question we need to solve the equation $\int_{-\infty}^X n(x)\,dx = 0.10$.

\verb|(int e^(-(x-35)^2/(2 3^2))/(sqrt(2 pi) 3) from -infinity to X) = 0.10| \href{https://www.wolframalpha.com/input/?i=%28int+e%5E%28-%28x-35%29%5E2%2F%282+3%5E2%29%29%2F%28sqrt%282+pi%29+3%29+from+-infinity+to+X%29+%3D+0.10}{\ALPHA}
  
   What is the cut-off determining the 10\% with the shortest lifespans? \underline{\quad 31.155 years old \quad}
  
\end{enumerate}

\end{document}
