Problem: Upon successfully running the “widget” plant at Pittsburgh, you are promoted to manage a newly established overseas manufacturing plant at Arica, the most northern city in Chile. Chile can be considered a linear country that is 2500 miles in length. A competitor has already built a similar plant at Punta Arenas, the most southern city of Chile. Both you and the competitor produce these widgets at a rate of $C = 8.00$ dollars per unit.

Population of potential customers is uniformly distributed with a density of $\delta = 5$ people/mi. The prevailing market price of these widgets is $P = 10.00$ dollars per unit everywhere, and every customer will buy $D = 2$ widgets per person per year. A public carrier is hired by both plants to deliver the widgets to the customers at a cost rate of $f = 10^{-3}$ dollars per widget-mile. To most effectively penetrate into the market, you may choose to offer variable discount prices to certain customers. Your competitor may use the same strategy.

\begin{itemize}
    \item [a.]  While maximizing your own profit, what are your plant’s total revenue, total cost, and total profit?
    \item [b.]  Due to recent technological innovations (confidential from your competitor), you are able to produce these widgets at a rate of $C = 7.00$ dollars per unit next year. Nothing else will change. What will be your total revenue, cost, and profit?
\end{itemize}
Solution: \begin{itemize}
    \item [a.] First, assume there is no competitor. The revenue, \(R\), will be

\[
R = L\delta PD
\]

where \(L\) is the distance to the end point of the serviced length (without the competitor). 
The total cost will be the sum of the transportation cost, \(T\), and manufacturing cost, \(Z\):

\[
Z = L\delta CD
\]

\[
T = f \delta D \int_0^L x \, dx
\]

\[
T = f\delta D \frac{L^2}{2}
\]

The profit for a given \(L\) can then be found by the equation:

\[
\text{Profit} = R - (Z + T)= L\delta PD - L\delta CD -  f\delta D \frac{L^2}{2}
\]

We can maximize the profit by optimizing the above formula over $L$. Specifically, we can take the derivative, setting it to 0, and solving for the optimal \(L^*\):

\[
L^* = \frac{P - C}{f}
\]

which yields a service length for both us and our competitor of 2000 miles. Alternatively, the value of $L^*$ can directly be found by equating the marginal revenue to the marginal cost per widget; i.e., 
$$ P = C+fL.$$

When the competitor is present, by symmetry, the competition will reduce your market to be between 0 and 1250 miles (past \(x = 1250\) our competitor can undercut us). From \(x = 0\) to \(x = 500\) (i.e., the area where our competitor does not serve --- farther than 2000 miles from its plant) we can charge the full price, but for \(x = 500\) to \(x = 1250\), we must offer a discount equal to the profit our competitor earns (in other words, our price must be equal to the competitor's cost). So, our profit will be the sum of profits from the ranges 0-500 and 500-1250:

\[
\text{Profit} = P_{0-500} + P_{500-1250}
\]

\[
\text{Profit}_{0-500} = 500P\delta D - 500C\delta D - f \delta D \frac{500^2}{2}
\]

\[
\text{Profit}_{500-1250} = 750P \delta D - 750C \delta D - f\delta D \int_{500}^{1250} x \, dx - \left( 750P\delta D - 750C\delta D - f\delta D \int_{1250}^{2000} x \, dx \right) 
\]

Simplifying,

\[
\text{Profit}_{500-1250} = -f\delta D \int_{500}^{1250} x \, dx + f\delta D \int_{1250}^{2000} x \, dx
\]

We get \(\text{Profit}_{0-500} = \$8,750.0\), \(\text{Profit}_{500-1250} = \$5,625.0\) for a total profit of \$14,375.0. Our cost, \(Z\), is given by:

\[
Z = 1250C \delta D + f\delta D \frac{1250^2}{2}
\]

which yields \$107,812.5. Our revenue, \(R\), is the sum of the revenue from the two ranges:

\[
R = 500P\delta D + 750P\delta D - \left( 750P\delta D - 750C\delta D - f\delta D \int_{1250}^{2000} x \, dx \right) 
\]

which yields a revenue of \$122,187.5.


    \item [b.] First we must find the point at which both of us would pay the same cost so we set the equations for marginal costs equal to one another. Solving $7 + f x = 8 + f (2500-x)$, we have the market boundary would be at $x=1750$ miles, and the total revenue would be 167,187.5. Total cost would be 137,812.5. The total profit would be 29,375.
\end{itemize}