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. You possess advanced production technology to produce the widgets at $C_1 = 6.00$ per unit. A public carrier is hired to deliver the widgets to the customers at a cost rate of $f = 10^{-3}$ dollars per widget-mile.

Population of potential customers is uniformly distributed with a density of $\delta = 5$ people/mi. Every customer will need to buy $D = 2$ widgets per person per year no matter how expensive the widgets are. However, the customers will choose to purchase widgets at the lowest available price. A competitor has already built a plant at Punta Arenas, the southern city of Chile. The competitor produces the widgets at a rate of $C_2 = 8.00$ dollars per unit, and its established rail transportation system allows negligible variable transportation cost.

\begin{itemize}
    \item [a.] While maximizing your own profit, what are your plant’s total revenue, total cost, and total profit?
    \item [b.] The northern part of the country (1000 miles) is mountainous. Due to fuel price increase, the public carrier now charges $1.5f$ per widget-mile for shipping in the mountainous area. Nothing else will change. What will be your total revenue, cost, and profit?
\end{itemize}
Solution: \begin{itemize}
    \item [a.] Let \(P_1\) be the price we sell at, and \(P_2\) be the price our competitor sells at. Were there no competitor, our profit per unit distance at any given distance \(x\) away from Arica would be given by

\[
\text{Profit}_{\text{US}} = P_1 \delta D - C_1 \delta D - f \delta D x
\]

The competitor’s unit profit is independent of the distance, and is given by

\[
\text{Profit}_{\text{COMPETITOR}} = P_2 \delta D - C_2 \delta D
\]

In order for us to maximize our profit, we must set our price at every unit length such that our competitor cannot reduce their price below ours without losing money. Intuitively, we should be able to profit by setting our price below our competitor's lowest possible price up until some point where due to the increasing transportation cost, we can no longer reduce our price. At this point, both us and our competitor will have an equal price since neither can reduce the price without losing money. It will also be at this point where neither can profit, since both of us will have reduced our price such that the profit is 0. We can rewrite the two equations above in terms of the profit instead of price.

\[
P_1 = \frac{\text{Profit}_{\text{US}}}{\delta D} + C_1 + fx
\]

\[
P_2 = \frac{\text{Profit}_{\text{COMPETITOR}}}{\delta D} + C_2
\]

Setting them equal, and setting both profits to 0, we can solve for \(x\):

\[
x = \frac{C_2 - C_1}{f}
\]

Solving for \(x\) yields 2000 miles. Thus, our price (and our competitor's) at 2000 miles from Arica must be set to \$8. Our competitor will set their price to \$8 from \(x = 0\) to \(x = 2000\), so our price will be set to \$8 from \(x = 0\) to \(x = 2000\). Our revenue is then

\[
R = 2000 \cdot \delta \cdot P_1 D = \$160{,}000
\]

Our total cost, \(Z\), is

\[
Z = 2000 \cdot \delta \cdot C_1 D + \delta \cdot f D \int_0^{2000} x \, dx= \$140{,}000
\]

So our profit is \$20{,}000.


    \item [b.] Because the transportation cost for the first 1000 miles increases, our extent of service will decrease. For widgets transported more than 1000 miles, the profit per unit distance is 

\[
\text{Profit}_1 = P_1 \delta D - C_1 \delta D - D \delta f \left(1.5 \times 1000 + x - 1000\right)
\]

Our competitor’s profit is still 

\[
\text{Profit}_2 = P_2 \delta D - C_2 \delta D
\]

Again, rewriting both equations above in terms of the profit instead of price:

\[
P_1 = \frac{\text{Profit}_{\text{US}}}{\delta D} + C_1 + f(x + 500)
\]

\[
P_2 = \frac{\text{Profit}_{\text{COMPETITOR}}}{\delta D} + C_2
\]

Setting the above equal to each other and setting the profits to 0 gives:

\[
x = \frac{C_2 - C_1}{f} - 500
\]

Solving for \(x\) yields 1500 miles. Thus, our price (and our competitor's) at 1500 miles from Arica must be set to \$8. Our competitor will set their price to \$8 from \(x = 0\) to \(x = 1500\), so our price will be set to \$8 from \(x = 0\) to \(x = 1500\). Our revenue is then:

\[
R = 1500\delta P_1 D = \$120{,}000
\]

Our total cost, \(Z\), is 

\[
Z = 1500\delta C_1 D + 1.5 \delta f D \int_0^{1000} x \, dx + 1.5 f \cdot 1000 \cdot 500 \cdot \delta \cdot D + \delta \cdot f D \int_{1000}^{1500} x \, dx
\]

\[
Z = \$111{,}250
\]

So our profit is \$8{,}750.

\end{itemize}