Problem 3 - Calculus Concepts

Question

Airline Profit The function \(f\) gives the weekly profit, in thousand dollars, that an airline makes on its flights from Boston to Washington D.C. when the ticket price is \(p\) dollars. Interpret the following: a. \(f(65)=15\) b. \(f^{\prime}(65)=1.5\) c. \(f^{\prime}(90)=-2\)

Step-by-Step Solution

Answer

a. $65 ticket yields $15,000 profit. b. Increasing price at $65 improves profit. c. Increasing price at $90 reduces profit.

Step 1: Evaluate Function Value

Interpret \(f(65) = 15\). This means that when the ticket price is \\(65, the airline's weekly profit is \\)15,000. The function value \(f(65)\) gives the profit in thousands of dollars.

Step 2: Analyze Positive Derivative Value

Interpret \(f'(65) = 1.5\). At a ticket price of \\(65, if the price increases by \\)1, the weekly profit will increase by \$1,500. A positive derivative indicates that raising ticket prices increases profit at this price.

Step 3: Analyze Negative Derivative Value

Interpret \(f'(90) = -2\). At a ticket price of \\(90, if the price increases by \\)1, the weekly profit will decrease by \\(2,000. A negative derivative suggests that further increasing the ticket price from \\)90 will decrease profit.

Key Concepts

Function Value Analysis

In calculus, the function value tells us the specific output of a function when given an input. Here, we have a function \(f(p)\) that describes profit in terms of ticket price. When we look at \(f(65) = 15\), it represents that setting the ticket price at \(65\) dollars results in a profit of \(15\,000\) dollars. This information is essential:

• It shows us the exact profit for a given price without any need for estimation.
• This profit is already optimized at this price, given no other changes in the situation.
• Understanding this helps in comparing with profits at other prices.

Having the function value allows airlines to make strategic pricing decisions to maximize profit. It serves as a confirmation point to ensure their pricing strategy is effective.

Positive Derivative

A positive derivative signifies growth. In the context of our problem, when \(f'(65) = 1.5\), it means small increases in the ticket price from \(65\) dollars lead to an increase in profit. For each dollar increase in ticket price, the weekly profit grows by \(1,500\) dollars. Here's what that means:

• If the airline increases prices slightly, they are likely to see an immediate profit boost.
• A positive derivative implies a range of ticket prices exists around \(65\) dollars where profits may continue to increase if prices are managed carefully.

This derivative helps airlines understand a point of pricing elasticity, where demand might endure higher prices while still driving profit.

Negative Derivative

A negative derivative tells us something is decreasing or is likely to decrease. With \(f'(90) = -2\), it indicates that profit declines when the ticket price is slightly raised from \(90\) dollars. Specifically, a one-dollar increase results in a \(2,000\) dollar decrease in profit. This suggests:

• The ticket price may have reached an upper limit where customers are less willing to purchase, resulting in decreased sales and profit.
• At this point, the airline must consider other pricing strategies, as pushing prices higher reduces profitability.

The negative derivative serves as a warning. It highlights a sensitive area of pricing where increasing prices may hurt business rather than improve it.