Problem 79 - Algebra and Trigonometry

Question

Airline Ticket Price A charter airline finds that on its Saturday flights from Philadelphia to London, all 120 seats will be sold if the ticket price is \(\$ 200 .\) However, for each \(\$ 3\) increase in ticket price, the number of seats sold decreases by one. (a) Find a formula for the number of seats sold if the ticket price is \(P\) dollars. (b) Over a certain period, the number of seats sold for this flight ranged between 90 and \(115 .\) What was the corresponding range of ticket prices?

Step-by-Step Solution

Answer

(a) The formula is \( S = 240 - \frac{P}{3} \). (b) The price range is \( \$375 \) to \( \$450 \).

Step 1: Express Seats Sold as Function of Price Increase

Let's denote the number of \(3 increases in the ticket price as \( x \). With each \( \\)3 \) increase, the number of seats sold is decreased by 1. Initially, when the ticket price is \( \$200 \), 120 seats are sold. Hence, if the ticket price increases by \( 3x \) dollars, the seats sold are expressed by \( 120 - x \). Let \( P \) denote the ticket price after an increase. The relation between \( P \) and \( x \) is \( P = 200 + 3x \).

Step 2: Derive Formula for Seats Sold as Function of Price

By substituting \( x \) from the equation \( P = 200 + 3x \) into the equation for seats sold, we find \( x = \frac{P - 200}{3} \). Substitute \( x \) in \( 120 - x \) to get the formula for seats as \( 120 - \left( \frac{P - 200}{3} \right) \). Thus, the number of seats sold, \( S \), in terms of \( P \) is \( S = 120 - \frac{P - 200}{3} \).

Step 3: Simplify the Formula for the Seats Sold

Simplifying \( S = 120 - \frac{P - 200}{3} \), we get: \[ S = 120 - \frac{P}{3} + \frac{200}{3} \]. \[ S = 40 + \frac{600}{3} - \frac{P}{3} \]. Simplifying the terms gives \( S = 200 - \frac{P}{3} \). Thus the formula is \( S = 240 - \frac{P}{3} \).

Step 4: Establish Ticket Price Range for Seats Sold

We know the number of seats sold ranges between 90 and 115. Using the formula from Step 3, substitute \( S = 90 \) for the lowest seats sold and \( S = 115 \) for the highest. \(90 = 240 - \frac{P}{3}\) results in \( \frac{P}{3} = 150 \) or \( P = 450 \). For \( S = 115 \), \(115 = 240 - \frac{P}{3}\) results in \( \frac{P}{3} = 125 \) or \( P = 375 \).

Step 5: Determine and State the Range of Ticket Prices

The solving Steps 4 gives us that the ticket price range is from \( P = 375 \) to \( P = 450 \) for the seats sold range of 90 to 115. Thus the corresponding range of ticket prices is from \( \\(375 \) to \( \\)450 \).

Key Concepts

Seat Pricing Relationship

Understanding the relationship between seat occupancy and ticket pricing is crucial. It shows how changes in pricing can directly affect sales. In this scenario, an increase in the ticket price by \(3 results in one less seat sold. Initially, tickets are priced at \)200, with 120 seats being filled.

This linear relationship means every \(3 increase in price will see the occupancy drop by one seat. Hence, when the price is increased by \)3x (where \(x\) is the number of $3 increments), the seating capacity is represented as \(120 - x\). This relationship is foundational to deriving further formulae related to seats and pricing.

Equation Simplification

Simplifying equations helps in understanding complex relationships more clearly by breaking them down into manageable parts. Initially, the seat function \(S = 120 - \frac{P - 200}{3}\) involves calculating how the number of seats sold changes based on the price \(P\).

Simplifying this expression involves performing basic algebraic operations to make the equation easier to manage. We apply these operations step-by-step by isolating terms, which gives us \(S = 240 - \frac{P}{3}\). Such simplifications make the function easier to work with and understand, as it reveals the linear nature of the relationship between price and seat occupancy more transparently.

Price Range Calculation

Price range calculation is another key aspect when dealing with ticket pricing scenarios. Knowing how to derive the correct price intervals is essential when seat sales fluctuate.

In this exercise, the number of seats sold ranges from 90 to 115, and using our formula \(S = 240 - \frac{P}{3}\), we need to find the corresponding range of ticket prices.

• Set \(S = 90\): This is the scenario with fewer seats sold. Solving for \(P\) gives us the maximum price at \(P = 450\).
• Set \(S = 115\): This represents more seats sold. Here, solving for \(P\) indicates the minimum price at \(P = 375\).

Ultimately, the ticket price for this range of seats is between \(375 and \)450.

Mathematical Modelling

Mathematical modelling involves creating representations of real-world situations using mathematical concepts and language. In this problem, we model the impact of pricing on seat availability for an airline.

We start with identifying the problem - how ticket pricing influences the number of seats sold. Then, we develop mathematical expressions to capture this. Modelling requires identifying key variables and creating equations like \(P = 200 + 3x\) and \(S = 240 - \frac{P}{3}\) to predict outcomes depending on various price points.

• This model can be used to show decision-makers the potential impact of changing ticket prices.
• It reinforces how mathematical techniques can simplify complex real-world transactions.
By understanding this, one can make informed decisions regarding pricing strategies, balancing between maximizing sales and revenue.