Problem 87 - 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

Seats sold formula: \((560 - P)/3\). Price range: $215 to $290.

Step 1: Understanding the Relationship Between Price and Seats Sold

The problem states that the number of seats sold decreases with an increase in price. Initially, all 120 seats sell at \(200. For every \)3 increase, 1 seat less is sold. We can express the change in seats sold as a function of how much the ticket price, \(P\), has increased compared to $200.

Step 2: Expressing Seats Sold as a Function of Price

Let \(x\) be the number of \(3 increments added to the initial \)200 price, thus \(P = 200 + 3x\). For each \(x\), 1 seat less than 120 is sold, so seats sold is \(120 - x\). Solving for \(x\) in terms of \(P\), we have \(x = \frac{P - 200}{3}\). Consequently, the formula for seats sold is \(120 - \frac{P - 200}{3}\).

Step 3: Simplifying the Formula for Seats Sold

Simplify \(120 - \frac{P - 200}{3}\):1. Distribute: \(120 - \frac{P}{3} + \frac{200}{3}\)2. Combine terms: \(\frac{360}{3} + \frac{200}{3} - \frac{P}{3} = \frac{560 - P}{3}\)Thus, the formula for seats sold as a function of \(P\) is \(\frac{560 - P}{3}\).

Step 4: Applying Seat Formula to Ranges

Given the number of seats sold ranges between 90 and 115, plug in these values into the formula to find \(P\).For 115 seats: \(\frac{560 - P}{3} = 115\)1. Multiply both sides by 3: \(560 - P = 345\)2. Solve for \(P\): \(P = 560 - 345 = 215\).For 90 seats: \(\frac{560 - P}{3} = 90\)1. Multiply both sides by 3: \(560 - P = 270\)2. Solve: \(P = 560 - 270 = 290\).Thus, the price range is \(215 to \)290.

Key Concepts

Linear Equations

Linear equations are fundamental in algebra and mathematical modeling. They consist of variables that represent unknown quantities and constants. A linear equation represents a straight line when graphed on a coordinate plane. In the exercise, we have a linear relationship between the ticket price and the number of seats sold. This is expressed as a change in seats sold when the price changes.

Consider the pricing model where the initial ticket price is \(200, and all 120 seats are sold at this price. It's given that for every \)3 increase in ticket price, one less seat is sold. This relationship can be set up as a linear equation. We define the number of $3 increments as a variable, let's say \(x\). Hence, the ticket price \(P\) can be written as \(P = 200 + 3x\).

To express the number of seats sold, we use the information that for each \(x\), there is one seat less than the original 120, so the number of seats sold is \(120 - x\). We rearrange \(x\) as \(x = \frac{P - 200}{3}\) to express the relationship between seats and price. Substituting back gives us a linear formula for seats sold based on the ticket price: \(\frac{560 - P}{3}\). Linear equations, like these, simplify complex relationships into understandable terms, making it easier to predict outcomes.

Functions of a Variable

A function of a variable in mathematics is a relation where each input is related to exactly one output. Functions are used to describe and predict patterns of change. In our problem, the number of seats that will be sold is a function of the ticket price.

Formally, a function can be noted as \(f(x)\), where \(x\) is the input variable. In the exercise, the output is the number of seats sold while the input is the change in prices due to the increments. Here, the expression \(\frac{560 - P}{3}\) can be seen as a function \(f(P)\) which maps ticket prices to the number of seats sold. This function captures the essence of how price variations impact sales, serving as a useful model for making predictions and adjustments.

Understanding functions and how to manipulate them is essential not just for academics but also for real-world applications like business and economics, where such functions help in decision-making processes.

Problem Solving in Algebra

Problem solving in algebra involves breaking down a complex problem into simpler steps. It requires setting up equations based on the problem's conditions and solving for unknowns. This systematic approach can be particularly effective in real-world scenarios involving mathematical modeling.

In the exercise, the problem is approached by first understanding the relationship between price and seats sold, then formulating this relationship as an equation and a function. Finally, these are manipulated to find specific values—to determine the range of ticket prices corresponding to various levels of seat occupancy between 90 and 115 seats.

Solving for these values requires step-by-step algebraic manipulation. For instance, dealing with the equation \(\frac{560 - P}{3} = 115\) involves isolating \(P\) to find the ticket price when 115 seats are sold. Similarly, solving \(\frac{560 - P}{3} = 90\) yields the price for 90 seats. This kind of algebraic problem solving is crucial in optimizing outcomes, whether in business, engineering, or other fields.