Problem 12 - Precalculus Essentials

Question

On a certain route, an airline carries 7000 passengers per month, each paying \(\$ 90 .\) A market survey indicates that for each \(\$ 1\) decrease in the ticket price, the airline will gain 60 passengers a. Express the number of passengers per month, \(N,\) as a function of the ticket price, \(x\) b. Express the monthly revenue for the route, \(R\), as a function of the ticket price, \(x\).

Step-by-Step Solution

Answer

The number of passengers (\(N\)) as a function of ticket price (\(x\)) is \(N = 7000 + 60x\). The monthly revenue (\(R\)) as a function of ticket price (\(x\)) is \(R = 60x^2 - 1300x + 630000\).

Step 1: Express the number of passengers as a function of ticket price

From the problem, it is known that for every $1 decrease in ticket price, the airline gains 60 passengers. Now, this relationship can be expressed as an equation. If we let \(N\) be the number of passengers per month and \(x\) be the decrease in ticket price, then \(N = 7000 + 60x\). Here, 7000 is the original number of passengers and 60x represents the increase in passengers for each $1 decrease in ticket price.

Step 2: Express the monthly revenue as a function of ticket price

The revenue (\(R\)) made by the airline per month can be expressed as the ticket price (\(P\)) times the number of passengers (\(N\)). But since \(P = 90 - x\) (the original ticket price minus the decrease) and \(N = 7000 + 60x\) as determined in step 1, the revenue can be expressed as: \(R = (90 - x)(7000 + 60x) = 630000 - 1300x + 60x^2\).

Step 3: Simplify the equation for revenue

The equation for revenue from step 2, \(R = 630000 - 1300x + 60x^2\), can be simplified by arranging it into standard quadratic form: \(R = 60x^2 - 1300x + 630000\). This equation represents the revenue based on ticket price.

Key Concepts

Economic Functions in Precalculus

Economic functions are mathematical tools used to model and analyze financial scenarios, such as costs, revenues, and profits. In precalculus, these functions can range from simple linear equations to more complex polynomial and exponential functions. Understanding such functions is vital for predicting economic outcomes and making informed financial decisions.

In the context of the exercise, the number of passengers (\(N\)) can be considered a dependent variable that's influenced by the independent variable, the ticket price. The ability to express this relationship mathematically is fundamental. The initial equation given, \(N = 7000 + 60x\), is a linear function derived from the real-world scenario that for each \$1 decrease in price, 60 additional passengers fly. It's important to recognize these relations as the basis for constructing economic models that predict behavior, such as consumer response to price changes. This practical application in the exercise not only drills precalculus skills but also reinforces the utility of mathematics in economic decision-making.

Modeling with Linear Functions

Linear functions, which have the form \(y = mx + b\), where \(m\) is the slope, and \(b\) is the y-intercept, are often used for modeling simple relationships between two variables. They have consistent rate of changes, signified by their constant slopes, and are graphically represented by straight lines.

At the core of linear modeling lies the concept of direct proportionality between variables. In our airline example, the linear equation \(N = 7000 + 60x\) suggests a direct, constant rate at which passengers increase relative to the ticket price reduction. Such models are useful for making predictions within the limits of the variables considered. For instance, the airline can estimate potential passenger count increases if ticket prices are decreased. However, it should be noted that models have limitations - they may not account for external factors like competition or market saturation.

Students can explore this further by graphing the function and analyzing the slope (\(60\)) as the passenger increase rate, and the y-intercept (\(7000\)) as the initial number of passengers when there's no price reduction (\(x=0\)).

Quadratic Revenue Functions

Quadratic functions extend the complexity of models beyond linear, introducing a parabolic relationship that can represent a variety of real-world scenarios, such as trajectory paths, object optimization problems, and economic situations – like maximizing revenue. These functions are often represented as \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\).

The significance of a quadratic revenue function is its ability to display increasing and then decreasing behavior, or vice versa. This is essential for identifying maximum or minimum revenue points. In our exercise, the simplified quadratic equation \(R = 60x^2 - 1300x + 630000\) represents the monthly revenue depending on the ticket price change. This equation reflects the non-linear impact of changing ticket prices on revenue, initially increasing with passenger numbers and eventually decreasing due to price cuts outpacing the added passenger revenue.

The airline could use this model to find the optimal ticket price for maximum revenue, which involves calculus concepts like finding the vertex of the parabola. The quadratic function is visually represented as an upside-down parabola since the leading coefficient (\(60\)) is positive, implying the revenue has a maximum point that can be precisely calculated.