Problem 74 - College Algebra with Corequisite Support

Question

A soccer stadium holds 62,000 spectators. With a ticket price of $$\$ 11,$$ the average attendance has been 26,000 . When the price dropped to $$\$ 9,$$ the average attendance rose to 31,000 . Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue?

Step-by-Step Solution

Answer

A ticket price of $10.70 would maximize revenue.

Step 1: Understand the Problem

The problem states that the relationship between ticket price and attendance is linear. Initially, at a ticket price of \( \\(11 \), the attendance was 26,000. When the price is \( \\)9 \), attendance increased to 31,000. We need to find the price that maximizes revenue.

Step 2: Set Up Revenue Equation

Revenue is calculated as the product of the number of attendees and the ticket price. Let \( p \) be the ticket price and \( a \) be the attendance. Then Revenue \( R = p \times a \). To maximize \( R \), we first express attendance \( a \) in terms of \( p \).

Step 3: Determine the Linear Equation for Attendance

Assume the attendance \( a \) varies linearly with price \( p \). Use the points (\(11, 26000\)) and (\(9, 31000\)) to find the equation. The slope \( m \) is \( \frac{31000 - 26000}{9 - 11} = -2500 \). The form is \( a = mp + b \). Substituting one point, say (11, 26000), \( 26000 = -2500 \times 11 + b \). Solving for \( b \), we get \( b = 53500 \). Thus, \( a = -2500p + 53500 \).

Step 4: Transform Revenue Equation

Substitute the expression for \( a \) into the revenue equation: \( R = p(-2500p + 53500) = -2500p^2 + 53500p \).

Step 5: Differentiate to Find Maximum Revenue

Differentiate \( R \) with respect to \( p \): \[ \frac{dR}{dp} = -5000p + 53500 \]Set \( \frac{dR}{dp} = 0 \) to find the critical points, giving:\[ -5000p + 53500 = 0 \]\[ 5000p = 53500 \]\[ p = \frac{53500}{5000} = 10.7 \].

Step 6: Validate Maximum Revenue with the Second Derivative Test

Calculate the second derivative \( \frac{d^2R}{dp^2} = -5000 \). Since \( \frac{d^2R}{dp^2} < 0 \), the point is a maximum.

Key Concepts

Linear Equations

Linear equations form the backbone of many mathematical representations in real-world scenarios. They are equations that create straight lines when plotted on a graph. In our exercise, we are dealing with a linear relationship between ticket price and attendance at a stadium. We are given two points:

• When the price is \(11, the attendance is 26,000.
• When the price drops to \)9, the attendance increases to 31,000.

To find the linear equation, we need the slope, which is calculated by the formula \( m = \frac{{y_2-y_1}}{{x_2-x_1}} \). Substituting the values:

• \( m = \frac{31000 - 26000}{9 - 11} = -2500 \).

This slope indicates a negative relationship, meaning as the price decreases, attendance increases. By using the slope-intercept form, \( a = mp + b \), we can apply one point to find the y-intercept (\(b\)): 26000 = -2500 \times 11 + b. Solving gives \( b = 53500 \, \). The complete linear equation for attendance \( (a) \) is: \( a =-2500p + 53500 \).

Optimization Problems

Optimization problems are about finding the best solution from a set of possible choices. The best solution can mean different things, such as maximum revenue, minimum cost, or highest efficiency. In this exercise, the goal is to maximize revenue from ticket sales.

Revenue is calculated by multiplying the number of attendees by the ticket price. Therefore, we define revenue \( (R) \) as:

• \( R = p \times a \)

where \( p \) is the ticket price, and \( a \) is attendance. We've expressed attendance in terms of price with our linear equation, \( a = -2500p + 53500 \). Substituting into the revenue equation gives:

• \( R = p(-2500p + 53500) = -2500p^2 + 53500p \)

This quadratic equation represents a parabolic shape on a graph, opening downwards. The challenge is to find the vertex of the parabola, which gives the maximum revenue.

Calculus - Derivatives

Derivatives are central to calculus and are used to find the rate at which one quantity changes with respect to another. In optimization problems, derivatives help us determine where a function reaches its maximum or minimum.

To find the price that maximizes revenue, we differentiate the revenue function with respect to \( p \):

• \( \frac{dR}{dp} = -5000p + 53500 \)

Setting this derivative equal to zero allows us to find critical points, which are potential points for maxima or minima related to optimization:

• \( -5000p + 53500 = 0 \)
• \( 5000p = 53500 \)
• \( p = 10.7 \)

We confirm it using the second derivative test. The second derivative is:

• \( \frac{d^2R}{dp^2} = -5000 \)

Since \( \frac{d^2R}{dp^2} < 0 \), we have a local maximum. Therefore, at $10.7, the revenue is maximized.

Linear Relationships in Economics

In economics, understanding linear relationships can simplify complex situations by focusing on the main contributing factors. This exercise uses the idea that attendance is a linear function of price, which helps in formulating strategic decisions about pricing tickets.

Economists often look for linear patterns due to their simplicity and clarity. Here, with every $1 drop in ticket price, approximately 2,500 more attendees show up, according to the linear equation we calculated. Such linear relationships allow businesses to predict outcomes effectively.

Businesses use this linear analysis to determine the point at which revenue is optimized. While linear assumptions can be limiting, they provide a foundational figure before considering more complex models with additional factors. This results in practical implications, guiding pricing strategies that can maximize profits while managing customer attention.