Problem 74 - College Algebra

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

The ticket price that maximizes revenue is $10.70.

Step 1: Define Variables and Find Relationship

Let the attendance be \( A \) and the price be \( P \). We assume a linear relationship \( A = mP + b \). Using the points \((11, 26000)\) and \((9, 31000)\), we can find the slope, \( m \), using the formula: \[m = \frac{A_2 - A_1}{P_2 - P_1} = \frac{31000 - 26000}{9 - 11} = -2500 \].Thus, the linear relationship is \( A = -2500P + b \).

Step 2: Find "b" Using a Known Point

Substitute one point into the line equation to find \( b \). Using \((P, A) = (11, 26000)\): \[\26000 = -2500(11) + b \b = 26000 + 27500 = 53500 \]So, the attendance function is \( A = -2500P + 53500 \).

Step 3: Define Revenue Function

Revenue, \( R \), is given by the product of attendance and price: \[\R = A \cdot P = (-2500P + 53500) \cdot P \R = -2500P^2 + 53500P \]

Step 4: Find Price Maximizing Revenue

The revenue function \( R = -2500P^2 + 53500P \) is a downward-opening parabola. The vertex formula \( P_{max} = -\frac{b}{2a} \) will give the maximum revenue. Here, \( a = -2500 \) and \( b = 53500 \): \[P_{max} = -\frac{53500}{2(-2500)} = \frac{53500}{5000} = 10.7.\]Therefore, the ticket price that maximizes revenue is \( \$10.70 \).

Key Concepts

Linear Relationship

In the context of the exercise, when we say attendance is linearly related to ticket price, it means there's a straight-line equation connecting the price of tickets to the number of spectators. This relationship is typically expressed in the form of a linear equation: \( A = mP + b \). Here, \( A \) is the attendance, \( P \) is the ticket price, \( m \) is the slope of the line, and \( b \) is the intercept.

To find this relationship, we need to identify how much attendance changes with each unit change in price. In our problem, as the ticket price drops from \\(11 to \\)9— which is two units — the attendance increases from 26,000 to 31,000 spectators — a change of 5,000 spectators. A linear relationship directly illustrates this dependency where you can express the change (slope) consistently across the range. This is fundamental in predicting attendance at different price levels.

Slope Calculation

The concept of slope in this exercise is about understanding how steep the line of price versus attendance really is. The slope can be calculated using the formula \( m = \frac{A_2 - A_1}{P_2 - P_1} \). This formula determines the rate at which the attendance varies with a corresponding change in ticket price.

• Using the exercise data: two price points, \( P_1 = 11, P_2 = 9 \), and their corresponding attendances \( A_1 = 26000, A_2 = 31000 \).


• The change in attendance \( (31000 - 26000 = 5000) \) occurred along with a drop in price of \( 2 \) units \((11 - 9) \) resulting in a slope of \( -2500 \).

Thus, a slope of \( -2500 \) suggests that for every dollar decrease in price, the attendance increases by 2500 spectators. Understanding the slope assists in forming the equation for the linear relationship and helps predict future attendance values at various price points.

Vertex Formula

The vertex formula is essential in maximizing or minimizing values in quadratic equations, such as when maximizing revenue in this context. A quadratic equation like the revenue function \( R = -2500P^2 + 53500P \) graphs as a parabola. For maximization problems, we use the vertex formula \( P_{max} = -\frac{b}{2a} \).

• Here \( a = -2500 \) and \( b = 53500 \).
• Substituting these into the vertex formula gives \( P_{max} = \frac{53500}{5000} = 10.7 \).

This value \( P_{max} \) represents the optimum ticket price of \$10.70, ensuring the highest revenue given the constraints of attendance shifts and ticket pricing. Simplified, the vertex of the parabola provides a focused point at which the revenue is at its peak.

Attendance Function

Creating an attendance function involves developing an equation that accurately reflects changes in attendance with varying ticket prices, which stems from the linear relationship and slope calculation. Taking the determined slope \( m = -2500 \) and one of the known points, we find the intercept \( b \) to complete our linear equation:\( A = -2500P + 53500 \).

To make this function, we used a known attendance level and price point to solve for \( b \) by rearranging the linear equation. Using the point \((P, A) = (11, 26000)\):

• Substitute into \( \, 26000 = -2500(11) + b \) to find \( b = 53500 \).

The resulting equation, \( A = -2500P + 53500 \), helps us predict attendance levels based on different ticket prices. By recognizing the relationship between the change in attendance and pricing, strategically setting ticket prices becomes feasible for maximizing spectator numbers.