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Chapter 2: Problem 12

The average ticket price for a concert at the opera house was \(550 .\) The average attendance was \(4000 .\) When the ticket price was raised to \(\$ 52,\) attendance declined to an average of 3800 persons per performance. What should the ticket price be to maximize the revenue for the opera house? (Assume a linear demand curve.)

Short Answer

Expert verified
The optimal ticket price is $45.

Step by step solution

01

- Set up the Demand Function

Assume the demand function is linear and represented as: \[ P = mx + b \] where \( P \) is the price and \( x \) is the attendance. Use given points \((4000, 50)\) and \((3800, 52)\) to find the slope \( m \).
02

- Calculate the Slope

Calculate the slope \( m \) using the formula: \[ m = \frac{P_2 - P_1}{x_2 - x_1} = \frac{52 - 50}{3800 - 4000} = \frac{2}{-200} = -0.01 \]
03

- Find the Demand Function

Substitute one point, say \((4000, 50)\), into the demand function to find \( b \): \[ 50 = -0.01 \cdot 4000 + b \] Solve for \( b \): \[ 50 = -40 + b \] \[ b = 90 \] Thus, the demand function is: \[ P = -0.01x + 90 \]
04

- Set up the Revenue Function

Revenue \( R \) is the product of price \( P \) and attendance \( x \): \[ R = x \cdot P = x(-0.01x + 90) \] Simplify to obtain: \[ R = -0.01x^2 + 90x \]
05

- Maximize the Revenue

To maximize revenue, find the vertex of the quadratic function: \[ x = \frac{-b}{2a} \] Here, \( a = -0.01 \) and \( b = 90 \): \[ x = \frac{-90}{2(-0.01)} = \frac{-90}{-0.02} = 4500 \]
06

- Find the Optimal Price

Substitute \( x = 4500 \) back into the demand function to find \( P \): \[ P = -0.01 \cdot 4500 + 90 = -45 + 90 = 45 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Demand Function
A linear demand function is used to describe the relationship between the quantity demanded and the price of a product. This relationship is typically linear when plotted on a graph. For this problem, we assume the demand function has the form: \[ P = mx + b \] Here, \( P \) stands for the price, and \( x \) represents the attendance or quantity. We start with the points given in the exercise: \((4000, 50)\) and \((3800, 52)\). These points help us determine the constants \( m \) (slope) and \( b \) (intercept) in the linear demand function.
Revenue Function
The revenue function is essential for determining how changes in price affect the overall earnings. Revenue \( R \) is calculated as the product of the price \( P \) and the quantity demanded \( x \). In equation form, it's: \[ R = P \times x \] Given the linear demand function \( P = -0.01x + 90 \), we substitute \( P \) into the revenue formula: \[ R = x(-0.01x + 90) = -0.01x^2 + 90x \] This quadratic equation represents the revenue function. To maximize revenue, we need to explore this quadratic relationship.
Quadratic Optimization
Quadratic optimization helps in finding the maximum or minimum value of a quadratic function. Since the revenue function is quadratic: \[ R = -0.01x^2 + 90x \] it opens downward (because the coefficient of \( x^2 \) is negative). The vertex of this parabola gives us the maximum revenue. To find the vertex, use the formula for the x-coordinate: \[ x = \frac{-b}{2a} \] For our equation, \( a = -0.01 \) and \( b = 90 \). Plug in these values to find \( x \): \[ x = \frac{-90}{2(-0.01)} = \frac{-90}{-0.02} = 4500 \] The attendance that maximizes revenue is 4500.
Slope Calculation
The slope \( m \) of a linear function indicates the rate of change between the dependent and independent variables. It can be calculated using two points on the line. The formula for the slope is: \[ m = \frac{P_2 - P_1}{x_2 - x_1} \] For the points \((4000, 50)\) and \((3800, 52)\), the calculation is: \[ m = \frac{52 - 50}{3800 - 4000} = \frac{2}{-200} = -0.01 \] This negative slope tells us that as the price increases, the attendance decreases.
Vertex of a Parabola
The vertex of a parabola represented by a quadratic equation \( ax^2 + bx + c \) is crucial for finding the function's maximum or minimum value. The x-coordinate of the vertex is found using: \[ x = \frac{-b}{2a} \] With \( a = -0.01 \) and \( b = 90 \), the calculation is: \[ x = \frac{-90}{2(-0.01)} = 4500 \] Substituting \( x = 4500 \) back into the demand function \( P = -0.01x + 90 \) gives us the optimal ticket price: \[ P = -0.01(4500) + 90 = -45 + 90 = 45 \] Therefore, the optimal ticket price to maximize revenue is $45.

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Most popular questions from this chapter

Until recently hamburgers at the city sports arena cost \(\$ 4\) each. The food concessionaire sold an average of 10,000 hamburgers on a game night. When the price was raised to \(\$ 4.40,\) hamburger sales dropped oft to an average of 8000 per night. (a) Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue. (b) If the concessionaire has fixed costs of \(\$ 1000\) per night and the variable cost is \(\$ .60\) per hamburger, find the price of a hamburger that will maximize the nightly hamburger profit.

Advertising for a certain product is terminated, and \(t\) weeks later, the weekly sales are \(f(t)\) cases, where $$f(t)=1000(t+8)^{-1}-4000(t+8)^{-2}.$$ At what time is the weekly sales amount falling the fastest?

In the planning of a sidewalk cafe, it is estimated that for 12 tables, the daily profit will be \(\$ 10\) per table. Because of overcrowding, for each additional table the profit per table (for every table in the cafe) will be reduced by \(\$ .50 .\) How many tables should be provided to maximize the profit from the cafe?

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