Problem 64 - Calculus Early Transcendentals: Pearson New International Edition

Question

A riverboat company offers a fraternal organization a Fourth of July excursion with the understanding that there will be at least 400 passengers. The price of each ticket will be \(\$ 12.00\), and the company agrees to discount the price by \(\$ 0.20\) for each 10 passengers in excess of \(400 .\) Write an expression for the price function \(p(x)\) and find the number \(x_{1}\) of passengers that makes the total revenue a maximum.

Step-by-Step Solution

Answer

500 passengers maximize total revenue.

Step 1 - Define Variables and Understand the Problem

Let \( x \) represent the number of passengers. We start with 400 passengers. Each ticket initially costs \( \\(12.00 \). The company discounts \( \\)0.20 \) for every 10 passengers over 400. Our aim is to express the ticket price \( p(x) \) as a function of the number of passengers \( x \) and find the number of passengers that maximizes total revenue.

Step 2 - Establish the Ticket Price Function

The price per ticket \( p(x) \) starts at \( \\(12.00 \) and is reduced by \( \\)0.20 \) for each additional 10 passengers over 400. Thus, the number of 10-passenger groups over 400 is \( \frac{x - 400}{10} \). Therefore, the price per ticket as a function of \( x \) is \( p(x) = 12 - 0.2\left(\frac{x - 400}{10}\right) \). Simplifying, \( p(x) = 12 - 0.02(x - 400) \). Thus, \( p(x) = 12 - 0.02x + 8 = 20 - 0.02x \).

Step 3 - Express Total Revenue Function

Total revenue \( R(x) \) is the product of the number of passengers \( x \) and the price per ticket \( p(x) \). Therefore, \( R(x) = x \cdot (20 - 0.02x) \). Simplifying the expression gives \( R(x) = 20x - 0.02x^2 \).

Step 4 - Find Maximum Revenue

To find the maximum total revenue, we need to solve for the vertex of this quadratic function. The vertex form solution for the quadratic equation \( ax^2 + bx + c \) gives the maximum or minimum point at \( x = \frac{-b}{2a} \). Here, \( a = -0.02 \) and \( b = 20 \), so \( x = \frac{-20}{2 \cdot (-0.02)} = \frac{20}{0.04} = 500 \). Thus, the number of passengers that maximizes revenue is 500.

Key Concepts

Quadratic Functions

When it comes to optimizing revenue and price functions, understanding quadratic functions is essential. These functions have the form of a parabola, which can either open upwards or downwards. In our riverboat company's example, we are dealing with a downward-opening parabola because the coefficient of the squared term is negative.
This means we are looking for the maximum point of the function, which is located at the vertex of the parabola.

Quadratic functions generally have the form \( ax^2 + bx + c \). Identifying this form will help you locate critical points like the maximum or minimum. In optimization problems, these points are vital to finding solutions.

• The coefficient \( a \) determines whether the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)).
• The vertex of the parabola helps find either the maximum or minimum of the function in question. Use \( x = \frac{-b}{2a} \) to locate the vertex of the function where \(x\) is the number of interest, such as passengers in this case.

The step-by-step approach to solving includes expressing the problem using this formula and then finding the value of \(x\) at the vertex.

Revenue Maximization

The ultimate goal of revenue maximization is to find the quantity — in this case, the number of passengers — that yields the highest possible revenue. For our riverboat example, we formulated the revenue function as \( R(x) = 20x - 0.02x^2 \).
Recognizing this as a quadratic function, it becomes clear that we need to identify the vertex to find the maximum revenue.

Revenue, expressed as a function of price and quantity (or number of passengers), is calculated with the expression \( R(x) = x \, \text{price per ticket} \). Therefore, understanding the interaction between these components is critical.

• The revenue function indicates how changes in passenger numbers (and linked ticket prices) influence revenue.
• By finding the vertex, we pinpoint the optimal number of passengers needed to optimize revenue.

Thus, solving for \( x \) in the vertex formula \( x = \frac{-b}{2a} \) leads us to the optimal number of passengers, which is 500 in this context.

Price Function

The price function helps determine how much should be charged per ticket based on passenger numbers. This involves taking initial pricing and incorporating any adjustments or discounts, like what our riverboat company does when passengers exceed 400.
In this case, the price of a ticket decreases by \( \$0.20 \) for every additional ten passengers beyond 400. To derive the price function, we use:

• Identify the initial price as \( 12.00 \) dollars per ticket.
• Calculate discounts based on the number of passenger groups above 400: \( p(x) = 20 - 0.02x \).

The price function \( p(x) \), incorporating the discount, clearly shows ticket pricing as passenger numbers vary.
This function models real-world scenarios where prices fluctuate based on demand or quantity, illustrating competitive pricing in practical settings.