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

3200 passengers maximize revenue.

Step 1: Define Variables

Let \( x \) represent the number of passengers on the excursion. Since there will be at least 400 passengers, \( x - 400 \) represents the number of passengers exceeding 400.

Step 2: Formulate the Price Function

The base price per ticket is \( \\(12.00 \). For every 10 additional passengers, the price is reduced by \( \\)0.20 \). Therefore, the price function is \( p(x) = 12.00 - 0.02 \times \left(rac{x - 400}{10}\right) \). Simplifying this, we get \( p(x) = 12 - 0.002(x - 400) \).

Step 3: Simplify the Price Function

Distribute the \( -0.002 \) across \( (x - 400) \) to simplify: \( p(x) = 12 - 0.002x + 0.8 \). Thus, \( p(x) = 12.8 - 0.002x \).

Step 4: Formulate the Revenue Function

The revenue \( R(x) \) is the product of the number of passengers and the price per ticket, \( R(x) = x \cdot p(x) = x \cdot (12.8 - 0.002x) \).

Step 5: Expand the Revenue Function

Distribute \( x \) in the revenue function: \( R(x) = 12.8x - 0.002x^2 \). This is a quadratic equation in the standard form \( ax^2 + bx + c \).

Step 6: Find the Number of Passengers for Maximum Revenue

The revenue function, \( R(x) = -0.002x^2 + 12.8x \), is a downward-opening parabola. The maximum value occurs at the vertex. The formula for the vertex of a parabola \( ax^2 + bx + c \) is \( x = -\frac{b}{2a} \). Here, \( a = -0.002 \) and \( b = 12.8 \). Thus, \( x = -\frac{12.8}{2(-0.002)} = 3200 \).

Step 7: Conclusion

The number of passengers that maximizes total revenue is \( x = 3200 \).

Key Concepts

Quadratic Functions

Quadratic functions are an essential concept to grasp when tackling calculus optimization problems. A quadratic function is a type of polynomial that can be written in the form of \( ax^2 + bx + c \). Here, "\(a\)", "\(b\)", and "\(c\)" are constants, with "\(a\)" not equal to zero. The graph of this function is a parabola, which can open upwards or downwards depending on the sign of "\(a\)". If "\(a\)" is positive, the parabola opens upwards, and if "\(a\)" is negative, it opens downwards.

Understanding quadratic functions is crucial in revenue maximization problems because they often model situations where there is a peak or maximum point, like the maximum revenue condition explored in our exercise. Quadratic equations are straightforward to solve and provide insight into many real-world scenarios, especially involving maximization or minimization.

Price Functions

A price function provides a mathematical way to represent how the price of an item or a service changes under different conditions. In our exercise, the price function, \( p(x) = 12.8 - 0.002x \), shows how ticket price varies based on the number of passengers.

Initially, the base price for each ticket is set at \(\\(12.00\). However, the company offers a discount of \(\\)0.20\) for every 10 passengers over the initial 400. In algebraic terms, this means that for every additional passenger, the price decreases by \(\$0.02\).

• This relationship gives us the idea of how sensitive the pricing is with respect to changes in demand, which is crucial for setting strategic pricing policies.
• The concept of price elasticity can be further explored by analyzing how changes in the price function affect demand and revenue.

Having a precise mathematical model like the price function helps businesses make informed decisions about pricing strategies.

Revenue Maximization

Revenue maximization is a common goal for businesses aiming to increase their financial performance. In the context of our exercise, achieving maximum revenue involves determining the number of passengers such that total ticket sales or revenue reaches its peak.

The revenue function in this problem is \( R(x) = -0.002x^2 + 12.8x \). This equation represents the relationship between the number of passengers \((x)\) and the total revenue, giving rise to a downward-opening parabola due to the negative coefficient of \(x^2\).

• To maximize revenue, the goal is to find the vertex of this parabola, which indicates the highest revenue point.
• By using the formula for the vertex of a parabola, \( x = -\frac{b}{2a} \), we can determine the optimal number of passengers, which, in this case, is 3200.

Reaching the maximum revenue is vital for ensuring business sustainability and profitability.

Vertex of a Parabola

The vertex of a parabola is significant because it represents the maximum or minimum point of a quadratic function, depending on the direction the parabola opens. In problems involving revenue maximization, the vertex indicates where maximum revenue occurs.

To find the vertex of the parabola given by a quadratic equation in the form \( ax^2 + bx + c \), we use the vertex formula \( x = -\frac{b}{2a} \). Here, the coefficients "\(a\)" and "\(b\)" come directly from the revenue function, \( R(x) = -0.002x^2 + 12.8x \).

• Calculating it with "\(a = -0.002\)" and "\(b = 12.8\)", we find that \(x = 3200\), identifying the number of passengers for maximizing revenue.
• This approach is applicable to any quadratic function, not only those associated with pricing or revenue, making it a versatile tool in algebra and calculus.

Understanding how to find and interpret the vertex is essential for optimization in various applications.