Problem 32 - Single Variable Calculus

Question

Maximizing Revenue If exactly 200 people sign up for a charter flight, the operators of a charter airline charge \(\$ 300\) for a round-trip ticket. However, if more than 200 people sign up for the flight, then each fare is reduced by \(\$ 1\) for each additional person. Assuming that more than 200 people sign up, determine how many passengers will result in a maximum revenue for the travel agency. What is the maximum revenue? What would the fare per person be in this case?

Step-by-Step Solution

Answer

To maximize revenue, the travel agency should have 250 passengers, which would result in a maximum revenue of \(\$62,500\) and a fare per person of \(\$250\).

Step 1: Create the revenue function

Let x be the number of additional persons above 200. The ticket price would then be \((300-x)\) since the price reduces by \(\$1\) for each additional person. The total number of passengers would be \((200+x)\). Therefore, the revenue function R(x) can be defined as the product of the number of passengers and the ticket price: R(x) = (number of passengers) × (ticket price) = (200+x)(300-x)

Step 2: Find the maximum of the parabola

Since the revenue function is a quadratic function, we can find its maximum by completing the square or finding the vertex of the parabola. The vertex of a parabola in the form \(f(x)=ax^2+bx+c\) is given by \((-b)/(2a)\). In our case, the revenue function is: R(x) = -(x^2 + (-100)x + 200×300) So, a = -1 and b = 100.

Step 3: Calculate the number of passengers to maximize revenue

To find the x-value at the maximum, we use the formula \((-b)/(2a)\): x = \((-100)/(2×(-1))\) = 50 Since x represents the number of additional persons above 200, there should be 200 + 50 = 250 passengers to maximize the revenue.

Step 4: Calculate the maximum revenue and fare per person

To find the maximum revenue, substitute the optimal number of passengers (x=50) into the revenue function: R(50) = (200+50)(300-50) = 250 × 250 = 62,500 Thus, the maximum revenue is \(\$62,500\). To find out the fare per person, calculate the ticket price when x=50: Fare = 300 - 50 = $250 So, the fare per person would be \(\$250\) in this case. In conclusion, to maximize the revenue, the travel agency should have 250 passengers, resulting in a maximum revenue of \(\$62,500\) and a fare per person of \(\$250\).

Key Concepts

Understanding Quadratic Functions

A quadratic function is a type of polynomial that can be expressed in the form \( f(x) = ax^2 + bx + c \). This form describes a parabola when graphed. Parabolas can open upwards or downwards depending on the sign of \( a \): upward if \( a > 0 \) and downward if \( a < 0 \). In this context, we are working with a quadratic function that models revenue, which has a specific shape (a parabola) indicating how revenue changes with the number of passengers. Understanding the nature of quadratic functions is crucial since the maximum or minimum values occur at the vertex of the parabola. These characteristics help us determine how to achieve maximum revenue based on the number of passengers, which is directly tied to how the quadratic function behaves.

Defining the Revenue Function

The revenue function in this exercise is derived from the relationship between the number of passengers and the adjusted ticket price. By introducing \( x \) as the additional number of passengers over 200, the new ticket price becomes \( 300 - x \), representing a \( \$1 \) decrease for each additional passenger. Thus, the revenue \( R(x) \) is calculated as:

• Number of Passengers: \( 200 + x \)
• Ticket Price: \( 300 - x \)

Multiply these elements to create the revenue function:\[ R(x) = (200 + x)(300 - x) \]This relationship signifies how revenue changes as more passengers purchase tickets, with price adjustments directly affecting the total income. In essence, we're examining a balance between price and quantity to find the optimal scenario for maximum revenue.

Finding the Vertex of a Parabola

To maximize the revenue function, we must determine the vertex of the parabola described by the function. In quadratic terms, the vertex provides the maximum or minimum value of the function, depending on its orientation. Since our revenue function is \( R(x) = -(x^2 - 100x + 60000) \), where \( a = -1 \) and \( b = 100 \), we find the vertex using the formula:\[ x = \frac{-b}{2a} \]Substituting in the values gives:\[ x = \frac{-100}{2(-1)} = 50 \]This tells us that the number of additional passengers should be 50, leading to a total of 250 passengers (200 + 50) to reach the maximum revenue. Identifying the vertex is vital because it pinpoints the exact point where revenue peaks.

Optimizing Passenger Numbers for Maximum Revenue

Passenger optimization revolves around adjusting ticket sales to hit the sweet spot for maximum revenue. Based on the problem, once we know from the vertex calculation that 50 additional passengers beyond 200 yield the highest revenue, we conclude that a total of 250 passengers will optimize profits. This is because both the reduction in fare and the increase in passenger numbers are perfectly balanced at this point.To calculate the resulting maximum revenue and understand the impact of fare changes, we evaluate:

• Total Passengers: 250
• Fare per Person: \\(250 (derived from \( 300 - 50 \))
• Maximum Revenue: \( R(50) = 250 \times 250 = \\)62,500 \)

This shows that by fine-tuning passenger numbers and understanding fare adjustments, the airline maximizes its earnings while offering an attractive price point for passengers.