Problem 17 - Calculus for Biology and Medicine

Question

Ticket Price Optimization Dalmatian Airlines flies a daily flight from Los Angeles to Minneapolis. Currently they sell each ticket for \(\$ 300\), and on average 100 people take the flight, so their revenue per flight is 100 tickets \(\times \$ 300 /\) ticket \(=\$ 30,000\). They are interested in seeing whether they can increase their revenue by changing the price of a ticket. Based on market research they discover that for every \(\$ 1\) increase in ticket price, one fewer person will buy a ticket. Similarly for every \(\$ 1\) decrease in ticket price, one more person will buy a ticket. (a) What ticket price would maximize Dalmatian Airlines' revenue? (Hint: Denote the number of extra people flying on the route due to a price change by \(x\), and the cost of a ticket by \(\$ 300-x\). Then explain why the revenue to be maximized is \(R(x)=\) \((300-x)(100+x)\). You should also explain what the domain of this function is. (b) The plane can seat a maximum of 150 people. How does this information change the domain of \(R(x) ?\) What is the new optimal ticket price?

Step-by-Step Solution

Answer

Original optimal price: \$200; Adjusted with capacity: \$250.

Step 1: Understand the Revenue Formula

The problem provides us with a formula for revenue based on a price change: \[ R(x) = (300 - x)(100 + x) \]Here, \(x\) represents the variation in ticket sales due to changes in price. The term \((300 - x)\) stands for the price of a ticket, and \((100 + x)\) is the expected number of tickets sold.

Step 2: Determine the Revenue to Maximize

To find the ticket price that maximizes revenue, we need to maximize the function \( R(x) = (300-x)(100+x) \). This equation is quadratic and can be expanded to:\[R(x) = 30000 + 200x - x^2\].A quadratic function in the form \( ax^2 + bx + c \) achieves a maximum at \( x = -\frac{b}{2a} \). In this case, \( a = -1 \), \( b = 200 \).

Step 3: Calculate the Vertex of the Parabola

Use the vertex formula for a quadratic \( x = -\frac{b}{2a} \) to find the optimal \(x\):\[ x = -\frac{200}{2(-1)} = 100 \]This means the maximum revenue occurs when \(x = 100\).

Step 4: Calculate Corresponding Ticket Price

Plug \(x = 100\) into the price formula \( (300 - x) \) to find the ticket price that maximizes revenue:\[ 300 - 100 = 200 \]Thus, the ticket price that maximizes revenue is \$200.

Step 5: Determine Original Domain of Revenue Function

Without constraints, \(x\) could increase infinitely or decrease until tickets cost \$0 or until demand ceases. However, logical constraints limit \(x\) such that the ticket price remains positive, so \(300-x > 0\) implies \(x < 300\).

Step 6: Adjust for Plane Capacity

Given the plane can seat a maximum of 150 people, \(100 + x\) must be \( \leq 150\). This implies:\[ x \leq 50 \].

Step 7: Find Adjusted Optimal Solution

Considering the maximum of \(x = 50\) due to plane capacity, set \(x = 50\) in the price formula \((300-x)\):\[ 300 - 50 = 250 \]Therefore, the maximum revenue ticket price when considering seat capacity is \$250.

Key Concepts

Understanding Quadratic Functions

A quadratic function is a type of mathematical expression that is widely used in different fields, particularly in economic modeling to optimize outcomes like revenue. Quadratic functions have the general form:

• \[ ax^2 + bx + c \]

In this expression, \(a\), \(b\), and \(c\) are constants where \(a eq 0\). The key feature of quadratic functions is that they produce a parabolic graph. This means that the curve shaped like a 'U' can open upwards or downwards depending on the sign of \(a\).
If \(a\) is positive, the parabola opens upwards, indicating that the graph has a minimum point. If \(a\) is negative, as in our ticket price optimization exercise, the graph opens downwards, signifying a maximum point, ideal for revenue maximization.
The maximum or minimum point of this curve is called the vertex, and it is crucial for determining the optimal value that maximizes or minimizes the function in question. The vertex can be calculated using the formula:

• \[ x = -\frac{b}{2a} \]

This precise point helps in determining the optimum price for maximized revenue, ensuring practical decisions in economic contexts.

The Concept of Revenue Maximization

Revenue maximization is a fundamental goal in business strategies, especially in pricing strategies for goods and services. It involves finding the perfect balance between the price of a product and the number of units sold.
In our airline ticket pricing example, maximizing revenue involves adjusting the ticket price to achieve the most total revenue collected from passengers. The revenue function here is expressed as:

• \[ R(x) = (300-x)(100+x) \]

Here, \(R(x)\) is revenue, \(300-x\) is the adjusted price of one ticket, and \(100+x\) is the respective number of tickets sold. When we expand this function, we acquire a quadratic expression that assists in determining the maximum revenue through its vertex.
By calculating the vertex using the formula \[ x = -\frac{b}{2a} \], we discover the optimal value of the variable 'x', which decides how much the price should change to maximize revenue. Thus, revenue maximization takes into account both the price elasticity of demand and capacity constraints like plane seating to discover the ideal pricing strategy.

Economic Modeling and Practical Constraints

Economic modeling involves the use of mathematical equations and functions to represent real-life economic situations to predict outcomes and optimize performance. In our problem, we use a revenue model to optimize ticket pricing to enhance income.
Practically, several constraints must be considered in real-world applications. For instance, in the ticket optimization scenario, seat capacity limits the number of passengers on a flight. Initially, the variable \(x\) indicated that under theoretical conditions, \(x\) could range without limits, only ensuring that tickets are not priced below zero.
However, with constraints like plane capacity, further adjustments are mandatory. The maximum number of 150 passengers redefines the function's domain, necessitating:

• \[ x \leq 50 \]

This critical aspect of economic modeling ensures that the solutions derived are both realistic and applicable. It also illustrates how theoretical models need to adapt to practical circumstances to ensure optimal strategies in business.