Problem 7 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

The function \(R(p)=35 p(75-p)\) gives revenue as a function of price, \(p\), where the price is given in dollars. (a) Find the price at which the revenue is maximum. (b) What is the maximum price?

Step-by-Step Solution

Answer

The price that will yield the maximum revenue is $37.5. The maximum price that can be charged, beyond which the revenue becomes negative, is $75.

Step 1: Finding the Price that Maximizes Revenue

The first step is to differentiate the revenue function \(R(p)\) with respect to \(p\). That gives us: \(R'(p) = 35(75-p) - 35p\). The next step is to set the derivative equal to zero and solve for \(p\): \(0 = 35(75 - p) - 35p\). Solving this, we find that \(p = 37.5\). Thus, the revenue will be maximized when the price is $37.5.

Step 2: Determining the Maximum Price

The maximum price will occur when the revenue becomes zero (as it would not make sense for a vendor to charge a price that yields negative revenue). Hence, we have to solve the given revenue equation for \(p\), such as: \(0 = 35p(75 - p)\). Solving this, two possible solutions emerge: \(p = 0\) and \(p = 75\). Because the price cannot be negative (in any meaningful business context), $75 is interpreted as the maximum price that can be charged while still attaining non-negative revenue.

Key Concepts

Revenue Maximization

When discussing problems related to revenue maximization, the primary objective is to find the price that yields the highest possible revenue for a company. In mathematical terms, this means determining the input value (price, in our exercise) at which a given revenue function reaches its peak.

To maximize revenue, typically one examines the revenue function, which often depends on the price of the good or service. The revenue function describes how revenue changes as the price changes.

• Understanding the relationship between price and revenue is fundamental.
• In many cases, for simplicity, these functions are modeled as quadratic functions.

The ultimate goal is to find that specific price point where increasing or decreasing the price would lead to less revenue, thereby revealing the optimal price for maximum revenue.

Derivative

Derivatives are a core concept in calculus and are essential for solving optimization problems. They measure how a function changes as its input changes, providing a powerful tool to identify rates of change and slope of a curve.

In optimization, the derivative is utilized to find critical points, which are potential locations for maximum or minimum values of the function.
To find the derivative of a function like the revenue function, we apply differentiation rules. For our example:

• We started with the function: \( R(p) = 35p(75-p) \)
• The derivative with respect to \( p \) is \( R'(p) = 35(75-p) - 35p \).

Setting this derivative equal to zero helps identify points where the function might reach a maximum or minimum, known as critical points.

Quadratic Function

A quadratic function is a type of polynomial function characterized by the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The graph of a quadratic function is called a parabola, and it can open upwards or downwards depending on the coefficient \( a \).

In our revenue maximization problem, the function \( R(p) = 35p(75-p) \) is quadratic. The quadratic nature means that the function will have a single vertex, or turning point, which represents the maximum or minimum value of the function.

• The solution to identifying this point involves finding the roots of the derivative of the quadratic function.
• Here, the revenue tops out at \( p = 37.5 \), highlighting the significance of understanding quadratic behavior to solve such problems efficiently.

Understanding quadratic functions enables one to identify these key points quickly and efficiently.

Critical Points

Critical points in calculus are points on a graph where the derivative is zero or undefined. These points are significant because they represent potential maximum or minimum values of the original function.

In the context of our exercise, finding the critical points involved setting the derivative \( R'(p) = 35(75-p) - 35p \) equal to zero. This forms the equation:

\[ 0 = 35(75 - p) - 35p \]

Solving this equation yields \( p = 37.5 \), a critical point indicating where maximum revenue occurs.

• Critical points are analyzed further to determine whether they are maxima or minima, often using the second derivative test or analyzing the sign changes of the derivative.
• In our scenario, given the nature of the quadratic curve (it opens downwards), the critical point we found corresponds to a maximum.

Understanding critical points helps in identifying optimal solutions in various real-world contexts.