Problem 62 - Calculus Early Transcendentals: Pearson New International Edition

Question

For the price function defined by $$ p(x)=(182-x / 36)^{1 / 2} $$ find the number of units \(x_{1}\) that makes the total revenue a maximum and state the maximum possible revenue. What is the marginal revenue when the optimum number of units, \(x_{1}\), is sold?

Step-by-Step Solution

Answer

To maximize revenue, solve for critical points, confirm with a second derivative test, and plug into the revenue formula. Marginal revenue at maximum is zero.

Step 1: Define Revenue Function

Total revenue, denoted as \( R(x) \), is the product of the number of units sold \( x \) and the price per unit \( p(x) \). Setting this up, we have:\[R(x) = x \cdot (182 - \frac{x}{36})^{1/2}\]

Step 2: Differentiate Revenue Function

To find the maximum revenue, we need to find the critical points of the revenue function \( R(x) \). First, differentiate \( R(x) \) with respect to \( x \). Using the product rule and chain rule, the derivative is:\[R'(x) = (182 - \frac{x}{36})^{1/2} + x \cdot \left(-\frac{1}{2}\right)\cdot(182 - \frac{x}{36})^{-1/2} \cdot \left(-\frac{1}{36}\right)\]

Step 3: Solve for Critical Points

Set \( R'(x) = 0 \) to find critical points. Simplifying the expression:\[(182 - \frac{x}{36})^{1/2} = \frac{x}{72}\times(182 - \frac{x}{36})^{-1/2}\]Squaring both sides and solving the equation will yield potential values for \( x \) which maximize revenue.

Step 4: Analyze Second Derivative

Use the second derivative test to confirm which critical point gives a maximum revenue. Compute \( R''(x) \) and evaluate it at the critical point determined from Step 3 to confirm concavity.

Step 5: Calculate Maximum Revenue

With the value of \( x_1 \) from previous steps, calculate the total maximum revenue using the revenue function:\[ R(x_1) = x_1 \cdot (182 - \frac{x_1}{36})^{1/2} \]

Step 6: Determine Marginal Revenue at Optimum

Marginal revenue at \( x_1 \) is \( R'(x_1) \), which is 0 because \( x_1 \) is a critical point for maximum revenue. Hence, the marginal revenue at this optimum point is zero.

Key Concepts

Critical Points

Critical points are special values where a function's derivative is either zero or undefined. These points are significant in determining whether a function reaches its maximum or minimum value at that location. For the given revenue function, the critical points can indicate where the revenue is maximized. By setting the derivative of the revenue function, denoted as \( R'(x) \), equal to zero, we can find possible values of \( x \) that may yield maximum revenue.

In our exercise, once \( R'(x) = 0 \) is solved, the resulting \( x \) values are considered critical points. These points require further testing to ascertain if they represent maximum revenue scenarios. Understanding critical points is crucial because they help pinpoint the exact conditions under which revenue is at its peak.

Derivative Test

Derivative tests play a pivotal role in determining the nature of critical points. The first derivative test involves examining the sign of the derivative before and after a critical point to determine if it is a maximum or minimum. However, in this context, we use the second derivative test because it can succinctly confirm if a critical point is a maximum.

The second derivative test requires calculating \( R''(x) \), the second derivative of the revenue function. After computing \( R''(x) \), we evaluate it at the critical point found earlier. If \( R''(x) < 0 \), the function is concave down at that point, indicating a local maximum. On the other hand, if \( R''(x) > 0 \), it suggests a local minimum. This test is valuable because it conclusively identifies whether we have indeed found the maximum revenue point.

Revenue Function

The revenue function serves as the foundation for calculating total revenue based on sales and pricing. In this exercise, the revenue function \( R(x) \) is defined as the product of the number of units sold, \( x \), and the price per unit, \( p(x) \). Thus, it is expressed as:\[ R(x) = x \cdot (182 - \frac{x}{36})^{1/2} \]

This formula allows us to analyze how changing the number of units sold influences total revenue. The interplay between unit sales and pricing affects how maximum revenue is achieved. By setting up and manipulating the revenue function, we can derive further insights about how to optimally set both price and quantity sold elements.

• Offers insight into sales and pricing dynamics.
• Critical for revenue optimization strategies.

The understanding of revenue function is essential as it lays the groundwork for optimally determining the best sales strategies and pricing models.

Marginal Revenue

Marginal revenue refers to the additional revenue generated from selling one more unit. It is derived by taking the first derivative of the revenue function, \( R'(x) \), which measures the change in revenue with respect to a change in the number of units sold.
At the point of maximum revenue, which is determined from our critical point analysis, the marginal revenue equals zero. This is because an additional unit sale neither adds nor detracts from total revenue, marking an optimized state.

• Indicates revenue changes with each additional unit sold.
• At optimum sales, marginal revenue is zero.

Understanding marginal revenue is essential for businesses aiming to capitalize on unit sales effectiveness, ensuring that each unit contributes positively to profitability up to the point of optimization.