Problem 39 - Calculus Early Transcendentals

Question

A manufacturer of athletic footwear finds that the sales of their ZipStride brand running shoes is a function \(f(p)\) of the selling price \(p\) (in dollars) for a pair of shoes. Suppose that \(f(120)=9000\) pairs of shoes and \(f^{\prime}(120)=-60\) pairs of shoes per dollar. The revenue that the manufacturer will receive for selling \(f(p)\) pairs of shoes at \(p\) dollars per pair is \(R(p)=p \cdot f(p) .\) Find \(R^{\prime}(120) .\) What impact would a small increase in price have on the manufacturer's revenue?

Step-by-Step Solution

Answer

A small increase in price raises revenue by about $1800.

Step 1: Understand Revenue Equation

The revenue function is given by \( R(p) = p \cdot f(p) \). This means the revenue generated is the product of the price per pair \( p \) and the number of pairs sold \( f(p) \).

Step 2: Differentiate the Revenue Function

To find the derivative of the revenue function \( R(p) \), use the product rule. The product rule states that \((uv)' = u'v + uv'\) for two functions \(u\) and \(v\). In this case, \( u = p \) and \( v = f(p) \), thus: \( R'(p) = 1 \cdot f(p) + p \cdot f'(p) = f(p) + p \cdot f'(p) \).

Step 3: Substitute Known Values into Derivative

Substitute the given values \( f(120) = 9000 \) and \( f'(120) = -60 \) into the derivative expression \( R'(p) = f(p) + p \cdot f'(p) \). Therefore: \( R'(120) = 9000 + 120 \cdot (-60) \).

Step 4: Compute the Derivative at the Specific Price

Calculate \( R'(120) = 9000 + 120 \cdot (-60) = 9000 - 7200 = 1800 \).

Step 5: Interpret the Result

\( R'(120) = 1800 \) indicates that a small increase in selling price from 120 dollars would result in an increase in revenue by approximately 1800 dollars. This is because \( R'(120) > 0 \).

Key Concepts

Revenue Function

In the context of this exercise, the revenue function plays a crucial role in understanding how the sales and pricing of a product affect the overall income for a business. For the manufacturer of the ZipStride shoes, the revenue function is expressed as \( R(p) = p \cdot f(p) \). This means:

• \( p \) represents the selling price of each pair of shoes.
• \( f(p) \) indicates the number of shoes sold at that price.
• The entire function \( R(p) \) captures the total revenue from selling shoes at price \( p \).

By calculating this function, the manufacturer can predict revenue changes based on different pricing strategies. If the price changes, the function can help forecast how many shoes may be sold and what revenue can be anticipated.

Product Rule

The solution to this problem heavily relies upon differential calculus, particularly the product rule. The product rule is essential for differentiating products of two or more functions.When you have two functions \( u \) and \( v \), the product rule tells us how to differentiate their product: \[ (uv)' = u'v + uv' \]In the scenario given, the revenue function \( R(p) = p \cdot f(p) \) requires using the product rule for differentiation. Here's how it applies:- Set \( u = p \), which means the derivative \( u' = 1 \).- Set \( v = f(p) \), where the derivative \( v' = f'(p) \).Applying the product rule gives us:\[ R'(p) = f(p) + p \cdot f'(p) \]This form lets you find how quickly revenue changes with a small change in price, allowing the manufacturer to make informed pricing decisions.

Derivative Evaluation

Evaluating the derivative at a specific point provides valuable insight into the effect of small changes in variables. In this exercise, once we differentiate our revenue function, we substitute known values to assess the result at the price point of \( p = 120 \).From the exercise:

• \( f(120) = 9000 \) indicating 9000 pairs of shoes are sold at \(120 each.
• \( f'(120) = -60 \) meaning for every dollar increase, 60 fewer shoes are sold.

Substitute these into the differentiated function:\[ R'(120) = 9000 + 120 \times (-60) \]Calculating this gives:\[ R'(120) = 9000 - 7200 = 1800 \]This outcome implies that a small increase in price around \)120 will lead to a revenue increase of approximately $1800. Understanding the sign and magnitude of \( R'(p) \) is crucial for making strategic pricing decisions and evaluating business profitability effectively.