Problem 44 - Applied Calculus

Question

The quantity, \(q\), of a certain skateboard sold depends on the selling price, \(p\), in dollars, so we write \(q=f(p)\). You are given that \(f(140)=15,000\) and \(f^{\prime}(140)=-100\). (a) What do \(f(140)=15,000\) and \(f^{\prime}(140)=-100\) tell you about the sales of skateboards? (b) The total revenue, \(R\), earned by the sale of skateboards is given by \(R=p q\). Find \(\left.\frac{d R}{d p}\right|_{p=140}\). (c) What is the sign of \(\left.\frac{d R}{d p}\right|_{p=140}\) ? If the skateboards are currently selling for \(\$ 140\), what happens to revenue if the price is increased to \(\$ 141 ?\)

Step-by-Step Solution

Answer

Revenue is increasing at \$140; increasing the price to \$141 will increase revenue.

Step 1: Interpret Given Function Values

The given value \( f(140) = 15,000 \) indicates that when the price \( p \) is \$140, the quantity of skateboards sold is 15,000 units. This is a snapshot of how many skateboards are sold at that specific price.

Step 2: Interpret Known Derivative

The given derivative \( f'(140) = -100 \) tells us the rate of change of quantity with respect to price. Specifically, for every dollar increase in price from \$140, 100 fewer skateboards are sold. This provides insight into consumer demand sensitivity to price changes.

Step 3: Differentiate Revenue Function

The total revenue \( R \) is given by \( R = p \cdot f(p) \). To find \( \frac{dR}{dp} \), use the product rule for differentiation: \( \frac{dR}{dp} = f(p) + p \cdot f'(p) \).

Step 4: Substitute and Calculate Derivative at Given Price

Substitute the known values into the differentiated equation: \( f(140) = 15,000 \) and \( f'(140) = -100 \). So, \[ \left. \frac{dR}{dp} \right|_{p=140} = 15,000 + 140(-100) = 15,000 - 14,000 = 1,000. \]

Step 5: Determine the Sign of the Derivative

The derivative \( \left. \frac{dR}{dp} \right|_{p=140} = 1,000 \) is positive, indicating that revenue increases when the price is increased from \\(140 to \\)141.

Key Concepts

Demand Sensitivity

Demand sensitivity refers to how the quantity demanded of a product changes in response to changes in price. This concept helps businesses understand how consumers will react to changes in the prices of their products. In the example given, we have a mathematical representation of this sensitivity through the function \( q = f(p) \) and its derivative. The interpretation of \( f'(140) = -100 \) is crucial here. It tells us that for each dollar increase in the price of the skateboard from $140, the quantity sold decreases by 100 units. This means there is a negative relationship between price and quantity sold at this particular price point, highlighting the responsiveness or sensitivity of the skateboard's demand to price changes. Understanding this sensitivity can help businesses set optimal pricing strategies to maximize revenue or market shares.

Product Rule

The product rule is a critical tool in calculus used when differentiating functions that are the product of two simpler functions. This rule states that if you have a function \( R = u \cdot v \), where \( u \) and \( v \) are functions of \( p \), then the derivative \( \frac{dR}{dp} \) is given by:

• \( \frac{dR}{dp} = u' \cdot v + u \cdot v' \)

In the context of revenue optimization, the revenue function \( R = p \cdot f(p) \) is the product of price \( p \) and quantity function \( f(p) \). Applying the product rule, we get:\

• \( \frac{dR}{dp} = f(p) + p \cdot f'(p) \)

This calculation allows us to explore how changes in price affect total revenue, combining the effects of price changes on both price itself and the quantity sold. Recognizing this relationship is essential for optimizing prices to enhance revenue outcomes.

Differentiation in Calculus

Differentiation is a fundamental concept in calculus that involves computing the rate at which one quantity changes with respect to another. It helps in analyzing how a small change in one variable affects another variable. By applying differentiation to the revenue function \( R = p \cdot f(p) \), we can determine how the revenue changes with respect to price changes.

• This skill is used to find maxima and minima of functions, such as finding optimal prices to maximize revenue.
• In our exercise, \\( \frac{dR}{dp} = 15,000 + 140(-100) \) allows us to quickly understand the impact of price adjustments on skateboard sales revenue.

Differentiation thus provides us with a precise tool to make informed economic decisions and perform consumer behavior analysis, helping in maximizing strategic objectives like profit or market penetration.

Consumer Behavior Analysis

Consumer behavior analysis involves studying the responses and preferences of consumers, which are often influenced by price changes, among other factors. The exercise here is a demonstration of how calculus, particularly demand sensitivity and differentiation, can be used to understand and forecast consumer behavior effectively.

• The given derivative \( f'(140) = -100 \) demonstrates not just demand sensitivity, but also consumer behavior, as it reflects a decrease in sales when prices increase.
• Understanding consumer behavior through mathematical analysis allows businesses to predict how price changes can impact demand, enabling more strategic decisions that consider consumer preferences.
• The resulting positive derivative \( \frac{dR}{dp} > 0 \) indicates that, despite the drop in quantity, the increased price might lead to higher revenue, aligning with profit maximization goals.

This kind of analysis helps businesses fine-tune pricing strategies according to consumer responses, ensuring that business objectives are met while still catering to consumer needs.