Problem 18 - Applied Calculus

Question

(a) Let \(p\) be the price and \(q\) be the quantity sold of a good with a high elasticity of demand, \(E .\) Explain intuitively (without formulas) the effect of raising the price on the revenue, \(R\) (b) Derive an expression for \(d R / d p\) in terms of \(q\) and E. Show all the steps and reasoning. (c) Explain how your answer to part (b) confirms your answer to part (a).

Step-by-Step Solution

Answer

For goods with high elasticity, increasing price lowers revenue. The derivative confirms revenue decreases as price increases.

Step 1: Understanding Elasticity of Demand

Elasticity of demand is a measure of how much the quantity demanded of a good changes when its price changes. If demand is highly elastic, it means that small changes in price lead to large changes in the quantity demanded.

Step 2: Revenue and Price Relationship

Revenue, \( R \), is calculated as \( R = p \times q \). For goods with high elasticity of demand, when price \( p \) is increased, the quantity sold \( q \) tends to decrease significantly. This large decrease in \( q \) usually results in a decrease in \( R \), despite the higher price.

Step 3: Setting up the Expression for Revenue

Revenue \( R \) is defined as \( p imes q \). In order to find how revenue changes with price, we need to find \( \frac{dR}{dp} \).

Step 4: Using the Product Rule for Differentiation

Differentiate \( R = p imes q \) with respect to \( p \):\[ \frac{dR}{dp} = q + p \times \frac{dq}{dp}. \]

Step 5: Relating Elasticity to Derivatives

Elasticity \( E \) is defined as \( E = \frac{dq/dp}{q/p} \), which simplifies to \( E = \frac{p}{q} \times \frac{dq}{dp} \). Rearrange to find \( \frac{dq}{dp} = \frac{E \cdot q}{p} \).

Step 6: Substitute Elasticity into the Derivative

Substitute \( \frac{dq}{dp} = \frac{E \cdot q}{p} \) into \( \frac{dR}{dp} \):\[ \frac{dR}{dp} = q + p \left( \frac{E \cdot q}{p} \right) = q(1 + E). \]

Step 7: Analyzing the Derived Expression

For goods with high elasticity \( E < -1 \), \( q(1 + E) < 0 \). This means \( \frac{dR}{dp} < 0 \), indicating that revenue decreases as price increases, confirming the intuition from part (a).

Key Concepts

Revenue

Revenue in economics is the total income generated from the sale of goods or services. Understanding revenue is crucial when discussing the impact of price changes on sales.
\[ R = p \times q \]
This formula calculates revenue where \( R \) represents revenue, \( p \) is the price per unit, and \( q \) is the quantity sold.
When considering goods with high elasticity of demand, if the price \( p \) is increased, the outcome can be counterintuitive.
Even though an increase in price might seem like it should increase revenue, when demand is elastic, customers react more sensitively to price changes, leading to a significant drop in quantity sold \( q \), which in turn can decrease overall revenue.

• For elastic goods, small price increases can lead to large revenue decreases.
• Marketers must consider how elastic demand affects total revenue.

Overall, to optimize revenue, it's vital to understand the elasticity and how it affects consumer buying behavior.

Differentiation

Differentiation in mathematics involves finding the derivative of a function. When applied to economics, it provides insights into how a variable changes concerning another. Let's consider how revenue changes as the price changes.
The goal is to determine \( \frac{dR}{dp} \), or how revenue \( R \) changes with respect to price \( p \).
This requires differentiating the revenue equation \( R = p \times q \) with respect to price.

• Differentiation tells us the rate at which revenue changes for a small change in price.
• It highlights the responsiveness of revenue to price variations.

In economics, understanding how to differentiate can provide pivotal insights into optimizing financial outcomes and understanding the behavior of graphs in relation to economic functions.

Product Rule

The product rule is a crucial method in calculus for finding the derivative of functions that are products of two or more variables. Let's see how it's applied in this context to understand changes in revenue.
For a function \( y = f(x)g(x) \), the derivative \( y' \) is calculated as \[ y' = f'(x)g(x) + f(x)g'(x) \].
In our case, when determining how \( R = p \times q \) changes, we apply the product rule because \( R \) is a product of \( p \) and \( q \). Applying the product rule gives:
\[ \frac{dR}{dp} = q + p \times \frac{dq}{dp} \]

• This method allows breaking down complex expressions.
• The product rule is vital to handle functions where independent and dependent variables change.

Using the product rule helps us systematically understand and predict changes in economic models, ensuring accurate results.

Derivative

A derivative represents the rate at which a function is changing at any given point. It is a fundamental concept in calculus, particularly useful in economic analysis for understanding changes in quantities.
In our revenue scenario, we are interested in how the revenue \( R \) changes as the price \( p \) changes, expressed as \( \frac{dR}{dp} \).
To determine this, understanding how the derivative connects to economic quantities is important. We previously derived:
\[ \frac{dR}{dp} = q(1 + E) \]
Where \( E \) is the elasticity of demand.

• The derivative helps assess whether increasing prices will increase or decrease revenue.
• Studying derivatives aids in anticipating market reactions.

When the derivative is negative, it suggests that as the price increases, revenue decreases for highly elastic goods, confirming intuition about consumer pricing sensitivity.