Problem 42 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Suppose that \(q=500-2 p\) units of a certain commodity are demanded when \(p\) dollars per unit are charged, for \(0 \leq p \leq 250\). a. Determine where the demand is elastic, inelastic, and of unit elasticity with respect to price. b. Use the results of part (a) to determine the intervals of increase and decrease of the revenue function and the price at which revenue is maximized. c. Find the total revenue function explicitly, and use its first derivative to determine its intervals of increase and decrease and the price at which revenue is maximized.

Step-by-Step Solution

Answer

The demand is elastic for \( p > 125 \), inelastic for \( p < 125 \), and unit elastic at \( p = 125 \). Revenue increases for \( 0 < p < 125 \) and decreases for \( p > 125 \). The price at which revenue is maximized is \( p = 125 \).

Step 1: Express Demand Function in Terms of p

Given the demand function is \[ q = 500 - 2p \]

Step 2: Find the Elasticity of Demand

Elasticity of demand, denoted by E(p), is given by the formula:\[ E(p) = \left| \frac{dQ/dP \cdot P}{Q} \right| \]First, differentiate the demand function concerning p:\[ \frac{dq}{dp} = -2 \]Then, substitute into the elasticity formula:\[ E(p) = \left| \frac{-2\cdot p}{500 - 2p} \right| = \frac{2p}{500 - 2p} \]

Step 3: Determine Elasticity Regions

Classify the elasticity regions:- Elastic (E(p) > 1): \( \frac{2p}{500 - 2p} > 1 \rightarrow 2p > 500 - 2p \rightarrow 4p > 500 \rightarrow p > 125 \)- Inelastic (E(p) < 1): \( \frac{2p}{500 - 2p} < 1 \rightarrow 2p < 500 - 2p \rightarrow 4p < 500 \rightarrow p < 125 \)- Unit Elasticity (E(p) = 1): \( \frac{2p}{500 - 2p} = 1 \rightarrow 2p = 500 - 2p \rightarrow 4p = 500 \rightarrow p = 125 \)So, elasticity regions are:- Elastic: \( p > 125 \)- Inelastic: \( p < 125 \)- Unit Elasticity: \( p = 125 \)

Step 4: Define Revenue Function

Revenue function R(p) is given by:\[ R(p) = p \cdot q \]Substituting \( q = 500 - 2p \) gives:\[ R(p) = p (500 - 2p) = 500p - 2p^2 \]

Step 5: Determine Intervals of Increase/Decrease of Revenue

Differentiate the revenue function with respect to p:\[ \frac{dR}{dp} = 500 - 4p \]Set the derivative equal to zero to find critical points:\[ 500 - 4p = 0 \rightarrow p = 125 \]To find the intervals of increase/decrease, analyze the sign of the derivative in the intervals:- For \( 0 < p < 125 \), \( \frac{dR}{dp} > 0 \) (Revenue is increasing)- For \( p > 125 \), \( \frac{dR}{dp} < 0 \) (Revenue is decreasing)Thus, Revenue increases when \( 0 < p < 125 \) and decreases when \( p > 125 \).The revenue is maximized at \( p = 125 \).

Step 6: Find Total Revenue Function

Reiterate the revenue function for confirmation:\[ R(p) = 500p - 2p^2 \]

Key Concepts

Revenue Function

The **Revenue Function** is crucial when dealing with demand and pricing. It helps us understand how changes in price affect total revenue. Generally, revenue \(R\(p\)\) is calculated as the product of price \(p\) and quantity demanded \(q\). In this exercise, the demand function is \[ q = 500 - 2p \], and the revenue function is thus: \[ R(p) = p(500 - 2p) = 500p - 2p^2 \]. This quadratic equation will help us analyze revenue behavior as price changes. \[ \[\[\begin{align*} \text{When } p = 0, & \ R(0) = 0 \ \text{When } p = 250, & \ R(250) = 0 \end{align*}\]\] \] For prices between 0 and 250, the revenue varies, and we'll delve deeper using derivatives to understand this variation better.

Critical Points

To find the **Critical Points** of the revenue function, we need to take its derivative and set it to zero. The derivative tells us the rate of change of revenue with respect to price. For \(R(p) = 500p - 2p^2\), differentiate it with respect to \(p\) to get: \[ \frac{dR}{dp} = 500 - 4p \] Setting the derivative to zero to find critical points: \[ 500 - 4p = 0 \] \[ 4p = 500 \] \[ p = 125 \] This means at **p = 125**, a critical point is found. At this price, the revenue function has a critical point, meaning it could be a maximum or minimum. Since it's a quadratic function opening downwards, \(p = 125\) is where the revenue is maximized. Beyond these calculations, examining the sign of the derivative helps in confirming that this point provides the maximum revenue.

Derivative Analysis

The **Derivative Analysis** is key to understanding the behavior of the revenue function over different price intervals. By differentiating \[ R(p) = 500p - 2p^2 \], we found \[ \frac{dR}{dp} = 500 - 4p \]. We obtained a critical point at \(p = 125 \).
Now let's interpret it further:

• For \: 0 < p < 125 \, \(\frac {dR}{dp}\) > 0 (Revenue is increasing)
• For \: p > 125 \, \(\frac {dR}{dp}\) < 0 (Revenue is decreasing)

Therefore, **Revenue increases** when \(\0 < p < 125\,\) and then **Revenue decreases** beyond \(\p > 125\).
This tells us that the revenue function reaches its peak, or in mathematical terms, its maximum at the critical point **p = 125**. Understanding these intervals helps businesses strategically price their products to maximize revenue.