Suppose that the market equilibrium price for printers is P*=100 and quantity traded of printers is Q=250. At the equilibrium point, the own price elasticity of demand is: Eqx,px = -4. a) Find the demand equation. b) What is the price that maximizes revenues? What are total revenues at that price?
Suppose that the market equilibrium price for printers is P*=100 and quantity traded of printers is Q=250. At the equilibrium point, the own price elasticity of demand is: Eqx,px = -4. a) Find the demand equation. b) What is the price that maximizes revenues? What are total revenues at that price?
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![### Equilibrium Price and Quantity Analysis
**Problem Statement:**
Suppose that the market equilibrium price for printers is \( P^* = 100 \) and the quantity traded of printers is \( Q = 250 \). At the equilibrium point, the own price elasticity of demand is:
\[ E_{Qx,Px} = -4 \]
**Tasks:**
a) Find the demand equation.
b) What is the price that maximizes revenues? What are total revenues at that price?
### Detailed Explanation:
**Elasticity and Demand Equation:**
The own price elasticity of demand ( \( E_{Q,P} \) ) is given by:
\[ E_{Q,P} = \frac{\partial Q}{\partial P} \cdot \frac{P}{Q} \]
Given:
- \( P = 100 \)
- \( Q = 250 \)
- \( E_{Qx,Px} = -4 \)
To find the demand equation, which is generally of the form \( Q = a - bP \), follow these steps:
1. Use the elasticity formula:
\[ -4 = \frac{\partial Q}{\partial P} \cdot \frac{100}{250} \]
2. Rearrange to find \( \frac{\partial Q}{\partial P} \):
\[ -4 = \frac{\partial Q}{\partial P} \cdot 0.4 \]
\[ \frac{\partial Q}{\partial P} = \frac{-4}{0.4} = -10 \]
So, \( -b = -10 \), thus \( b = 10 \).
3. Use the equilibrium condition \( Q = a - bP \):
\[ 250 = a - 10 \cdot 100 \]
\[ 250 = a - 1000 \]
\[ a = 1250 \]
Thus, the demand equation is:
\[ Q = 1250 - 10P \]
**Revenue Maximization:**
Revenue (\( R \)) is given by:
\[ R = P \cdot Q \]
To maximize revenue, we use the relationship between revenue and elasticity:
- Revenue is maximized when the price elasticity of demand equals -1.
Given the initial elasticity and the linear demand equation, compute:
\[ E_{Q,P} = -1 = \frac{\partial Q}{\partial P} \cdot \frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F300cbca4-f6ea-4709-a050-de562527f506%2F932a2883-feb1-4bbb-ab15-3b68350b5ba8%2Fi280wo9_processed.png&w=3840&q=75)
Transcribed Image Text:### Equilibrium Price and Quantity Analysis
**Problem Statement:**
Suppose that the market equilibrium price for printers is \( P^* = 100 \) and the quantity traded of printers is \( Q = 250 \). At the equilibrium point, the own price elasticity of demand is:
\[ E_{Qx,Px} = -4 \]
**Tasks:**
a) Find the demand equation.
b) What is the price that maximizes revenues? What are total revenues at that price?
### Detailed Explanation:
**Elasticity and Demand Equation:**
The own price elasticity of demand ( \( E_{Q,P} \) ) is given by:
\[ E_{Q,P} = \frac{\partial Q}{\partial P} \cdot \frac{P}{Q} \]
Given:
- \( P = 100 \)
- \( Q = 250 \)
- \( E_{Qx,Px} = -4 \)
To find the demand equation, which is generally of the form \( Q = a - bP \), follow these steps:
1. Use the elasticity formula:
\[ -4 = \frac{\partial Q}{\partial P} \cdot \frac{100}{250} \]
2. Rearrange to find \( \frac{\partial Q}{\partial P} \):
\[ -4 = \frac{\partial Q}{\partial P} \cdot 0.4 \]
\[ \frac{\partial Q}{\partial P} = \frac{-4}{0.4} = -10 \]
So, \( -b = -10 \), thus \( b = 10 \).
3. Use the equilibrium condition \( Q = a - bP \):
\[ 250 = a - 10 \cdot 100 \]
\[ 250 = a - 1000 \]
\[ a = 1250 \]
Thus, the demand equation is:
\[ Q = 1250 - 10P \]
**Revenue Maximization:**
Revenue (\( R \)) is given by:
\[ R = P \cdot Q \]
To maximize revenue, we use the relationship between revenue and elasticity:
- Revenue is maximized when the price elasticity of demand equals -1.
Given the initial elasticity and the linear demand equation, compute:
\[ E_{Q,P} = -1 = \frac{\partial Q}{\partial P} \cdot \frac{
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