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{
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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