Question 1200290: Please help me solve the following question--
Suppose that the demand equation for a monopolist is p = 100 -0.01x and the cost function is C(x) = 50x + 10,000. Find the value of x that maximizes the profit and determine the corresponding price and total profit for this level of production. Then, also find it under the condition that the government has imposed an excise tax of $10 per unit.
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **1. Find Profit Function (Without Tax)**
* **Revenue Function (R(x))**:
* Revenue = Price * Quantity
* R(x) = p * x
* R(x) = (100 - 0.01x) * x
* R(x) = 100x - 0.01x²
* **Profit Function (π(x))**:
* Profit = Revenue - Cost
* π(x) = R(x) - C(x)
* π(x) = (100x - 0.01x²) - (50x + 10,000)
* π(x) = -0.01x² + 50x - 10,000
**2. Find Profit-Maximizing Quantity (Without Tax)**
* **Find the derivative of the profit function:**
* π'(x) = -0.02x + 50
* **Set the derivative to zero and solve for x:**
* -0.02x + 50 = 0
* 0.02x = 50
* x = 50 / 0.02
* x = 2500 units
* **Find the corresponding price:**
* p = 100 - 0.01x
* p = 100 - 0.01 * 2500
* p = 100 - 25
* p = $75
* **Calculate the profit:**
* π(x) = -0.01x² + 50x - 10,000
* π(2500) = -0.01 * (2500)² + 50 * 2500 - 10,000
* π(2500) = $52,500
**Therefore, without the tax:**
* **Profit-maximizing quantity (x): 2500 units**
* **Price (p): $75**
* **Total profit: $52,500**
**3. Find Profit-Maximizing Quantity (With Tax)**
* **New Cost Function (C'(x))**:
* C'(x) = 50x + 10,000 + 10x
* C'(x) = 60x + 10,000
* **New Profit Function (π'(x))**:
* π'(x) = R(x) - C'(x)
* π'(x) = 100x - 0.01x² - (60x + 10,000)
* π'(x) = -0.01x² + 40x - 10,000
* **Find the derivative of the new profit function:**
* π'(x) = -0.02x + 40
* **Set the derivative to zero and solve for x:**
* -0.02x + 40 = 0
* 0.02x = 40
* x = 40 / 0.02
* x = 2000 units
* **Find the corresponding price:**
* p = 100 - 0.01x
* p = 100 - 0.01 * 2000
* p = 100 - 20
* p = $80
* **Calculate the profit:**
* π'(x) = -0.01x² + 40x - 10,000
* π'(2000) = -0.01 * (2000)² + 40 * 2000 - 10,000
* π'(2000) = $30,000
**Therefore, with the $10 excise tax:**
* **Profit-maximizing quantity (x): 2000 units**
* **Price (p): $80**
* **Total profit: $30,000**
**Key Observations:**
* The excise tax reduces the profit-maximizing output level.
* The excise tax leads to an increase in the price charged to consumers.
* The excise tax significantly reduces the total profit for the monopolist.
|
|
|
