Math Problem Statement
If, in a monopoly market, the demand function for a product is p = 160 − 1.00x and the revenue function is R = px, where x is the number of units sold and p is the price per unit, what price will maximize revenue?
Solution
To find the price that maximizes revenue in a monopoly market, we need to follow these steps:
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Revenue function: The revenue is given by: where and is the number of units sold.
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Substitute the demand function into the revenue function: Substitute into the revenue equation: Simplifying the expression: This is the revenue function in terms of .
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Differentiate the revenue function: To maximize revenue, we need to take the derivative of with respect to and set it equal to zero:
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Find the critical point: Set to find the value of that maximizes revenue: Solving for :
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Find the corresponding price: Now that we know , substitute this value into the demand function to find the price that maximizes revenue:
Thus, the price that maximizes revenue is $80.
Would you like further details, or have any questions?
Here are 5 related questions to explore this topic further:
- How does the elasticity of demand affect the revenue-maximizing price?
- What would happen to the revenue-maximizing price if the demand curve slope changes?
- How can you find the profit-maximizing price if costs are included?
- Can the revenue-maximizing price be higher or lower than the profit-maximizing price? Why?
- How do you calculate marginal revenue from the demand function?
Tip: Remember that in a monopoly, the price set by the firm directly affects the quantity demanded and total revenue.
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Math Problem Analysis
Mathematical Concepts
Algebra
Calculus
Revenue Maximization
Formulas
Revenue function: R = px
Demand function: p = 160 - 1.00x
Maximization by derivative: dR/dx = 160 - 2x
Theorems
Critical points from derivative
Revenue maximization
Suitable Grade Level
Undergraduate Economics or Advanced High School Math (Grades 11-12)
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