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:

  1. Revenue function: The revenue RR is given by: R=pxR = px where p=1601.00xp = 160 - 1.00x and xx is the number of units sold.

  2. Substitute the demand function into the revenue function: Substitute p=1601.00xp = 160 - 1.00x into the revenue equation: R=(1601.00x)xR = (160 - 1.00x) \cdot x Simplifying the expression: R=160x1.00x2R = 160x - 1.00x^2 This is the revenue function in terms of xx.

  3. Differentiate the revenue function: To maximize revenue, we need to take the derivative of RR with respect to xx and set it equal to zero: dRdx=16021.00x=1602x\frac{dR}{dx} = 160 - 2 \cdot 1.00x = 160 - 2x

  4. Find the critical point: Set dRdx=0\frac{dR}{dx} = 0 to find the value of xx that maximizes revenue: 1602x=0160 - 2x = 0 Solving for xx: 2x=160x=802x = 160 \quad \Rightarrow \quad x = 80

  5. Find the corresponding price: Now that we know x=80x = 80, substitute this value into the demand function p=1601.00xp = 160 - 1.00x to find the price that maximizes revenue: p=1601.00(80)=16080=80p = 160 - 1.00(80) = 160 - 80 = 80

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:

  1. How does the elasticity of demand affect the revenue-maximizing price?
  2. What would happen to the revenue-maximizing price if the demand curve slope changes?
  3. How can you find the profit-maximizing price if costs are included?
  4. Can the revenue-maximizing price be higher or lower than the profit-maximizing price? Why?
  5. 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)