Question 3 (a)
Price discrimination between two markets
9 marks
- Time to write
- About 18 minutes
Question
Question 3 (a) · 9 marksAssume a monopolist sells his product in two separate markets, with demand functions p₁ = 16 - x₁ and p₂ = 28 - 3 x₂. His total cost function is C = 3 + 4 x, where x = x₁ + x₂. Determine the prices he will charge and the quantities he will sell in each market so as to maximize his profit.
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- Assume a monopolist sells his product…The paper fixes this: use it as given, and say so in your answer.
- DetermineFormula first, then the working, then the answer with its unit (₹, %, units).
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Profit function 2 marks 4 min
- 2 First-order conditions 3 marks 5 min
- 3 Second-order conditions 2 marks 4 min
- 4 Prices, quantities and profit 2 marks 3 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Try it first. About 18 minutes for 9 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Profit function. Revenue in market 1: R₁ = p₁x₁ = 16x₁ − x₁²; in market 2: R₂ = p₂x₂ = 28x₂ − 3x₂². Cost C = 3 + 4(x₁ + x₂). Profit π = R₁ + R₂ − C:
π = 16x₁ − x₁² + 28x₂ − 3x₂² − 3 − 4x₁ − 4x₂ = 12x₁ − x₁² + 24x₂ − 3x₂² − 3.
First-order conditions (maxima of a function of two variables without constraints)
∂π/∂x₁ = 12 − 2x₁ = 0 ⟹ x₁ = 6 ∂π/∂x₂ = 24 − 6x₂ = 0 ⟹ x₂ = 4
(Equivalently MR₁ = 16 − 2x₁ = 4 = MC and MR₂ = 28 − 6x₂ = 4 = MC.)
Second-order conditions. π₁₁ = −2 < 0, π₂₂ = −6 < 0, π₁₂ = 0, so π₁₁π₂₂ − π₁₂² = 12 > 0: a maximum.
| Market 1 | Market 2 | Total | |
|---|---|---|---|
| Quantity | 6 | 4 | 10 |
| Price (₹) | 16 − 6 = 10 | 28 − 12 = 16 | |
| Revenue (₹) | 60 | 64 | 124 |
Total cost = 3 + 4 × 10 = ₹43; maximum profit = 124 − 43 = ₹81.
He should sell 6 units at ₹10 in market 1 and 4 units at ₹16 in market 2.
Check (elasticities at the optimum): η₁ = (dx₁/dp₁)·(−p₁/x₁) = 10/6 = 1.67 and η₂ = (1/3)(16/4) = 1.33. The higher price is charged in the market where demand is less elastic, as discrimination requires (p(1 − 1/η) = MC = 4 in both: 10(1 − 0.6) = 4 and 16(1 − 0.75) = 4).
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Profit function | 2 |
| 2 | First-order conditions | 3 |
| 3 | Second-order conditions | 2 |
| 4 | Prices, quantities (and profit) | 2 |
| Total | 9 |
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