Question 2 OR (b)
Monopolist: elasticity of demand, profit-maximising output and the AC-TC elasticity identity
9 marks
- Time to write
- About 18 minutes
Question
Question 2 OR (b) · 9 marksA monopolist faces the following demand and average cost functions:
p = 58 - x/4 and = 0.5 x + 10 + 400/X, where p is the price and x is the quantity.
Find the elasticity of demand when p = 40.
Determine the output level at which profit will be maximum.
At the output level found in part (ii), prove that the elasticity of average cost equals the elasticity of total cost minus one.
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
- Find…Determine…Two asks in one question: answer both.
- prove thatThe result is given: the marks are for the steps that reach it.
- FindFormula first, then the working, then the answer with its unit (₹, %, units).
What the paper leaves open
As printed, the average cost function is written without its symbol ("= 0.5 x + 10 + 400/X"), and its last term has a capital X; both are kept as printed and read as the average cost function and x.
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Elasticity of demand at p = 40 2 marks 4 min
- 2 Profit function and maximising output 4 marks 7 min
- 3 Elasticity identity at that output 3 marks 5 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.
Demand p = 58 − x/4, so x = 4(58 − p) = 232 − 4p. Average cost (the function printed without its symbol, with X read as x): AC = 0.5x + 10 + 400/x, so total cost TC = x · AC = 0.5x² + 10x + 400.
(i) Elasticity of demand at p = 40
At p = 40, x = 232 − 160 = 72. dx/dp = −4.
η = −(dx/dp)(p/x) = 4 × 40/72 = 160/72 = 2.22 (= 20/9). Demand is elastic at this price (η > 1).
TR = px = 58x − x²/4; profit π = TR − TC = 58x − 0.25x² − 0.5x² − 10x − 400 = 48x − 0.75x² − 400.
dπ/dx = 48 − 1.5x = 0 ⟹ x = 32; d²π/dx² = −1.5 < 0, so profit is a maximum there.
Price p = 58 − 32/4 = ₹50; TR = ₹1,600; TC = 0.5(1,024) + 320 + 400 = ₹1,232; maximum profit = ₹368. (MR = 58 − x/2 = 42 and MC = x + 10 = 42 at x = 32.)
(iii) Elasticity of AC = elasticity of TC − 1, at x = 32
Elasticity of a cost function with respect to output: e = (x/cost) × d(cost)/dx.
At x = 32: AC = 16 + 10 + 12.5 = 38.5; d(AC)/dx = 0.5 − 400/x² = 0.5 − 400/1,024 = 0.109375.
e_AC = (32/38.5) × 0.109375 = 3.5/38.5 = 1/11 = 0.0909.
TC = 1,232; MC = d(TC)/dx = x + 10 = 42. e_TC = (32/1,232) × 42 = 1,344/1,232 = 12/11 = 1.0909.
e_TC − 1 = 12/11 − 1 = 1/11 = e_AC. Proved.
Why it holds in general: TC = x · AC, so d(TC)/dx = AC + x · d(AC)/dx (product rule). Hence e_TC = (x/(x·AC)) × [AC + x · d(AC)/dx] = 1 + (x/AC) · d(AC)/dx = 1 + e_AC.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Elasticity of demand at p = 40 | 2 |
| 2 | Profit function and maximising output | 4 |
| 3 | Elasticity identity at that output | 3 |
| Total | 9 |
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