Question 3 OR (b)

Profit function from marginal revenue and marginal cost

9 marks

Time to write: About 18 minutes
Topic: Integration: elasticities and surplus

Question

Question 3 OR (b) · 9 marks

The marginal revenue for a commodity is MR = 60 - 4 x. The marginal cost is MC = 2 x + 12, and the cost of producing 50 units is ₹3,200.

Find:

The profit function.
The profit or loss on selling 100 units.
The profit-maximizing output.
The maximum 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

18 minfor 9 marks

Read from the question

FindFormula 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 Revenue and cost functions constants found 3 marks 5 min
2 Profit function 1 mark 2 min
3 Profit or loss at 100 units 2 marks 4 min
4 Profit-maximising output 2 marks 3 min
5 Maximum profit 1 mark 2 min
Check: arithmetic, units (₹, %), final answer underlined 1 min

12345

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.

Total revenue. R = ∫MR dx = ∫(60 − 4x) dx = 60x − 2x² + k₁. Revenue is zero when nothing is sold, so k₁ = 0 and R = 60x − 2x².

Total cost. C = ∫MC dx = ∫(2x + 12) dx = x² + 12x + k, where k is the fixed cost. The cost of 50 units is ₹3,200: 2,500 + 600 + k = 3,200 ⟹ k = 100, so C = x² + 12x + 100.

(i) Profit function

π = R − C = (60x − 2x²) − (x² + 12x + 100) = 48x − 3x² − 100.

(ii) Profit or loss on selling 100 units

π(100) = 4,800 − 30,000 − 100 = −₹25,300, a loss of ₹25,300.

At x = 100	₹
Total revenue 60(100) − 2(100)²	−14,000
Total cost (100)² + 12(100) + 100	11,300
Profit (loss)	(25,300)

(The revenue function turns negative beyond 30 units because MR is negative beyond 15 units; the question's figures are used as given.)

(iii) Profit-maximising output

dπ/dx = 48 − 6x = 0 ⟹ x = 8 units (equivalently MR = MC: 60 − 32 = 28 = 16 + 12). d²π/dx² = −6 < 0, a maximum.

(iv) Maximum profit

π(8) = 48(8) − 3(64) − 100 = 384 − 192 − 100 = ₹92 (R = ₹352, C = ₹260).

Where the 9 marks sit

#	Component	Marks
1	Revenue and cost functions (constants found)	3
2	Profit function	1
3	Profit or loss at 100 units	2
4	Profit-maximising output	2
5	Maximum profit	1
	Total	9

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