Where should the candle unit stop producing? Profit maximum vs minimum average cost
A Kolhapur candle unit sells Diwali gift packs at a price dictated by demand: for x packs sold in a month the price obtainable is p = 400 − 2x rupees per pack. The monthly cost of making and packing x packs is C(x) = 2,000 + 40x + x² rupees, where the ₹2,000 covers shed rent and supervision and is incurred whether or not anything is produced. The owner currently makes 50 packs a month, the most her two moulding tables can handle, and she is content because 'at 50 packs my average cost is close to its lowest'. A consultant has told her that profit would rise if she could produce 60 packs, and that this is proved by marginal revenue equalling marginal cost at 60. The owner wants both claims checked, and to know how much a third moulding table would actually be worth, before she commits to buying one.
1. Determine the profit-maximising output and the maximum monthly profit.
Read the worked answer
Revenue: R(x) = p·x = (400 − 2x)x = 400x − 2x².
Profit: P(x) = R(x) − C(x) = (400x − 2x²) − (2,000 + 40x + x²) = −3x² + 360x − 2,000.
First-order condition: P′(x) = −6x + 360 = 0 → x = 60 packs.
Second-order condition: P″(x) = −6 < 0, so x = 60 is a maximum, not a minimum or a point of inflexion.
Maximum profit: P(60) = −3(3,600) + 21,600 − 2,000 = −10,800 + 21,600 − 2,000 = ₹8,800 per month.
Check by first principles: at x = 60 the price is 400 − 120 = ₹280, revenue = ₹16,800, cost = 2,000 + 2,400 + 3,600 = ₹8,000, profit = ₹8,800. Consistent.
Note the ₹2,000 fixed cost does not affect the optimal output (its derivative is zero) but it does reduce the profit figure; a candidate who maximises revenue or ignores the fixed cost will get the wrong answer to the second part.
2. Test the consultant's claim that marginal revenue equals marginal cost at 60, and evaluate the owner's present position of 50 packs.
Read the worked answer
MR = dR/dx = 400 − 4x; MC = dC/dx = 40 + 2x.
At x = 60: MR = 400 − 240 = 160 and MC = 40 + 120 = 160. They are equal, so the consultant's arithmetic is right and his conclusion happens to be correct. But MR = MC is only a necessary condition - it also holds at a profit minimum. The second-order condition (slope of MR, −4, must be less than slope of MC, +2, i.e. MR falling faster than MC) is what confirms a maximum, and it does here: −4 < 2.
Current position, x = 50: price = 400 − 100 = ₹300; revenue = ₹15,000; cost = 2,000 + 2,000 + 2,500 = ₹6,500; profit = ₹8,500. At the margin, MR(50) = 200 and MC(50) = 140, so the 51st pack adds about ₹60 of profit; the owner is leaving money on the table, and the move from 50 to 60 packs is worth ₹8,800 − ₹8,500 = ₹300 a month - a small gain that will not by itself justify a third table.
3. Check the owner's belief that producing near the level of minimum average cost is the right target, and explain the conceptual error if there is one.
Read the worked answer
AC(x) = C(x)/x = 2,000/x + 40 + x.
d(AC)/dx = −2,000/x² + 1 = 0 → x² = 2,000 → x ≈ 44.7 packs. AC″ = 4,000/x³ > 0, so this is a minimum: minimum AC ≈ 2,000/44.72 + 40 + 44.72 ≈ ₹129.44 per pack.
So the owner is also mistaken on her own terms: minimum average cost occurs near 45 packs, not 50.
More important is the conceptual error. Minimising average cost means maximising the profit margin per pack, not total profit. Profit is x(p − AC), and pushing output beyond the AC-minimising point lowers the margin but adds volume; because price falls with output, the two objectives diverge. At x ≈ 44.7, price = 400 − 89.4 = ₹310.6 and profit = 44.72 × (310.6 − 129.4) ≈ ₹8,100 - about ₹700 less than at 60 packs. 'Efficient' production and 'profitable' production are different questions, and the exam usually rewards the candidate who states clearly which one is being asked.