Fee revision at a capacity-constrained coaching institute
A coaching institute in Cuttack charges a monthly fee of ₹p and estimates enrolment q = 300 − 3p, with p between ₹40 and ₹100. Its monthly cost is ₹1,200 fixed plus ₹40 per enrolled student. The classroom can seat at most 80 students, so any plan with q > 80 is infeasible. The director currently charges ₹80 and is considering a cut to ₹75 to attract more students. A local parents' association argues that because demand is elastic at ₹80, a fee cut will raise revenue and therefore profit. The accountant warns that the capacity limit and variable costs change the picture. The institute wants to know the unconstrained profit-maximising fee, the best feasible fee given the 80-seat cap, and whether the elasticity argument is sound.
1. Formulate monthly profit as a function of p and find the unconstrained profit-maximising fee. Check it against the 80-seat capacity.
Read the worked answer
Revenue is R = p q = p(300 - 3p) = 300p - 3p^2.
Cost is C = 1,200 + 40q = 1,200 + 40(300 - 3p) = 13,200 - 120p.
Profit is π(p) = R - C = -3p^2 + 420p - 13,200.
Differentiate: π'(p) = -6p + 420. Setting π' = 0 gives p = 70. Since π'' = -6 < 0, this is a maximum. At p = 70, q = 300 - 210 = 90.
But q = 90 exceeds the 80-seat capacity, so the unconstrained optimum is infeasible. Its paper profit would be π(70) = -3(4,900) + 29,400 - 13,200 = ₹1,500, but the institute cannot enrol 90 students.
2. Find the best feasible fee under the 80-seat cap, allowing fractional rupees; then check whether a whole-rupee fee of ₹74 or ₹75 is better.
Read the worked answer
The cap requires q = 300 - 3p ≤ 80, so p ≥ 220/3 ≈ ₹73.33. Since π(p) is decreasing for p > 70, the constrained maximum occurs at the boundary p = 220/3, q = 80.
At p = 220/3, revenue is (220/3)(80) = 17,600/3 ≈ ₹5,866.67. Cost is 1,200 + 40(80) = ₹4,400. Profit is about ₹1,466.67.
For whole-rupee fees:
- p = 74 gives q = 78. Revenue = 74(78) = ₹5,772. Cost = 1,200 + 3,120 = ₹4,320. Profit = ₹1,452.
- p = 75 gives q = 75. Revenue = ₹5,625. Cost = 1,200 + 3,000 = ₹4,200. Profit = ₹1,425.
So ₹74 is better than ₹75 and is very close to the fractional optimum.
3. At ₹80, q = 60. The manager says demand is elastic, so cutting the fee to ₹75 must raise revenue and therefore profit. Evaluate the logic using computed values.
Read the worked answer
At p = 80, elasticity is (dq/dp)(p/q) = -3(80/60) = -4, so demand is elastic. A fee cut does raise revenue: R(80) = 80(60) = ₹4,800, while R(75) = 75(75) = ₹5,625.
Profit also rises: π(80) = 4,800 - (1,200 + 2,400) = ₹1,200; π(75) = 5,625 - (1,200 + 3,000) = ₹1,425.
But the 'therefore' is not generally valid. Profit equals revenue minus cost, and the extra 15 students add ₹600 of variable cost. Here revenue rises by ₹825 but profit rises by only ₹225. Elasticity tells us the direction of revenue, not of profit. The capacity limit and the ₹40 marginal cost must also be checked before recommending a fee cut.