Fee Optimisation for a Coaching Course
An IOE aspirant helps a local coaching institute set its fee for a six-month physics course. The institute currently charges NPR 8,000 per student and enrols 120 students. A small survey suggests that for every NPR 500 increase in the fee, 6 students would leave; for every NPR 500 decrease, 6 more would join. The institute's fixed cost for the course is NPR 300,000, and each enrolled student costs an additional NPR 2,000 in materials and tutoring. The owner wants to know whether raising the fee could actually increase total profit, and at what point the course would merely break even. The student is asked to model profit as a function of the number of NPR 500 changes, keeping the number of students non-negative and treating that number as an integer. The owner also insists that the fee cannot exceed NPR 15,000 because of local competition.
1. Build a profit function in terms of x, the number of NPR 500 fee increases, and determine the fee and enrolment that maximise profit. State the maximum profit.
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Let x be the number of NPR 500 increases. Fee = 8000 + 500x; students = 120 - 6x. Contribution per student = fee - 2000 = 6000 + 500x. Profit P(x) = (6000 + 500x)(120 - 6x) - 300000. Expanding gives P(x) = 420000 + 24000x - 3000x². This is a downward parabola; its vertex is at x = -24000/(2 × -3000) = 4. Fee = 8000 + 2000 = NPR 10,000; students = 120 - 24 = 96. P(4) = 420000 + 96000 - 48000 = NPR 468,000. Check neighbouring integers: P(3) = 465,000 and P(5) = 465,000, so x = 4 is indeed the integer maximum. The owner should not assume that the current NPR 8,000 fee is best: a moderate increase raises profit because the lost students are fewer than the extra contribution per remaining student.
2. Find the approximate break-even fee(s) and explain whether the owner's NPR 15,000 cap prevents the institute from reaching the profit-maximising fee. Include the integer-feasibility check.
Read the worked answer
Break-even means P(x) = 0: 420000 + 24000x - 3000x² = 0, or x² - 8x - 140 = 0. Roots are x = (8 ± √(64 + 560))/2 = (8 ± √624)/2. Since √624 ≈ 24.98, x ≈ 16.49 or x ≈ -8.49. Only non-negative enrolments make sense, so the first break-even is between x = 16 and x = 17. At x = 16, fee = NPR 16,000, students = 24, P = 36,000; at x = 17, fee = NPR 16,500, students = 18, P = -39,000. Thus a fee around NPR 16,000–16,500 is the upper break-even boundary. The NPR 15,000 cap corresponds to x = 14: P(14) = 420000 + 336000 - 588000 = NPR 168,000, still positive, so the cap does not prevent the optimum at x = 4. It does, however, rule out the upper break-even region and limits how far the fee can be raised. Integer feasibility: x must be 0–20 for non-negative students, and 0–14 under the cap; x = 4 lies well inside.