Hotel Costa: revenue rate, peak season and a proposed extension
Hotel Costa models its daily revenue rate during a 40-day summer season. Let t be the day, with 0 ≤ t ≤ 40, and let R(t) be revenue in euros per day. For the first 20 days the model is R(t) = -3t^2 + 120t + 300. From day 20 onward a large festival ends and the rate follows R(t) = 1500 - 50(t - 20). The manager wants to know whether the change on day 20 is smooth, when daily revenue peaks, and how much total revenue the season generates. The hotel's daily operating cost is €300. A tour operator proposes extending the season by 10 more days, assuming the same linear decline continues after day 40. No other costs are modelled.
1. Study the continuity and differentiability of R(t) at t = 20. Interpret the result in terms of daily revenue and its rate of change.
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At t = 20 the first branch gives R(20) = -3(400) + 120(20) + 300 = -1200 + 2400 + 300 = 1500. The second branch gives 1500 - 50(0) = 1500, so R(t) is continuous at t = 20: there is no jump in daily revenue. For t < 20, R'(t) = -6t + 120, so the left derivative at t = 20 is 0. For t > 20, R'(t) = -50. Since 0 ≠ -50, R(t) is not differentiable at t = 20. The graph has a kink: just before day 20 revenue is momentarily flat, and just after day 20 it falls at €50 per day per day. Because there is no single derivative at t = 20, the maximum must be checked by comparing the two branches and the endpoint, not by solving R'(t) = 0 at the joining point.
2. Find the day of maximum daily revenue and the maximum daily revenue. Then calculate the total revenue generated over the 40-day season.
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On [0,20], R'(t) = -6t + 120, which is positive for t < 20 and zero at t = 20, so R increases up to R(20) = 1500. On (20,40], R(t) = 1500 - 50(t - 20) is decreasing with slope -50. Therefore the maximum daily revenue is €1,500 per day, attained on day 20. Total revenue is the area under the rate curve: ∫0^20 (-3t^2 + 120t + 300) dt + ∫20^40 [1500 - 50(t - 20)] dt. First integral: [-t^3 + 60t^2 + 300t] from 0 to 20 = -8000 + 24000 + 6000 = 22,000. Second integral, with u = t - 20: ∫0^20 (1500 - 50u) du = [1500u - 25u^2] from 0 to 20 = 30,000 - 10,000 = 20,000. Total revenue = €42,000.
3. The tour operator proposes adding 10 days after day 40, with the same linear decline. Calculate the additional revenue and compare it with the hotel's operating cost. Should the hotel accept?
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For t between 40 and 50, continue R(t) = 1500 - 50(t - 20). At t = 40, R = 1500 - 50(20) = 500 euros per day; at t = 50, R = 1500 - 50(30) = 0. The additional revenue is the area of the triangle with base 10 days and height €500 per day: ∫40^50 [1500 - 50(t - 20)] dt = (1/2)(10)(500) = €2,500. Average additional revenue is €250 per day, while daily operating cost is €300. The extension therefore loses about €50 per day, or €500 over the 10 days. Under the stated model the hotel should reject the extension unless it brings other revenue not represented by R(t).