Pricing with a Production Cap
A student club is selling commemorative T-shirts to fund a trip. The printer charges a fixed setup fee of 30,000 yen plus 800 yen for each shirt produced. A survey suggests that if the shirts are priced at 2,000 yen each, the club can sell 120, but every 100-yen increase in price reduces sales by 5 shirts. The supplier can produce at most 80 shirts for this order, and the club has decided the price must be between 1,500 yen and 3,000 yen inclusive, in multiples of 100 yen. Let x be the number of 100-yen increases above 2,000 yen. The club wants to choose a price that maximizes profit, not revenue, and must respect the production limit. Some members argue that the algebraic vertex is automatically the best choice; others point out that the limited supply may change the answer. The treasurer also asks what happens if the supplier can provide only 70 shirts.
1. Write the profit as a function of x, state the feasible range of x, and identify the unconstrained vertex.
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Let x be the number of 100-yen increases. Then price p = 2000 + 100x and quantity q = 120 - 5x. Profit is (p - 800)q - 30000 = (1200 + 100x)(120 - 5x) - 30000 = 114000 + 6000x - 500x². The price restriction gives -5 ≤ x ≤ 10. The supply restriction q ≤ 80 gives 120 - 5x ≤ 80, so x ≥ 8. Also q ≥ 0 gives x ≤ 24, which is weaker here. Hence the feasible interval is 8 ≤ x ≤ 10. The unconstrained vertex is at x = -6000/(2·(-500)) = 6, with profit 132000 yen; this is outside the feasible interval.
2. Using the production limit, find the price that maximizes profit and the maximum profit. Explain why the vertex is not usable.
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On 8 ≤ x ≤ 10 the profit function is a concave quadratic whose vertex x=6 lies to the left of the interval, so profit decreases throughout the interval. Therefore the maximum is at x=8. Price = 2000+800 = 2800 yen, quantity = 120-40 = 80 shirts. Profit = (2800-800)·80 - 30000 = 2000·80 - 30000 = 130000 yen. The vertex price of 2600 yen would sell 90 shirts, but the supplier can provide only 80, so that plan is not feasible. This is why one must check the vertex against the production constraint rather than accepting it automatically.
3. If the supplier can provide only 70 shirts, does the optimal price change? Compute the new best profit and compare.
Read the worked answer
If only 70 shirts are available, q ≤ 70 gives 120 - 5x ≤ 70, so x ≥ 10. Together with x ≤ 10 from the price ceiling, the only feasible choice is x=10. Then price = 3000 yen, quantity = 70, profit = (3000-800)·70 - 30000 = 2200·70 - 30000 = 154000 - 30000 = 124000 yen. The optimal price rises to 3000 yen, and maximum profit falls by 6000 yen compared with the 80-shirt case. The binding constraints, not the algebraic vertex, determine the answer.