Tune-Up Pricing and the Profit Trap
A bicycle repair shop charges $30 per tune-up and currently completes about 15 tune-ups per week. The owner models weekly demand as q = 60 − 1.5p, where p is the price in dollars and q is the number of tune-ups, for 20 ≤ p ≤ 40. Variable cost is $12 per tune-up for parts and labour, and fixed weekly costs are $280 for rent, insurance, and utilities. The owner is considering a price increase and wants to know whether the shop is actually profitable at $30, where profit is greatest, and whether revenue can be used as a shortcut for profit. A neighbouring shop charges $28 and is often busy. The owner has reliable sales records only for prices between $24 and $32; the linear demand model was fitted to those records and then extended to $20–$40 without further checking.
1. Find the break-even price or prices implied by the model, and state whether the current $30 price is profitable.
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Profit (\Pi=(p-12)q-280), with (q=60-1.5p). Expand: ((p-12)(60-1.5p)=60p-1.5p^2-720+18p=-1.5p^2+78p-720). So (\Pi=-1.5p^2+78p-1000). Set to zero and multiply by -1: (1.5p^2-78p+1000=0), or (p^2-52p+666.67=0). Discriminant (52^2-4(666.67)=2704-2666.67=37.33), root (\sqrt{37.33}=6.11); (p=(52\pm6.11)/2=22.94) or 29.06. Thus profit is positive only for prices roughly between $22.95 and $29.06. At the current $30, profit is ((30-12)(15)-280=270-280=-$10), a loss. Break-even at the upper root is close, which matters for the owner's plan.
2. Compare the revenue-maximizing price with the profit-maximizing price. What does the difference imply about using revenue as a proxy for profit?
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Revenue (R=pq=60p-1.5p^2). This concave quadratic has vertex at (p=60/(2\times1.5)=20), which is the lower endpoint of the model's domain; revenue is $600 at p=20. Profit at p=20 is ((8)(30)-280=240-280=-$40). Profit is maximized where marginal revenue equals marginal cost, or by differentiating (\Pi=-1.5p^2+78p-1000): (\Pi'=-3p+78=0), so p=26. At p=26, q=21, profit=((14)(21)-280=294-280=$14), and revenue is $546, less than $600. Revenue maximization ignores the $12 variable cost, so it selects a price that sells more units at a loss. Revenue is not a safe proxy for profit here.
3. The owner proposes charging $40. Evaluate this using the model and explain why the model's answer should not be accepted without qualification.
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At p=40 the model gives q=60−60=0, revenue 0, profit −$280. The arithmetic is correct but the conclusion “demand is zero at $40” should not be accepted directly. The model was fitted to data from $24 to $32 and extended beyond that range. A linear demand curve cannot be trusted at an endpoint where it predicts zero; demand may be small but nonzero, or the shop may lose all customers to the $28 competitor long before $40. Also q must be nonnegative, so the formula is invalid above p=40 unless truncated. The owner should treat $40 as outside the model's evidence and test a smaller increase, such as $31–$33, before making a decision.