Marginal Revenue at a Market Stall
A student-run stall sells hand-painted mugs. Weekly demand is modelled as q = 240 − 6p, where p is the price in pounds and q is mugs sold, valid for 5 ≤ p ≤ 40. Each mug costs 4 to make, plus a fixed weekly stall fee of 90. The treasurer wants to raise the price from 15 to 18 to 'increase revenue per mug'. A second member argues that raising price will reduce total profit because fewer mugs sell. The group needs a rate-of-change argument, not just arithmetic, to settle the dispute before printing new price labels.
1. Express weekly profit as a function of price and use a derivative to find the profit-maximising price.
Read the worked answer
Revenue R(p) = p·q = p(240 − 6p) = 240p − 6p^2. Variable cost = 4q = 4(240 − 6p) = 960 − 24p. Profit P(p) = R − variable cost − 90 = 240p − 6p^2 − 960 + 24p − 90 = −6p^2 + 264p − 1050. Differentiate: P'(p) = −12p + 264. Set P' = 0 gives p = 22. Second derivative P'' = −12 < 0, so p = 22 is a maximum. Check endpoints: P(5) = −150 + 1320 − 1050 = 120; P(22) = −2904 + 5808 − 1050 = 1854; P(40) = −9600 + 10560 − 1050 = −90. So the optimal price is 22, giving profit 1854 per week. Both 15 and 18 are below the optimum, so raising price from 15 to 18 does increase profit, but the treasurer should go further, to 22.
2. Interpret P'(15) as a marginal rate and use it to estimate the profit change from raising price from 15 to 16. Compare with the exact change.
Read the worked answer
P'(p) = −12p + 264, so P'(15) = −180 + 264 = 84. Interpretation: at a price of 15, each additional pound of price raises weekly profit by roughly 84, i.e. the marginal profit with respect to price is positive and large. Linear estimate for a 1 increase: ΔP ≈ 84·1 = 84. Exact: P(15) = −1350 + 3960 − 1050 = 1560. P(16) = −1536 + 4224 − 1050 = 1638. Exact change = 78. The estimate (84) overshoots the exact value (78) because P is concave (P'' = −12), so the tangent line lies above the curve. This is a useful check: linearisation over-estimates gains when the function is concave, and the error grows with the step size.