Pricing Cookie Boxes: Revenue vs. Profit
A student club runs a small bakery and sells custom cookie boxes. Market research suggests that if the club charges p dollars per box, it can sell q = 200 − 4p boxes per week, where 12 ≤ p ≤ 50. Ingredients and packaging cost $12 per box, and the club pays $300 per week to rent a commercial kitchen. The faculty advisor argues that because revenue is price times quantity, the club should simply charge $50 and 'maximize the price.' A student treasurer replies that profit, not revenue, is the goal and that the advisor's plan would sell zero boxes. The club wants to set a single price for the fall semester and needs to know both the break-even prices and the profit-maximizing price before presenting a budget. The club has no other revenue and cannot change the recipe or the per-box cost in the short run.
1. Write the weekly profit function P(p) in terms of the price p, and find the price that maximizes weekly profit. Show the vertex calculation and state the maximum weekly profit.
Read the worked answer
Revenue is R(p) = p(200 − 4p). The variable cost is 12(200 − 4p), and the fixed cost is 300, so
P(p) = p(200 − 4p) − 12(200 − 4p) − 300 = (p − 12)(200 − 4p) − 300.
Expanding: P(p) = 200p − 4p² − 2400 + 48p − 300 = −4p² + 248p − 2700. This is a downward-opening parabola, so the maximum occurs at p = −b/(2a) = −248/(2·−4) = 31. Then P(31) = −4(31²) + 248(31) − 2700 = −3844 + 7688 − 2700 = 1144. The club should charge $31 per box for a maximum weekly profit of $1,144, provided 31 is in the demand domain (it is, since 12 ≤ 31 ≤ 50).
2. Find the break-even prices, and explain what they mean in the context of the club's price range. Round to the nearest cent and identify the integer prices that yield a positive profit.
Read the worked answer
Set P(p) = 0: −4p² + 248p − 2700 = 0, or p² − 62p + 675 = 0 after dividing by −4. The quadratic formula gives
p = (62 ± √(62² − 4·675))/2 = (62 ± √(3844 − 2700))/2 = (62 ± √1144)/2.
Since √1144 ≈ 33.82, p ≈ (62 − 33.82)/2 ≈ 14.09 and p ≈ (62 + 33.82)/2 ≈ 47.91. Profit is positive between these prices because the parabola opens downward. Within the stated domain 12 ≤ p ≤ 50, the club must charge more than about $14.09 and less than about $47.91. For whole-dollar prices, $15 through $47 yield positive weekly profit. At $50, demand is zero and profit is −$300, the fixed kitchen cost.
3. The advisor notes that revenue is maximized at $25 per box and proposes charging $25. Compare revenue and profit at $25 and at the profit-maximizing price, and explain why the two prices differ.
Read the worked answer
Revenue is R(p) = p(200 − 4p) = −4p² + 200p, whose vertex is p = −200/(2·−4) = 25. At p = 25, R(25) = 25(200 − 100) = 2500. Profit at p = 25 is P(25) = (25 − 12)(100) − 300 = 1300 − 300 = 1000. At p = 31, revenue is R(31) = 31(200 − 124) = 31·76 = 2356, which is less than 2500, but profit is 1144, which is greater than 1000. The difference is the $12 per-box cost: raising the price from 25 to 31 loses some sales but keeps more dollars per box, and the lost sales reduce variable cost. Revenue maximization ignores the cost side, so it is not the same as profit maximization.