Poster Pricing with a Restricted Domain
A small studio sells hand-printed posters. Market testing suggests that if the price is p dollars per poster, the number sold each week is q = 120 − 4p. The studio's weekly fixed cost is $150, and each poster costs $8 to print. The owner currently charges $18, sells 48 posters, and makes a modest profit. A friend solves the equation P(p)=190 and finds two prices, $13 and $25, then argues that either price is equally good because both give the same profit. The owner knows that the online shop contract requires a listed price of at least $15, and that the studio cannot produce more than 90 posters per week because of drying space. The owner wants to choose a price that maximises weekly profit and to understand why one of the friend's two prices is not actually usable under these constraints.
1. Show that the weekly profit can be written as P(p) = −4p^2 + 152p − 1110, and state the restricted domain for p from the online shop contract and the production limit.
Read the worked answer
Revenue is price × quantity: R(p) = p(120−4p) = 120p − 4p^2.
Cost is fixed cost plus printing cost: C(p) = 150 + 8(120−4p) = 150 + 960 − 32p = 1110 − 32p.
Profit is R−C:
P(p) = 120p − 4p^2 − (1110 − 32p) = −4p^2 + 152p − 1110.
The contract requires p ≥ 15. The production limit gives q = 120−4p ≤ 90, so 120−90 ≤ 4p, or p ≥ 7.5. Since p ≥ 15 is stronger, the production upper limit does not bind here. Also q ≥ 0 gives p ≤ 30. Thus the restricted domain is 15 ≤ p ≤ 30.
2. Find the price that maximises weekly profit, the maximum profit, and the number of posters sold. Check that your answer satisfies the constraints.
Read the worked answer
The profit is a downward-opening quadratic, so the maximum occurs at the vertex:
p = −b/(2a) = −152/(2×−4) = 19.
At p = 19, q = 120−4(19) = 44 posters.
P(19) = −4(19)^2 + 152(19) − 1110 = −1444 + 2888 − 1110 = 334.
So the maximum weekly profit is $334 at a price of $19, selling 44 posters. This satisfies 15 ≤ 19 ≤ 30 and 44 ≤ 90.
3. The friend claims p = 13 and p = 25 are equally good because P(13) = P(25) = 190. Evaluate this claim under the owner's constraints. Include the number of posters sold at each price and explain which price is usable.
Read the worked answer
The algebra is correct: both prices give profit $190. At p = 13, q = 120−4(13) = 68 posters. At p = 25, q = 120−4(25) = 20 posters. However, p = 13 violates the online shop contract requiring p ≥ 15, so it is not a feasible choice. The price p = 25 is feasible and gives the same $190 profit, but it is not the profit-maximising price. The friend's error is solving the profit equation without imposing the restricted domain. Under the constraints, only p = 25 is usable, and the owner should prefer p = 19 for maximum profit.