Pricing a Student T-Shirt Run
A student entrepreneurship group is producing T-shirts for a charity run. Fixed costs are $850 for screens and setup; each shirt costs $4.25 in materials and printing. A survey suggests that at $15 per shirt they can sell 120 shirts, and that for each $1 increase in price, sales drop by 8 shirts. The school requires the price to be between $15 and $30 inclusive. The group will print only the number they expect to sell, and storage limits them to 240 shirts. They expect no leftover inventory and no other costs. Before presenting to the teacher, they need a profit model and a defensible pricing recommendation. They are also aiming to donate at least $500 from profit. The teacher asks them to justify any price they recommend and to say whether the $500 target is realistic under this survey model.
1. Write an expression for the expected number of shirts sold, q, in terms of price p. Then write the profit P(p) and state the restricted domain.
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Let p be the price in dollars. The demand model is q = 120 - 8(p - 15) = 240 - 8p. Since price is restricted to 15 ≤ p ≤ 30 and q ≥ 0, this gives 15 ≤ p ≤ 30. Revenue: R(p)=p(240-8p)=240p-8p^2. Cost: C(p)=850+4.25(240-8p)=1870-34p. Therefore P(p)=R-C=-8p^2+274p-1870, with domain 15 ≤ p ≤ 30.
2. Determine the price that maximizes profit under this model, and calculate the maximum profit. Does the group reach its $500 target? Justify.
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The vertex of P(p) is at p=-274/(2×-8)=17.125. This lies in the domain. P(17.125)=-8(17.125)^2+274(17.125)-1870=476.125. So maximum profit is about $476.13 at $17.13 per shirt. Check endpoints: P(15)=440 and P(30)=-850, so the vertex is the maximum on the domain. The $500 target is not reached; the model predicts a shortfall of about $23.88. The group should not claim $500 without changing costs, prices outside the survey range, or adding sales.
3. The group wants profit of at least $400. Solve and interpret the interval of prices that satisfies this under the model.
Read the worked answer
Solve P(p) ≥ 400: -8p^2+274p-1870 ≥ 400, so 8p^2-274p+2270 ≤ 0. The roots are p=(274 ± sqrt(274^2-4·8·2270))/(16)=(274 ± sqrt2436)/16, giving p≈14.04 and p≈20.21. Because the quadratic opens upward, the inequality holds between the roots. Intersecting with 15 ≤ p ≤ 30 gives 15 ≤ p ≤ 20.21. So any price from $15.00 to about $20.21 yields at least $400 profit under the model; the lower endpoint is set by the survey range, not by the inequality.