Pricing T-Shirts for a Class Fundraiser
The sophomore class sells school-spirit T-shirts to fund a field trip. Last year they charged $12 per shirt and sold 200 shirts. A student survey suggests that for each $1 increase in price, they will sell 10 fewer shirts, and that this pattern holds for increases up to $20. The class advisor wants to raise the price but is worried about losing sales. The class needs at least $2,300 in revenue to cover the trip deposit, and they can order at most 220 shirts. Before voting, the class wants a model for revenue as a function of the price increase, the price that maximizes revenue, and an honest assessment of whether a $15 price is good enough or whether they should go higher.
1. Define a variable for the price increase, write a revenue function, and state a reasonable domain with justification.
Read the worked answer
Let x be the price increase in dollars. Then price = 12 + x and shirts sold = 200 - 10x. Revenue is price times quantity:
R(x) = (12 + x)(200 - 10x) = 2400 - 120x + 200x - 10x^2 = -10x^2 + 80x + 2400.
Domain: x >= 0 because the survey describes increases, not discounts. Also 200 - 10x >= 0, so x <= 20. The survey itself only supports increases up to $20, so 0 <= x <= 20 is the defensible domain. At x = 20 the price is $32 and sales fall to 0, giving $0 revenue, so the upper endpoint is a boundary rather than a useful price. The 220-shirt order limit is not binding under this model because sales never exceed 200.
2. Find the price increase that maximizes revenue and the corresponding price, quantity, and revenue. Show why the vertex gives a maximum rather than a minimum.
Read the worked answer
For R(x) = -10x^2 + 80x + 2400, the vertex occurs at x = -b/(2a) = -80/(2(-10)) = -80/(-20) = 4.
So the best increase is $4, giving price 12 + 4 = $16. Quantity is 200 - 10(4) = 160 shirts. Revenue is 16(160) = $2,560.
Check with the expanded form: R(4) = -10(16) + 80(4) + 2400 = -160 + 320 + 2400 = 2560.
The graph is a parabola opening downward because a = -10 < 0, so the vertex is a maximum. Endpoint checks confirm it: R(0) = 2400 and R(20) = 0, both below 2560. The maximum is therefore $2,560 at a $16 price.
3. A treasurer proposes $20 per shirt, arguing that a higher price must bring in more money. Compare $15, $16, and $20, and recommend a price using revenue evidence.
Read the worked answer
Compute each option from the model.
$15: x = 3, shirts = 200 - 30 = 170, revenue = 15(170) = $2,550. $16: x = 4, shirts = 160, revenue = 16(160) = $2,560. $20: x = 8, shirts = 200 - 80 = 120, revenue = 20(120) = $2,400.
All three clear the $2,300 deposit, but they are not equally good. The $15 price misses the maximum by only $10. The $20 price brings in $160 less than the maximum even though each shirt costs $4 more, because 40 fewer shirts are sold. From $16 to $20, the price rises by $4 but quantity falls by 40, and the lost sales outweigh the higher price.
Recommendation: $16 maximizes revenue at $2,560. If the class wants a safer, lower price for students, $15 is acceptable because it still clears the goal and earns $2,550. The $20 proposal is the weakest of the three despite sounding like the highest-priced option.