Sweatshirt Fundraiser Break-Even and Safety Margin
The student council at École secondaire des Rivières is ordering hooded sweatshirts for a spring fundraiser. The supplier charges a fixed setup fee of $180 plus $9.25 per sweatshirt, and orders must be placed in multiples of 12. The council has $500 in its activity fund that it can use upfront, but the treasurer insists the fund must be back to at least $900 after the fundraiser to cover the end-of-year barbecue. The sweatshirts will sell for $16 each. Based on last year's sales, the council expects to sell at most 140. A local business has offered to buy any unsold sweatshirts for $5 each, but only if the council orders at least 120. The council wants to know the smallest order that keeps the fund safe even if only 75% of the sweatshirts sell at full price, with the rest sold to the business.
1. Model the fund balance after the fundraiser as a function of the number n of sweatshirts ordered, assuming 75% sell at full price and the rest go to the local business. State the restrictions on n and determine the smallest allowable order.
Read the worked answer
Let n be the number ordered. Because orders are in multiples of 12, n=12k. If n≥120, the local business buys the unsold sweatshirts. At 75% full-price sales, 0.75n sell for $16 and 0.25n sell for $5. Revenue = 16(0.75n)+5(0.25n)=12n+1.25n=13.25n. Cost = 180+9.25n. Final fund = 500 − (180+9.25n) + 13.25n = 320+4n. Require 320+4n ≥ 900 ⇒ 4n ≥ 580 ⇒ n ≥ 145. The smallest multiple of 12 that is at least 145 is 156 (144 is too small). Check: n=156 gives 320+624=$944, and 75% of 156 =117 full-price sales, within the expected maximum of 140. If n<120, the business will not buy leftovers, so revenue = 16(0.75n)=12n and final fund = 500−(180+9.25n)+12n = 320+2.75n. Then 320+2.75n≥900 ⇒ n≥210.9, impossible for n<120. Therefore the smallest allowable order is 156 sweatshirts.
2. If the council orders 156 sweatshirts, what minimum number must sell at full price to reach the $900 target? Express the answer as a fraction and a percent, and check it against the target.
Read the worked answer
With n=156, cost = 180+9.25(156)=180+1443=1623. Let s be the number sold at $16; the remaining 156−s sell for $5. Revenue = 16s+5(156−s)=780+11s. Final fund = 500−1623+780+11s = −343+11s. Set −343+11s ≥900 ⇒ 11s ≥1243 ⇒ s ≥113. Since s is a whole number, at least 113 sweatshirts must sell at full price. 113/156 ≈0.724, so about 72.4%, not 75%. Check: s=113 gives revenue=16(113)+5(43)=1808+215=2023; final fund=500−1623+2023=900.
3. The treasurer proposes ordering 144 instead because it is also a multiple of 12 and closer to expected sales. Using the 75% assumption, show whether this meets the target and explain the risk.
Read the worked answer
At n=144, cost=180+9.25(144)=180+1332=1512. At the 75% assumption, full-price sales =108 and leftovers=36, so revenue=16(108)+5(36)=1728+180=1908. Final fund=500−1512+1908=896, which is $4 short of $900. To reach $900 with n=144, solve 500−1512+16s+5(144−s)≥900. This simplifies to −292+11s≥900 ⇒ 11s≥1192 ⇒ s≥108.36, so 109 full-price sales are needed. That is 109/144≈75.7%, slightly above the 75% assumption. The 144 order is riskier: a small drop below expectation misses the target, while 156 gives a $44 cushion.