Numeracy: The Fundraising Stall

Let c be the number of cupcakes and l be the number of cups of lemonade. The cost equation is 0.80c + 0.30l = 60. To find one combination, set c = 30: cost of cupcakes = 30 × 0.80 = $24. Remaining budget = 60 − 24 = $36. Number of lemonades = 36 ÷ 0.30 = 120. But the expected maximum for lemonade is 80, so 120 exceeds the expected sales. That combination is possible to make but may not sell. Another combination: set c = 40: cost = 40 × 0.80 = $32. Remaining = $28. Lemonades = 28 ÷ 0.30 ≈ 93.33, so they could make 93 lemonades (cost $27.90) with $0.10 left, but that also exceeds 80. To stay within expected sales, try c = 20: cost = $16. Remaining = $44. Lemonades = 44 ÷ 0.30 ≈ 146.67, too many. Try c = 40 and l = 80: cost = 32 + 24 = $56, which is under budget. So one possible combination within budget and expected sales is 40 cupcakes and 80 lemonades, costing $56. Another: c = 30, l = 80: cost = 24 + 24 = $48. So 30 cupcakes and 80 lemonades is another. These are two valid combinations that use at most $60 and do not exceed expected sales. The question asks for combinations that spend the entire $60, but if we restrict to expected sales, we cannot spend the entire $60 because the maximum expected sales cost $56. So the answer should note that spending the entire $60 would require making more than expected sales. Therefore, two possible combinations within expected sales are (40 cupcakes, 80 lemonades) and (30 cupcakes, 80 lemonades). If the question insists on spending exactly $60, then one combination is 30 cupcakes and 120 lemonades (cost $60), and another is 0 cupcakes and 200 lemonades (cost $60), but these exceed expected sales. The student should explain the constraint.

Revenue from cupcakes: 40 × $2.00 = $80. Revenue from lemonade: 80 × $1.00 = $80. Total revenue = $160. Cost of cupcakes: 40 × $0.80 = $32. Cost of lemonade: 80 × $0.30 = $24. Total cost = $56. Profit = revenue − cost = $160 − $56 = $104. So the profit is $104. This is within the $60 budget because the cost is $56, leaving $4 unspent. The class could use the $4 to buy more ingredients, but they expect to sell at most 80 lemonades, so making more may not be wise. They could reduce the number of cupcakes to spend less, but that would likely reduce profit. The maximum profit within expected sales might be at 40 cupcakes and 80 lemonades, but we should check other combinations.

Profit per cupcake = selling price − cost = $2.00 − $0.80 = $1.20. Profit per cup of lemonade = $1.00 − $0.30 = $0.70. So each cupcake gives more profit than each lemonade. However, the expected sales limits are 40 cupcakes and 80 lemonades. If they make the maximum of both, profit = 40 × 1.20 + 80 × 0.70 = 48 + 56 = $104. If they make fewer cupcakes, say 30, and still 80 lemonades, profit = 30 × 1.20 + 80 × 0.70 = 36 + 56 = $92, which is less. If they make 40 cupcakes and 70 lemonades, profit = 48 + 49 = $97, also less. So within the expected sales limits, making the maximum of both gives the highest profit. But if they had to choose due to budget, cupcakes give more profit per item, so they should prioritise cupcakes up to the limit of 40, then lemonades. However, the budget is not binding at the maximum expected sales because cost is $56 < $60. So they should make 40 cupcakes and 80 lemonades. The answer should note that profit per item alone does not determine the best mix; the sales limits and budget also matter.