📚 Advanced Maths in Action: A Case Study | 进阶数学实战：案例分析演练
Mathematics extends far beyond textbooks and examination papers. In this case study, you will step into the role of a project manager for a school charity fair. Your mission is to design the layout, manage a budget, predict income, analyse data and make informed decisions — all by applying the advanced mathematical skills you have developed in Year 7. This real‑world challenge weaves together measurement, ratio, percentages, equations, statistics and logical reasoning, showing you how each topic becomes a powerful tool when used with confidence.
数学远远超出课本和试卷的范畴。在这个案例分析中，你将扮演学校慈善嘉年华的项目经理。你的任务是设计场地布局、管理预算、预测收入、分析数据并做出明智决策——整个过程都将运用你在七年级进阶数学中学到的技能。这个现实挑战将度量、比和比例、百分数、方程、统计和逻辑推理融为一体，让你看到每一个知识点都能在自信运用时成为强大的工具。
1. Case Background | 案例背景
The school has given you a rectangular playground measuring 40 metres by 25 metres. You need to create a charity fair layout that includes a stage (10 m × 8 m), ten stalls (each 3 m × 3 m), a refreshment zone and clear walkways. Your total budget for materials and decorations is £500. You are also expected to propose ticket prices and estimate how much money the fair can raise for the local animal shelter. All plans must be supported by precise calculations and clear reasoning.
学校给了你一块长 40 米、宽 25 米的长方形操场。你需要设计一个慈善嘉年华布局，包括一个舞台（10 米 × 8 米）、十个摊位（每个 3 米 × 3 米）、一个餐饮休息区以及畅通的通道。材料和装饰的总预算是 500 英镑。你还需要提出票价方案并估计这次嘉年华能为当地动物收容所筹集多少善款。所有计划都必须有精确的计算和清晰的推理作为支撑。
2. Task 1: Layout and Measurement | 任务一：布局与度量
Begin by calculating the total area of the playground. The area of a rectangle is length × width, so 40 × 25 = 1000 m². You must then work out the total footprint of the structures. The stage occupies 10 × 8 = 80 m². Each stall covers 3 × 3 = 9 m², and ten stalls will occupy 10 × 9 = 90 m². The combined structure area is 80 + 90 = 170 m². However, regulations require at least 35 % of the playground to remain as open walkway space. Find 35 % of 1000 m², which is 350 m². Subtracting the 170 m² of structures, the remaining space is 1000 – 170 = 830 m², well above the minimum. You can now sketch a scale plan, perhaps using a scale of 1 cm to represent 2 m, and position the stage and stalls to ensure smooth flow of visitors.
首先计算操场总面积。长方形的面积等于长乘宽，即 40 × 25 = 1000 平方米。接着需要算出所有设施的占地面积。舞台占 10 × 8 = 80 平方米。每个摊位占 3 × 3 = 9 平方米，十个摊位共占 10 × 9 = 90 平方米。设施占地总计 80 + 90 = 170 平方米。然而，根据规定，至少要有 35% 的操场保留为开放通道区域。1000 平方米的 35% 是 350 平方米。减去 170 平方米的设施占地后，剩余空间为 1000 – 170 = 830 平方米，远高于最低要求。现在你可以画一幅按比例缩放的平面图，例如用 1 厘米代表 2 米，并安排好舞台和摊位的位置，确保参观者顺畅流动。
3. Task 2: Calculating Material Costs | 任务二：计算材料成本
You decide to create bunting that runs along the perimeter of the playground. The perimeter of the rectangle is 2 × (40 + 25) = 130 metres. Each pack of bunting covers 20 metres and costs £4.50. To find how many packs are needed, divide the perimeter by the coverage per pack: 130 ÷ 20 = 6.5. Since you cannot buy half a pack, you must purchase 7 packs. The cost is 7 × 4.50 = £31.50. Next, you need tablecloths for the ten stalls. Each tablecloth covers a 3 m by 1.2 m front table and costs £6.80 per metre of length. The total tablecloth length required is 10 × 3 = 30 metres, so the cost is 30 × 6.80 = £204.00. So far, decorations and materials have reached £31.50 + £204.00 = £235.50, leaving £264.50 from the £500 budget for other items.
你决定制作围绕操场周边的彩旗装饰。长方形操场的周长是 2 × (40 + 25) = 130 米。每包彩旗可覆盖 20 米，价格为 4.50 英镑。计算需要多少包：130 ÷ 20 = 6.5。因为不能买半包，必须购买 7 包。费用为 7 × 4.50 = 31.50 英镑。接下来需要为十个摊位购买桌布。每块桌布覆盖一张 3 米长、1.2 米宽的前桌，布料按长度计价，每米 6.80 英镑。所需桌布总长度为 10 × 3 = 30 米，所以成本为 30 × 6.80 = 204.00 英镑。至此，装饰和材料费用合计 31.50 + 204.00 = 235.50 英镑，500 英镑预算中还剩下 264.50 英镑可用于其他物品。
4. Task 3: Work Allocation and Ratios | 任务三：工作分配与比和比例
You have a team of 24 volunteers. They need to be split among three teams: Set‑up, Refreshments and Games. The ratio of Set‑up to Refreshments to Games is 3 : 1 : 2. First, find the total number of parts: 3 + 1 + 2 = 6 parts. Each part is 24 ÷ 6 = 4 volunteers. Therefore, Set‑up gets 3 × 4 = 12 volunteers, Refreshments gets 1 × 4 = 4 volunteers, and Games gets 2 × 4 = 8 volunteers. Set‑up also needs to work in pairs for safety while lifting heavy items. How many pairs can the 12 volunteers form? 12 ÷ 2 = 6 pairs. This ensures all tasks are safely covered.
你有一支由 24 名志愿者组成的团队。他们需要被分配到布置组、餐饮组和游戏组这三个组中。布置组、餐饮组和游戏组的人数比是 3 : 1 : 2。首先求出总份数：3 + 1 + 2 = 6 份。每份代表 24 ÷ 6 = 4 名志愿者。因此，布置组需要 3 × 4 = 12 人，餐饮组需要 1 × 4 = 4 人，游戏组需要 2 × 4 = 8 人。出于安全考虑，布置组在搬运重物时需要两人一组。12 名志愿者能组成几对？12 ÷ 2 = 6 对。这确保了所有任务都能安全覆盖。
5. Task 4: Setting Ticket Prices and Revenue Prediction | 任务四：设定票价与收入预测
You consider two pricing schemes. Scheme A: a single entrance fee of £3.00 per person, with all activities free inside. Scheme B: free entry but each activity costs £1.50, and you expect each visitor to try on average four activities. To decide, you predict attendance. By surveying other events, you estimate that in good weather, about 350 people will attend; in poor weather, only 200. Under Scheme A, revenue = attendance × £3. Under Scheme B, revenue = attendance × (4 × £1.50) = attendance × £6. Clearly Scheme B brings in more per person, but free entry might attract more visitors. For now, calculate the expected revenue under good weather for each: Scheme A: 350 × 3 = £1050; Scheme B: 350 × 6 = £2100. For planning, you decide to use Scheme B as it gives a stronger fundraising potential, but you will also build a contingency for lower attendance.
你考虑了两种票价方案。方案 A：每人 3.00 英镑单次入场费，所有内部活动免费。方案 B：免费入场，但每个活动收费 1.50 英镑，你预计每位参观者平均会参与四项活动。为了做决定，你需要预测参与人数。通过调查其他活动，你估计天气好时大约有 350 人参加，天气差时只有 200 人。在方案 A 下，收入 = 人数 × 3 英镑。在方案 B 下，收入 = 人数 × (4 × 1.50 英镑) = 人数 × 6 英镑。显然方案 B 的人均收入更高，但免费入场可能会吸引更多游客。现在，先计算好天气下两种方案的预期收入：方案 A：350 × 3 = 1050 英镑；方案 B：350 × 6 = 2100 英镑。为保险起见，你决定采用方案 B，因为它筹款潜力更大，但你也会为参加人数较低的情况制定应急计划。
6. Task 5: Break‑Even Analysis | 任务五：盈亏平衡分析
Your total fixed costs (materials, decorations, licences) come to £480. Each visitor costs the fair £0.80 for refreshments and prizes regardless of ticket scheme. Under Scheme B, the net income per visitor is £6.00 – £0.80 = £5.20. The break‑even point is the number of visitors needed so that total net income equals fixed costs. Set up the equation: 5.20 × N = 480. Solve for N: N = 480 ÷ 5.20 = 92.307… You must round up to 93 visitors because you cannot have a fraction of a person. This means once 93 people attend, the fair begins to make a profit. What if attendance only reaches 200? Then profit = 200 × 5.20 – 480 = 1040 – 480 = £560. This analysis helps you see that even in poor‑weather scenario (200 visitors), you can still raise a significant amount.
你的固定成本（材料、装饰、许可证等）总计为 480 英镑。每位参观者还会为嘉年华带来 0.80 英镑的茶点和奖品成本，与票价方案无关。在方案 B 下，每位参观者的净收入为 6.00 – 0.80 = 5.20 英镑。盈亏平衡点就是需要多少参观者才能使总净收入等于固定成本。列出方程：5.20 × N = 480。解出 N：N = 480 ÷ 5.20 = 92.307… 必须向上取整到 93 人，因为不能有零头的人数。这意味着一旦有 93 人参加，嘉年华就开始盈利了。如果参与人数只达到 200 人会怎样？那么利润 = 200 × 5.20 – 480 = 1040 – 480 = 560 英镑。这个分析让你明白，即使在天气差的情景下（200 人），你仍然能筹集到一笔可观的善款。
7. Task 6: Data Collection and Statistics | 任务六：数据收集与统计
During the fair, you record the number of activities each visitor completes from a random sample of 50 attendees. The results are: four people did 2 activities, twelve did 3, eighteen did 4, ten did 5, five did 6, and one did 7. The mean number of activities = (4×2 + 12×3 + 18×4 + 10×5 + 5×6 + 1×7) ÷ 50 = (8 + 36 + 72 + 50 + 30 + 7) ÷ 50 = 203 ÷ 50 = 4.06. The mode is 4, as it has the highest frequency. The median lies between the 25th and 26th values: after ordering, both are 4, so median = 4. The data confirms your initial estimate of four activities per person was realistic, giving you confidence in your revenue projections.
在嘉年华进行中，你从一个 50 名参观者的随机样本中记录了每人完成的活动数量。结果如下：4 人参与了 2 项活动，12 人参与了 3 项，18 人参与了 4 项，10 人参与了 5 项，5 人参与了 6 项，1 人参与了 7 项。平均活动数量 = (4×2 + 12×3 + 18×4 + 10×5 + 5×6 + 1×7) ÷ 50 = (8 + 36 + 72 + 50 + 30 + 7) ÷ 50 = 203 ÷ 50 = 4.06。众数是 4，因为它的出现频率最高。中位数位于第 25 和 26 个值之间：排序后，这两个值都是 4，所以中位数 = 4。数据证实了你最初预估每人四项活动是符合实际的，这让你对自己的收入预测更有信心。
8. Task 7: Profit Maximisation Strategies | 任务七：利润最大化策略
With real data, you can explore ways to boost profit. Notice that if you could increase the average activities per person to 5, net income per visitor becomes 5 × 1.50 – 0.80 = £6.70. For 350 visitors, profit = 350 × 6.70 – 480 = 2345 – 480 = £1865, compared with £1340 (when average was 4). How to make this happen? You could introduce a loyalty card where five activities earn a free drink. Mathematically, you set up an inequality: profit goal ≥ £1500. Using the original £5.20 net per visitor, solve 5.20 × N – 480 ≥ 1500 → 5.20N ≥ 1980 → N ≥ 380.8, so you would need 381 visitors. If you boost net income to £6.70 by raising the activity rate, then 6.70N – 480 ≥ 1500 → 6.70N ≥ 1980 → N ≥ 295.5, requiring only 296 visitors. This demonstrates the power of small improvements in customer behaviour.
有了真实数据，你可以探索提高利润的方法。注意，如果你能将人均活动数量提高到 5，那么每位参观者的净收入就变成 5 × 1.50 – 0.80 = 6.70 英镑。350 名参观者对应的利润 = 350 × 6.70 – 480 = 2345 – 480 = 1865 英镑，相比原来（平均为 4 时）的 1340 英镑提高了不少。如何做到这一点呢？你可以推出一种积分卡，参与五项活动即可免费获得一杯饮料。从数学上看，你可以设立不等式：利润目标 ≥ 1500 英镑。利用原先每名参观者 5.20 英镑的净收入，解 5.20 × N – 480 ≥ 1500 → 5.20N ≥ 1980 → N ≥ 380.8，也就是需要 381 名参观者。如果通过提高活动率将净收入提升到 6.70 英镑，那么 6.70N – 480 ≥ 1500 → 6.70N ≥ 1980 → N ≥ 295.5，只需 296 名参观者即可。这展示了顾客行为上微小改进的巨大力量。
9. Task 8: Risk Assessment and Contingency Planning | 任务八：风险评估与应急计划
Every project carries risks. Here, two main risks are poor weather and lower than expected attendance. You can build a simple probability model. Suppose the chance of rain is 30 %. If it rains, attendance drops to 180, and net income per person stays £5.20. Expected attendance = 0.7 × 350 + 0.3 × 180 = 245 + 54 = 299 visitors. Expected profit = 299 × 5.20 – 480 = 1554.8 – 480 = £1074.80. To protect against the worst case, you negotiate with suppliers that you can return unopened bunting packs for a refund. This reduces potential loss if attendance is lower. You also create a formula for the minimum attendance to avoid loss: Minimum attendance = Fixed costs ÷ Net income per visitor = 480 ÷ 5.20 ≈ 93. Knowing these numbers helps you sleep better before the fair.
每个项目都存在风险。这里的主要风险是天气不佳和参与人数低于预期。你可以建立一个简单的概率模型。假设下雨的概率是 30%。如果下雨，参与人数下降到 180 人，每位参观者净收入仍为 5.20 英镑。预期参与人数 = 0.7 × 350 + 0.3 × 180 = 245 + 54 = 299 人。预期利润 = 299 × 5.20 – 480 = 1554.8 – 480 = 1074.80 英镑。为防范最坏情况，你与供应商协商，未开封的彩旗包可以退货退款，从而在参与人数较低时减少潜在损失。你还为保本最低人数建立了一个公式：最低人数 = 固定成本 ÷ 每位参观者净收入 = 480 ÷ 5.20 ≈ 93。了解这些数字能让你在嘉年华开始前睡得更安稳。
10. Task 9: Final Report and Graphical Display | 任务九：最终报告与图表展示
You present your findings using clear tables and graphs. A cost breakdown table helps visualise spending:
| Item | Calculation | Cost (£) |
|---|---|---|
| Bunting | 7 × 4.50 | 31.50 |
| Tablecloths | 30 × 6.80 | 204.00 |
| Licence & Misc. | Fixed | 244.50 |
| Total Fixed | 480.00 |
You also draw a break‑even chart with number of visitors on the horizontal axis and pounds on the vertical axis, plotting the total cost line (horizontal at £480) and the net income line (slope = 5.20). Their intersection at N = 93 indicates the break‑even point. Additionally, a pie chart shows the proportion of total funds raised that goes to the animal shelter after covering costs. Your report concludes that under expected conditions, the fair should raise around £1074.80, comfortably exceeding the initial target of £800.
你使用清晰的表格和图表来展示你的发现。一份成本细分表有助于直观地看到支出：
| 项目 | 计算 | 费用（英镑） |
|---|---|---|
| 彩旗 | 7 × 4.50 | 31.50 |
| 桌布 | 30 × 6.80 | 204.00 |
| 许可证及其他 | 固定 | 244.50 |
| 总固定成本 | 480.00 |
你还画了一张盈亏平衡图，横轴为参观人数，纵轴为金额，绘制了总成本线（在 480 英镑处的水平线）和净收入线（斜率为 5.20）。两条线在 N = 93 处的交点代表了盈亏平衡点。此外，一张饼图展示了筹集到的总资金在扣除成本后，捐给动物收容所的比例。你的报告总结道，在预期条件下，这次嘉年华应能筹集约 1074.80 英镑，轻松超过最初设定的 800 英镑目标。
11. Summary and Reflection | 总结与反思
This case study has demonstrated how algebra, ratio, measurement, statistics and logical thinking merge to solve a genuine problem. You have learned to calculate areas and perimeters, to work within a budget, to apply ratio to team organisation, to use equations for break‑even analysis, and to interpret data with mean, median and mode. Most importantly, you have seen that mathematics is not about isolated drills — it is a connected language that empowers you to plan, predict and persuade in real life. Reflect on which part of the challenge stretched your thinking the most, and consider how you might apply similar reasoning to a future project, such as organising a sports day or planning a trip.
这个案例分析展示了代数、比和比例、度量、统计以及逻辑思维如何融合起来解决一个真实的问题。你学会了计算面积和周长，在预算内运作，运用比来组织团队，使用方程进行盈亏平衡分析，以及用平均数、中位数和众数来解读数据。最重要的是，你看到了数学并非孤立的操练——它是一种相互关联的语言，赋予你在现实生活中规划、预测和说服的能力。回顾这个挑战中哪一部分最拓展你的思维，并思考如何将类似的推理应用到未来的项目中，比如组织一次运动日或策划一次旅行。
Published by TutorHao | Advanced Mathematics Revision Series | aleveler.com
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