📚 School Fair Planning: A Case Study in Applied Mathematics | 校园义卖会策划：应用数学案例实战演练
This case study walks you through the real-world planning of a Year 9 school charity fair, applying core CCEA Further Mathematics concepts such as algebra, statistics, geometry, probability and financial maths. You will see how mathematical modelling helps make informed decisions, from budgeting to layout design and risk assessment.
本案例带你走进一场真实的九年级校园慈善义卖会策划，应用代数、统计、几何、概率和财务数学等 CCEA 进阶数学核心知识点。你会看到数学建模如何帮助我们在预算、场地设计和风险评估等方面做出明智决策。
1. Introduction to the Case Study | 案例介绍
St Catherine's College plans to hold an outdoor charity fair. The mathematics department has tasked Year 9 pupils with using their further mathematics skills to produce a complete planning proposal. The goal is to raise at least £800 for a local hospice while keeping costs under control, designing a safe and attractive layout, and managing uncertainties such as weather. This integrated problem draws on multiple branches of mathematics and mimics professional project management.
圣凯瑟琳学院计划举办一场户外慈善义卖会。数学部门要求九年级学生运用进阶数学技能，完成一份完整的策划方案。目标是为当地临终关怀机构筹集至少 800 英镑，同时控制成本、设计安全美观的布局，并管理天气等不确定性。这个综合性问题涉及多个数学分支，模拟了专业的项目管理过程。
2. Understanding the Problem | 理解问题
Before any calculations, we must define the scope. Key questions include: How many visitors are expected? What activities will be offered? What are the fixed and variable costs? What ticket price is both attractive and profitable? How much space is available in the school field? Breaking down a large problem into smaller, manageable parts is a critical mathematical skill known as decomposition. We will use a structured approach: gather data, build models, solve equations, interpret results and refine.
在任何计算之前，我们必须界定范围。关键问题包括：预计有多少访客？将提供哪些活动？固定成本和可变成本是多少？怎样的票价既有吸引力又能盈利？学校操场有多少可用空间？将大问题分解成更小、可管理的部分是一种重要的数学技能，称为分解。我们将采用结构化方法：收集数据、建立模型、解方程、解释结果并优化。
3. Data Collection and Survey Design | 数据收集与调查设计
To estimate attendance, a stratified sample survey of 120 pupils across all year groups is conducted. Pupils are asked: 'Would you attend a Saturday charity fair if tickets cost £3–£5?' and 'How much would you typically spend on food and games?' The questionnaire avoids leading questions and offers multiple‑choice bands. The response rate is 92%, giving 110 valid replies. Stratified sampling ensures each year group is represented proportionally, reducing bias.
为了估算参与人数，我们在全校各年级中进行了一项分层抽样调查，共调查了 120 名学生。问题包括：“如果票价在 3 至 5 英镑之间，你会参加周六的慈善义卖会吗？”以及“你通常会在食物和游戏上花费多少？”问卷避免了引导性问题，并提供了多项选择区间。回复率为 92%，共获得 110 份有效回答。分层抽样确保每个年级按比例被代表，从而减少偏差。
4. Statistical Analysis of Survey Results | 调查结果的统计分析
The survey yields the following data on intended attendance: 78% of respondents said 'yes' ated number of attendees is 0.78 × 950 = 741. However, to be cautious, we calculate the 95% confidence interval using the formula for a proportion: margin of error ≈ 1.96 × √(p(1–p)/n) = 1.96 × √0.78×0.22/110 use the lower bound of 670 to avoid overestimating revenue. The mean intended spend on extras was £4.20 with a standard deviation of £1.90, suggesting a skewed distribution where a few might spend much more.
调查得出了以下关于出行意向的数据：78% 的受访者表示“会参加”。学校总人数为 950 人，预计参加人数为 0.78 × 950 = 741 人。但为了谨慎起见，我们使用比例置信区间公式计算 95% 的置信区间：误差幅度 ≈ 1.96 × √(p(1–p)/n) = 1.96 × √0.78×0.22/110 ≈ 0.075。因此，参加人数可能在 (0.78 − 0.075)×950 ≈ 670 到 (0.78+0.075)×950 ≈ 812 之间。为保险起见，我们采用下限 670 人，以免高估收入。额外消费的平均意向金额为 4.20 英镑，标准差为 1.90 英镑，表明分布偏斜，少数人可能消费更高。
5. Setting Ticket Prices: Algebra in Action | 定价策略：代数应用
led linearly: estimated attendance = 1000 − 70x. This equation is based on the observed 78% at £3 and a drop to about 55% at £ vertex of this quadratic: x = −b/(2a) = −1000/(2×−70) ≈ 7.14. But that suggests a price of £7.14 would maximise ticket income, which contradicts our £3–£ se the derivative 1000 − 140x is positive (e.g. at x=5, 1000−700=300>0). So the highest acceptable price, £5, maximises ticket revenue while remaining within the range.
设票价为 x 英镑。根据调查，价格与参加人数之间的关系可以线性建模：预计参加人数 = 1000 − 70x。这个方程基于观察结果：票价 3 英镑时 78% 参加，票价 5 英镑时约 55% 参加。仅门票收入为 R(x) = x(1000 − 70x) = −70x² + 1000x。为了使收入最大化，我们求出二次函数的顶点：x = −b/(2a) = −1000/(2×−70) ≈ 7.14。但这表明 7.14 英镑的票价能使门票收入最大化，而与 3–5 英镑的调查范围及可负担性相矛盾。该模型仅在定义域 3 ≤ x ≤ 5 内有效。在此区间内，收入随价格增加而增加，因为导数 1000 − 140x 为正（例如 x=5 时，1000−700=300>0）。因此，可接受的最高票价 5 英镑在既定范围内最大化了门票收入。
6. Budgeting and Cost Calculation | 预算与成本计算
Fixed costs include marquee hire £120, public liability insurance £45, printing and advertising £30. Variable costs per attendee are estimated at £ 50p for a raffle ticket. With an expected attendance of 670, total variable cost = 670 × (£1.50 + £0.50) = 670 × £2.00 = £1340. Total costs = fixed costs + variable costs = £195 + £1340 = £1535. Revenue streams: ticket sales (670 × £5 = £3350), food stall commissions (estimated 30% of average extra spend £4.20 × 670 = £2814, commission at 20% = £562.80), and direct game takings (estimated £2 per person × 670 = £1340, less prize costs £300 = £1040). Total projected revenue = £3350 + £562.80 + £1040 = £4952.80. Net profit = £4952.80 − £1535 = £3417.80, well above the £800 target.
固定成本包括帐篷租赁 120 英镑、公众责任险 45 英镑、印刷和广告 30 英镑。每位参加者的可变成本估计为礼品包 1.50 英镑和抽奖券 0.50 英镑。按预计参加人数 670 人计算，总可变成本 = 670 × (1.50+0.50) = 670 × 2.00 = 1340 英镑。总成本 = 固定成本 + 可变成本 = 195 + 1340 = 1535 英镑。收入来源：门票销售 670×5=3350 英镑；食品摊位佣金（估计平均额外消费 4.20 英镑 ×670=2814 英镑，按 20% 抽成 = 562.80 英镑）；直接游戏收入（估计每人 2 英镑 ×670=1340 英镑，减去奖品成本 300 英镑 = 1040 英镑）。预计总收入 = 3350 + 562.80 + 1040 = 4952.80 英镑。净利润 = 4952.80 − 1535 = 3417.80 英镑，远高于 800 英镑的目标。
7. Geometry of the Venue Layout | 场地布局的几何设计
256 m². The main marquee is a rectangle 15 m × 10 m, and each of the 12 game stalls requires a 3 m × 3 m square. To ensure smooth crowd flow, at least 25% ent area = 0.25 × 3256 = 814 m². Thus minimum total area needed = 258 + 814 = 1072 m², which is less than 3256 m², so the layout is feasible. We design a scaled plan using a 1:200 scale, where 1 cm represents 2 m, and arrange stalls along two straight paths forming a T-shape for easy access.
可用矩形场地尺寸为 80 米 × 50 米。安全规定要求四周留有 3 米的缓冲带，因此可用面积缩小为 74 米 × 44 米 = 3256 平方米。主场篷为 15 米 × 10 米的矩形，12 个游戏摊位每个需要 3 米 × 3 米的正方形。为保证人流顺畅，至少 25% 的可用面积必须留作开放通道。建筑物总面积 = 150 平方米（帐篷） + 12 × 9 平方米（摊位） = 150 + 108 = 258 平方米。通道需求面积 = 0.25 × 3256 = 814 平方米。因此所需最小总面积 = 258 + 814 = 1072 平方米，小于 3256 平方米，所以布局是可行的。我们使用 1:200 的比例尺（1 厘米代表 2 米）设计按比例缩小的平面图，并将摊位沿两条直线路径布置成 T 字形，以便通行。
8. Probability of Weather Impact | 天气影响的概率分析
planned for July, but rain is a risk. Historical Met Office data for the region show a 30% in on any given day in mid-July. If it rains, outdoor attendance is expected to drop by 40% and the food stalls' e earnings will halve because fewer people stay. We model the financial impact using expected monetary value (EMV). Profit in fine weather = £ (as calculated). Profit in rain: attendance = 0.6 × 670 = 402; ticket revenue = 402 × 5 = £2010; food commission becomes (402 × 4.20 × 0.5) × 0.2 = £168.84; game takings (402 × 2) − 300 = £504; total revenue = £2682.84; variable costs drop to 402 × 2 = £804, so total costs = 195 + 804 = £999; net profit = £2682.84 − 999 = £1683.84. EMV of profit = 0.7 × 3417.80 + 0.3 × 1683.84 = £2392.46 + £505.15 = £2897.61. This is still substantially above the £800 target, so the project is robust to weather risk.
义卖会计划在 7 月举行，但雨天是个风险。该地区历史气象局数据显示，七月中旬任何一天下雨的概率为 30%。如果下雨，预计户外参加人数将减少 40%，食品摊位收入将减半，因为逗留的人更少。我们使用预期货币价值（EMV）模拟财务影响。晴天利润 = 3417.80 英镑（如上计算）。雨天利润：参加人数 = 0.6 × 670 = 402；门票收入 = 402×5=2010 英镑；食品佣金变为 (402×4.20×0.5)×0.2 = 168.84 英镑；游戏收入 (402×2)−300 = 504 英镑；总收入 = 2682.84 英镑；可变成本降至 402×2 = 804 英镑，因此总成本 = 195+804 = 999 英镑；净利润 = 2682.84 − 999 = 1683.84 英镑。利润的预期货币价值 = 0.7×3417.80 + 0.3×1683.84 = 2392.46 + 505.15 = 2897.61 英镑。这仍然远高于 800 英镑的目标，因此该项目对天气风险具有稳健性。
9. Break-Even Analysis and Graphing | 盈亏平衡分析与图表
n + 0.84n + 2n − 300 = 7.84n − 300. To find the break-even point, set R = C: 7.84n − 300 = 195 + 2n → 5.84n = 495 → n = 495 / 5.84 ≈ 84.8. So we need at least 85 attendees to avoid a loss. Plotting both functions on a graph with n on the horizontal axis (0 to 800) and pounds on the vertical axis shows two straight lines intersecting at n ≈ 85. The area between the lines to the right of the intersection represents profit. This visual tool is particularly useful when presenting to the school finance committee, as it clearly demonstrates the safety margin.
我们定义一个简化的总成本函数 C = 195 + 2n，其中 n 为参加人数，总收入函数 R = 5n + 0.2×(4.20n) + (2n − 300) = 5n + 0.84n + 2n − 300 = 7.84n − 300。为求盈亏平衡点，令 R = C：7.84n − 300 = 195 + 2n → 5.84n = 495 → n = 495 / 5.84 ≈ 84.8。因此，至少需要 85 人参加才能避免亏损。将这两个函数绘制在图表上，横轴为 n（0 至 800），纵轴为金额（英镑），会显示两条直线相交于 n ≈ 85 处。交点右侧两线之间的区域表示利润。这种直观的工具在向学校财务委员会展示时特别有用，因为它清晰地表明了安全边际。
10. Optimising Profit with Inequalities | 用不等式优化利润
Suppose we can offer a discount if a group of friends buy tickets together. A 'family bundle' of 4 tickets costs £16 instead of £20, saving £ an inequality. Let y be the number of bundles sold. Each bundle reduces ticket revenue by £ bundles are sold (replacing 200 individual tickets) and 10 extra groups (40 people) attend. New ticket income = (670 − 200 + 40) individual tickets × £5 + 50 × £16 = 510×5 + 800 = 2550 + 800 = £3350. Wait, gross ticket revenue remains the same (£ l tickets are replaced, but we gain 40 attendees. However, variable costs increase by 40 × £2 = £80, and extra spending on food/games adds revenue: 40 × £4.20 × 0.2 (food commission) + 40 × £2 (games gross) = £33.60 + £80 = £113.60, but game prizes cost an extra 40 × (300/670) ≈ £17.91. Net profit change = £113.60 − £80 − £17.91 = £ quality analysis shows that deeper discounts erode profit unless they bring a substantial crowd increase.
假设我们可以为团体购票的朋友提供折扣。“家庭套票”4 张售价 16 英镑，而非 20 英镑，节省 4 英镑。为了判断这是否能增加利润，我们建立不等式。设 y 为售出的套票数量。每套票使门票收入每团减少 4 英镑，但可能额外吸引 10 个原本不会来的团体。先前无套票时利润为 3417.80 英镑。有套票时，假设售出 50 套（替代 200 张单人票），并额外吸引 10 个团体（40 人）参加。新门票收入 = (670 − 200 + 40) 张单人票 × 5 英镑 + 50 × 16 = 510×5 + 800 = 2550 + 800 = 3350 英镑。注意，毛门票收入保持不变（3350 英镑），因为取代了 200 张单人票，但增加了 40 名参加者。然而，可变成本增加了 40 × 2 = 80 英镑，额外消费在食品和游戏上的收入增加：40 × 4.20 × 0.2（食品佣金） + 40 × 2（游戏毛收入） = 33.60 + 80 = 113.60 英镑，但游戏奖品成本增加了约 40 × (300/670) ≈ 17.91 英镑。净利润变化 = 113.60 − 80 − 17.91 = 15.69 英镑的增益。我们可以将利润增加的条件表示为不等式：额外参加者带来的收入 > 额外可变成本 + 额外奖品成本。这一不等式分析表明，更深的折扣除非带来大量人群增加，否则会侵蚀利润。
11. Presenting Findings to Stakeholders | 向利益相关者展示结果
A clear presentation is vital. We prepare a report with a structured summary: (1) Executive summary: projected profit over £2800 even in rainy scenarios. (2) Key assumptions and data sources clearly stated. (3) A table summarising costs and revenue under three scenarios: sunny, rainy, and average EMV. (4) A scale layout diagram. (5) A break-even graph with commentary. We use accessible language and avoid jargon, so governors and parent association members can follow the logic. We also discuss limitations: the linear attendance model is an approximation, and the probability of rain is a historical average that may not reflect climate change trends.
清晰的展示至关重要。我们准备了一份报告，包含结构化摘要：（1）执行摘要：即使在雨天情景下，预计利润也超过 2800 英镑。（2）明确列出关键假设和数据来源。（3）一个表格，总结晴天、雨天和平均预期货币价值三种情景下的成本和收入。（4）按比例的布局图。（5）带有说明的盈亏平衡图。我们使用通俗易懂的语言，避免术语，以便校董和家长协会成员能跟上逻辑。我们还讨论了局限性：线性参加人数模型是近似值，降雨概率是历史平均值，可能无法反映气候变化趋势。
12. Reflection and Mathematical Communication | 反思与数学交流
This case study has shown that Year 9 further mathematics is not a collection of isolated topics but a toolkit for solving real problems. We applied stratified sampling, linear and quadratic models, profitability analysis, confidence intervals, expected value, inequalities, scale drawing, and graphical representation. Each mathematical decision was tied to a practical consequence. For example, choosing the lower confidence bound prevented over‑optimistic budgeting. The process also highlighted the importance of validating models—our revenue‑maximising quadratic gave a price outside the valid domain, teaching us to always consider context constraints. Effective communication of mathematical thinking is as important as the computation itself, because decisions must be justified to non‑specialists.
本案例表明，九年级进阶数学并非孤立的主题集合，而是解决现实问题的工具箱。我们应用了分层抽样、线性和二次模型、盈利能力分析、置信区间、预期值、不等式、比例绘图和图形表示。每一个数学决策都与实际后果相关联。例如，选择置信下限避免了过于乐观的预算。该过程也凸显了验证模型的重要性——我们的收入最大化二次函数给出的价格超出了有效定义域，这教会我们要始终考虑情境约束。数学思维的有效沟通与计算本身同样重要，因为决策必须向非专业人士解释清楚。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
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