📚 Year 9 Cambridge Further Maths: Case Study Masterclass | Year 9 剑桥进阶数学：案例分析实战演练
In Year 9 Cambridge Further Maths, students move beyond routine exercises and learn to apply mathematical concepts to real-world scenarios. Case study practical exercises are essential for developing problem-solving skills and modelling competence. This article presents a series of interconnected case studies covering topics such as optimisation, financial mathematics, statistics, probability, trigonometry, kinematics, and sequences. Each task is broken down into manageable steps, with clear explanations and computational walkthroughs.
在Year 9剑桥进阶数学中，学生超越常规练习题，学习将数学概念应用于真实情境。案例分析的实战演练对于培养解决问题的能力和建模素养至关重要。本文呈现一系列相互关联的案例，涵盖优化、金融数学、统计、概率、三角学、运动学和数列等主题。每个任务被分解成可操作的步骤，并配有清晰的解释和计算示范。
1. Case Study Introduction: The School Fair Project | 案例导入：学校嘉年华项目
Imagine your school is organising a charity fair. As part of the further maths team, you are tasked with solving several practical problems: designing a stall layout, planning an investment for fundraising, analysing visitor survey data, creating a fair game with balanced odds, measuring the distance for a display, predicting the path of a water balloon launch, arranging VIP seating, and designing a cost-effective packaging box for prizes. These interconnected challenges form the core of this case study masterclass.
假设你的学校正在组织一场慈善嘉年华。作为进阶数学团队的一员，你的任务是解决几个实际问题：设计摊位布局、规划筹款投资方案、分析访客调查数据、设计一个公平的游戏、测量展示区的距离、预测水球发射的轨迹、安排贵宾座位，以及为奖品设计成本最优的包装盒。这些相互关联的挑战构成了本次案例分析实战演练的核心内容。
2. Case Study 1: Maximising Stall Area (Quadratic Optimisation) | 案例1：摊位面积最大化（二次优化）
The school fair committee has 40 metres of fencing to create a rectangular stall area against a wall. The wall forms one side, so fencing is needed for only three sides. Let the width perpendicular to the wall be x metres. Then the length parallel to the wall is (40 − 2x) metres. The area A is given by A = x(40 − 2x) = 40x − 2x².
校嘉年华筹委会有40米长的围栏，想要靠墙围出一个矩形的摊位区。墙作为其中一边，因此只需围三边。设垂直于墙的宽度为 x 米，则平行于墙的长度为 (40 − 2x) 米。面积 A 表示为 A = x(40 − 2x) = 40x − 2x²。
To find the maximum area, we complete the square or use the vertex formula. The quadratic is A = −2x² + 40x. The x-coordinate of the vertex is −b/(2a dth is 10 m, giving a length of 40 − 2(10) = 20 m, and a maximum area of 10 × 20 = 200 m².
要找到最大面积，我们可以配方或使用顶点公式。二次函数为 A = −2x² + 40x。顶点的 x 坐标为 −b/(2a) = −40/(2 × −2) = 10。因此最优宽度为10米，长度为40 − 2(10) = 20米，最大面积为10 × 20 = 200平方米。
3. Case Study 2: Investment for Fundraising (Compound Interest & Sequences) | 案例2：筹款投资（复利与数列）
The fair committee plans to invest £500 of initial funds in an account offering 4% annual compound interest, credited at the end of each year. The amount after n years can be modelled by a geometric sequence: A Calculate the balance after 5 years and determine how many whole years it takes to exceed £600.
筹委会计划将500英镑的启动资金存入一个年利率为4%、每年末计息的复利账户。第 n 年后的金额可以用等比数列建模：Aₙ = 500 × (1.04)ⁿ。计算5年后的余额，并判断需要多少整年才能超过600英镑。
After 5 years: A₅ = 500 × (1.04)⁵ ≈ 500 × 1.2166529 = £608.33. To exceed £600, solve 500 × (1.04)ⁿ > 600 → (1.04)ⁿ > 1.2. Taking logs: n ln(1.04) > ln(1.2) → n > ln(1.2)/ln(1.04) ≈ 0.18232/0.03922 ≈ 4.65. So after 5 years the balance exceeds £600.
5年后：A₅ = 500 × (1.04)⁵ ≈ 500 × 1.2166529 = 608.33英镑。要超过600英镑，解 500 × (1.04)ⁿ > 600 → (1.04)ⁿ > 1.2。取对数：n ln(1.04) > ln(1.2) → n > ln(1.2)/ln(1.04) ≈ 0.18232/0.03922 ≈ 4.65。因此5年后余额超过600英镑。
This application shows how geometric sequences and logarithms help answer 'how long' questions in finance.
此应用展示了等比数列和对数如何帮助回答金融中的“需要多久”类问题。
4. Case Study 3: Survey Analysis (Statistics & Data Representation) | 案例3：调查分析（统计与数据呈现）
To understand visitor preferences, a survey collected the ages of 50 attendees. The data are grouped as follows:
为了解访客偏好，我们调查收集了50位参与者的年龄。数据分组如下：
| Age Range (years) | Frequency |
|---|---|
| 0 – 9 | 4 |
| 10 – 19 | 14 |
| 20 – 29 | 18 |
| 30 – 39 | 10 |
| 40 – 49 | 4 |
Calculate an estimate of the mean age and determine the modal class and median class.
计算年龄平均值的估计值，并确定众数区间和中位数区间。
) ÷ 50 = (18 + 203 + 441 + 345 + 178) ÷ 50 = 1185 ÷ 50 = 23.7 years. The modal class is 20– uencies: 4, 18, 36, 46, 50. The median position is the 25.5th value, which falls in the 20–29 class.
使用组中值（4.5, 14.5, 24.5, 34.5, 44.5）：估计平均值 .5 + 18×24.5 + 10×34.5 + 4×44.5) ÷ 50 = (18 + 203 + 441 + 345 + 178) ÷ 50 = 1185 ÷ 50 = 23.7岁。众数区间为20–29（频率最高）。累积频数：4, 18, 36, 46, 50。中位数位置为第25.5个值，落在20–29区间内。
5. Case Study 4: Probability of Game Success (Tree Diagrams & Expected Value) | 案例4：游戏成功的概率（树形图与期望值）
A fairground game involves spinning a wheel twice. The wheel has three equal sectors: Win (W), Lose (L), and Try Again (T). If a player gets two Wins in a row, they receive a prize worth £5. If they get exactly one Win and one Lose (in any order), they get a consolation prize worth £1. All other outcomes give nothing. The entry fee is £2. Draw a tree diagram and calculate the expected profit per game for the school.
一个游园游戏需转动一个转盘两次。转盘有三个等大的扇区：赢 (W)、输 (L) 和再转一次 (T)。如果玩家连续两次转到赢，可获得价值5英镑的奖品。如果恰好一次赢一次输（不论顺序），可获得价值1英镑的安慰奖。其他结果无奖。入场费为2英镑。画出树形图并计算学校每局游戏的期望利润。
P(WW) = (1/3)² = 1/9, prize £5; P(WL or LW) = 2 × (1/3)² = 2/9, prize £1; other outcomes have probability 1 − 1/9 − 2/9 = 6/9, prize £0. Expected payout = (1/9)×5 + (2/9)×1 + (6/9)×0 = 7/9 ≈ £0.78. Expected profit per game = entry fee − expected payout = 2 − 0.78 = £1.22. The game is financially favourable for the school.
P(WW) = (1/3)² = 1/9，奖品5英镑；P(WL 或 LW) = 2 × (1/3)² = 2/9，奖品1英镑；其他结果概率为 1 − 1/9 − 2/9 = 6/9，奖品0英镑。期望支出 = (1/9)×5 + (2/9)×1 + (6/9)×0 = 7/9 ≈ 0.78英镑。每局期望利润 = 入场费 − 期望支出 = 2 − 0.78 = 1.22英镑。对学校来说该游戏在财务上有利。
6. Case Study 5: Measuring Distances (Trigonometry & Sine/Cosine Rules) | 案例5：测量距离（三角学与正弦余弦定理）
BC = 60 m are measured, and angle ACB = 70°. Find the distance AB between the attractions using the cosine rule.
两个嘉年华景点A（主舞台）和B（美食广场）被一个池塘隔开。从售票亭C可以看到它们。测得距离 AC = 80 米，BC = 60 米，角 ACB = 70°。用余弦定理求出景点A与B之间的距离。
Cosine rule: c² = a² + b² − 2ab 3283.2 = 6716.8. So AB = √6716.8 ≈ 81.96 m. The two attractions are about 82 metres apart.
余弦定理：c² = a² + b² − 2ab cos C。此处，AB² = 80² + 60² − 2×80×60×cos 70°。cos 70° ≈ 0.3420。因此 AB² = 6400 + 3600 − 9600×0.3420 = 10000 − 3283.2 = 6716.8。所以 AB = √6716.8 ≈ 81.96 米。两个景点相距约82米。
7. Case Study 6: Projectile Motion (Quadratic Models & Kinematics) | 案例6：抛射运动（二次模型与运动学）
A water balloon is launched upward from a height of 1.5 m with an initial vertical speed of 14 m/s. Its height h (in metres) after t seconds is h = 1.5 + 14t − 4.9t². Determine the maximum height reached and the time when the balloon hits the ground.
一个水球从1.5米高处以14米/秒的初始竖直速度向上发射。其 t 秒后的高度 h（米）为 h = 1.5 + 14t − 4.9t²。求达到的最大高度和水球落地的时间。
The quadratic has vertex at t = −b/(2a) = −14/(2 × −4.9) = 14/9.8 ≈ 1.43 s. Max height h = 1.5 + 14(1.43) − 4.9(1.43)² ≈ 1.5 + 20.02 − 10.0 ≈ 11.5 m. It hits the ground when h = 0: 1.5 + 14t − 4.9t² = 0. Rearrange: 4.9t² − 14t − 1.5 = 0. Discriminant Δ = 14² − 4×4.9×(−1.5) = 196 + 29.4 = 225.4. t = [14 ± √225.4] / (9.8). Positive root: (14 + 15.01)/9.8 ≈ 2.96 s.
该二次函数的顶点在 t = −b/(2a) = −14/(2 × −4.9) = 14/9.8 ≈ 1.43 秒。最大高度 h = 1.5 + 14(1.43) − 4.9(1.43)² ≈ 1.5 + 20.02 − 10.0 ≈ 11.5 米。落地时 h = 0：1.5 + 14t − 4.9t² = 0。整理得 4.9t² − 14t − 1.5 = 0。判别式 Δ = 14² − 4×4.9×(−1.5) = 196 + 29.4 = 225.4。t = [14 ± √225.4] / (9.8)。正根为 (14 + 15.01)/9.8 ≈ 2.96 秒。
8. Case Study 7: Seating Arrangements (Permutations & Combinations) | 案例7：座位安排（排列与组合）
There are 8 VIP guests, and the committee needs to select 3 of them to sit on the front row of a presentation. Then, those 3 chosen guests must be ordered into 3 specific seats. How many different arrangements are possible? Also, if two particular guests refuse to sit next to each other, how many valid arrangements remain?
有8位贵宾，筹委会需要从中选出3位坐在演讲的前排。然后，这3位被选中的贵宾需安排到3个特定的座位上。问共有多少种不同的安排方式？此外，如果其中两位特定贵宾拒绝相邻而坐，还剩下多少种有效安排？
rangements (336) and subtract the arrangements where A and B are both selected and seated together. Number of ways to select the third guest from the remaining 6 is 6. For each such selection, treat A and B as a block: there are 2! ways to order the block and the third guest, and 2! seating orders. So invalid arrangements = 6 × 4 = 24. Valid arrangements = 336 − 24 = 312.
从8位中选出并排列3位的方法数为排列 ⁸P₃ = 8 × 7 × 6 = 336。若两位贵宾 A 和 B 拒绝相邻，首先计算总安排数 (336)，然后减去 A 和 B 都被选中且相邻而坐的安排数。从剩余6人中选出第三位的方法数为6。对每种这样的选择，将 A 和 B 视为一个整体：有2!种方式安排这个整体与第三位，同时内部有2!种排列，共计2 × 2 = 4种座位顺序。因此无效安排为6 × 4 = 24。有效安排 = 336 − 24 = 312。
9. Case Study 8: Packaging Design (3D Shapes & Volume/Surface Area) | 案例8：包装设计（三维形状与体积/表面积）
The fair needs a rectangular box with a square base to hold prizes. The box must have a volume of 8000 cm³. Let the side of the square base be x cm and the height be h cm. The material cost is proportional to the surface area. Express h in terms of x, then find the dimensions that minimise the surface area.
嘉年华需要一个底面为正方形的长方体盒子来装奖品。盒子体积必须为8000立方厘米。设底面正方形的边长为 x 厘米，高为 h 厘米。材料成本与表面积成正比。用 x 表示 h，然后求出使表面积最小的尺寸。
Volume V = x²h = 8000, so h = 8000/x². Surface area S = 2x² + 4xh = 2x² + 4x(8000/x²) = 2x² + 32000/x. To minimise, differentiate: dS/dx = 4x − 32000/x² = 0 → 4x³ = 32000 → x³ = 8000 → x = 20 cm. Then h The optimal box is a cube with side 20 cm, giving minimum surface area 6 × 20² = 2400 cm².
体积 V = x²h = 8000，故 h = 8000/x²。表面积 S = 2x² + 4xh = 2x² + 4x(8000/x²) = 2x² + 32000/x。求最小值时求导：dS/dx = 4x − 32000/x² = 0 → 4x³ = 32000 → x³ = 8000 → x = 20 厘米。则 h = 8000/400 = 20 厘米。最优盒子是一个边长为20厘米的立方体，最小表面积为6 × 20² = 2400 平方厘米。
10. Conclusion: Reflecting on Mathematical Modelling | 结论：数学建模反思
Through these eight case studies, you have seen how topics from Year 9 Cambridge Further Maths come together to solve authentic problems. Quadratic optimisation helped design the most spacious stall, geometric sequences predicted investment growth, statistics extracted meaning from survey data, probability and tree diagrams evaluated game fairness, trigonometry measured inaccessible distances, kinematic equations modelled projectile motion, permutations managed seating constraints, and calculus-based optimisation minimised packaging costs. The key skill is recognising which mathematical tool fits each real-world scenario.
通过这八个案例，你看到了Year 9剑桥进阶数学中的各个主题如何协同解决真实问题。二次优化帮助设计了最宽敞的摊位，等比数列预测了投资增长，统计从调查数据中提取信息，概率和树形图评估了游戏的公平性，三角学测量了不可达距离，运动学方程模拟了抛射运动，排列组合处理了座位限制，基于微积分的优化最小化了包装成本。核心技能在于识别哪种数学工具适合每种现实情境。
When approaching a case study, always define variables clearly, draw diagrams where possible, and verify your answers against the context. Mathematical modelling is not just about getting an answer—it is about interpreting results and making informed decisions. Continue practising with similar integrated tasks to build confidence and fluency for both coursework and examinations.
在处理案例分析时，始终清晰地定义变量，尽可能绘制图示，并对照实际情境验证答案。数学建模不仅仅是得出一个答案——更在于解释结果并做出明智的决策。继续通过类似的综合任务进行练习，以增强信心和熟练度，为课程作业和考试打下坚实基础。
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