📚 Year 11 CCEA Further Mathematics: Case Study Practical Exercises | CCEA 11年级进阶数学：案例分析实战演练
In CCEA Year 11 Further Mathematics, being able to apply abstract techniques to real‑world problems is a skill that separates good students from excellent ones. This article presents a series of case studies that mirror the style of longer exam questions and project work, each blending pure maths, mechanics and statistics. Work through them methodically – the paired English and Chinese explanations will help you master both the language of mathematics and the logical flow of a solution.
在 CCEA 11年级进阶数学中，能否将抽象的数学技巧应用于实际问题是区分良好与优秀学生的关键能力。本文提供了一系列与考试长问题和项目作业风格相近的案例，每个案例都融合了纯数、力学和统计学知识。请有条不紊地逐一演练——配对的英文和中文解释将帮助你掌握数学语言和解题的逻辑脉络。
1. Introduction to Case Studies in Further Maths | 进阶数学案例学习导论
A mathematical case study usually presents a realistic scenario, a set of data or constraints, and a clear question. Your task is to translate the words into equations, diagrams or statistical models, solve the resulting problem using techniques from the CCEA specification, and then interpret the answer back in the context of the original situation.
数学案例分析通常会呈现一个现实情景、一组数据或约束条件，以及一个明确的问题。你的任务是将文字转化为方程、图表或统计模型，运用 CCEA 大纲中的技巧求解，然后将答案放回原始情境中进行解释。
To succeed, always begin by identifying the relevant topic – is it linear programming, calculus, matrices, sequences, trigonometry or probability? Next, list every variable and write down what you know. Only then should you set up the mathematical machinery. Finally, check that your numerical result makes sense in the real world; a negative length or a probability above 1 is a warning sign.
要成功解题，务必先确定相关主题——是线性规划、微积分、矩阵、数列、三角学还是概率？接着，列出每一个变量并写下已知条件。然后再建立数学模型。最后，检查数值结果在现实中是否合理；负的长度或大于1的概率都是警示信号。
2. Case Study 1: Optimising Production with Linear Programming | 案例一：运用线性规划优化生产
A small furniture workshop makes two types of chairs, Classic (x) and Modern (y). Each Classic chair needs 2 hours of cutting and 3 hours of assembling; each Modern chair requires 4 hours of cutting and 2 hours of assembling. The workshop has a maximum of 40 cutting hours and 30 assembling hours available per week. The profit is £25 per Classic chair and £30 per Modern chair. Find how many of each type should be made weekly to maximise profit.
一家小型家具作坊生产两种椅子：经典型 (x) 和现代型 (y)。每张经典椅需 2 小时切割和 3 小时组装；每张现代椅需 4 小时切割和 2 小时组装。作坊每周最多可用切割工时 40 小时、组装工时 30 小时。每张经典椅利润为 25 英镑，现代椅利润为 30 英镑。求每周两种椅子各应生产多少才能实现利润最大化。
Constraints become inequalities: 2x + 4y ≤ 40 (cutting), 3x + 2y ≤ 30 (assembling), with x ≥ 0, y ≥ 0. The objective function to maximise is P = 25x + 30y. Plot the feasible region, find the vertices: (0,0), (0,10), (10,0) and the intersection of 2x + 4y = 40 and 3x + 2y = 30.
约束条件转化为不等式：2x + 4y ≤ 40（切割），3x + 2y ≤ 30（组装），同时 x ≥ 0，y ≥ 0。目标函数为求 P = 25x + 30y 的最大值。画出可行域，找出顶点：(0,0)、(0,10)、(10,0) 以及 2x + 4y = 40 与 3x + 2y = 30 的交点。
Solve simultaneously: multiply the second equation by 2 to get 6x + 4y = 60, subtract the first equation to give 4x = 20, so x = 5, then y = 7.5. Vertices are (0,0), (0,10), (10,0) and (5, 7.5). Evaluate P: at (0,10) P = £300; at (10,0) P = £250; at (5,7.5) P = 25(5) + 30(7.5) = 125 + 225 = £350. The maximum profit is £350 by making 5 Classic and 7.5 Modern chairs – since you cannot make half a chair, test integer points (5,7) giving P=335, (5,8) gives 5*25+8*30=365 but 2*5+4*8=42>40, so try 4,8 gives 4*25+8*30=100+240=340 but cutting 2*4+4*8=8+32=40 ok, assembling 12+16=28 ok. The integer optimum is (5,7) with profit £335, or check (6,7) 12+28=40 ok, assembling 18+14=32>30 no. So the real-world answer: 5 Classic and 7 Modern chairs.
联立方程求解：将第二个方程乘以2得 6x + 4y = 60，减去第一个方程得 4x = 20，因此 x = 5，y = 7.5。顶点为 (0,0)、(0,10)、(10,0) 和 (5, 7.5)。计算收益：在 (0,10) 处 P = 300英镑；在 (10,0) 处 P = 250英镑；在 (5,7.5) 处 P = 25×5 + 30×7.5 = 125 + 225 = 350英镑。最大理论利润为350英镑，需生产5张经典和7.5张现代椅——由于不能生产半张椅子，测试整数点：(5,7) 得335；(5,8) 得365但切割工时42>40；试用 (4,8) 利润340，切割40小时满足，组装12+16=28满足；最佳整数解为 (5,7)，利润335英镑。因此现实答案为每周生产5张经典椅和7张现代椅。
3. Case Study 2: Projectile Motion and Differentiation | 案例二：抛体运动与微分
A firework shell is launched vertically and its height h metres after t seconds is modelled by h(t) = −5t² + 40t + 2. Use calculus to find the maximum height and the time when it reaches that height. Also determine the velocity when it hits the ground (assume h = 0).
一枚烟花弹垂直发射，t 秒后高度 h 米，其运动模型为 h(t) = −5t² + 40t + 2。运用微积分求最大高度及到达该高度的时间，并求烟花落地时的速度（设 h = 0）。
Maximum height occurs when dh/dt = 0. Differentiate: dh/dt = −10t + 40. Set −10t + 40 = 0 → t = 4 s. Substitute into h: h(4) = −5(16) + 160 + 2 = −80 + 162 = 82 m. To find ground impact, solve −5t² + 40t + 2 = 0. Use quadratic formula: t = [−40 ± √(1600 + 40)]/(−10) = [−40 ± √1640]/(−10). √1640 ≈ 40.5, so t ≈ (−40 + 40.5)/(−10) = 0.5/(−10) = −0.05 (reject) or t ≈ (−40 − 40.5)/(−10) = −80.5/−10 = 8.05 s. Velocity at impact: v(t) = dh/dt = −10t + 40, at t ≈ 8.05, v ≈ −10(8.05) + 40 = −80.5 + 40 = −40.5 m/s (negative means downward).
当 dh/dt = 0 时高度最大。求导：dh/dt = −10t + 40。令 −10t + 40 = 0 → t = 4 秒。代入 h：h(4) = −5×16 + 160 + 2 = 82 米。求落地时间：解 −5t² + 40t + 2 = 0。用二次公式：t = [−40 ± √(1600 + 40)]/(−10)。√1640 ≈ 40.5，故 t ≈ 8.05 秒（舍去负根）。落地时的速度：v(t) = −10t + 40，代入 t ≈ 8.05，得 v ≈ −40.5 m/s（负号表示方向向下）。
4. Case Study 3: Financial Modelling with Geometric Sequences | 案例三：几何数列金融建模
A bank offers a savings account with 4% annual interest compounded yearly. A customer deposits £2000 at the start of each year for 10 years. What is the total balance immediately after the 10th deposit?
某银行提供年利率4%、每年复利的储蓄账户。一位客户每年年初存入2000英镑，连续存10年。问第10次存款后账户立即结余为多少？
This is a geometric series where the first deposit earns interest for 10 years, the second for 9 years, and so on. The balance after 10 deposits is S = 2000(1.04^10 + 1.04^9 + … + 1.04) = 2000 × (1.04(1.04^10 − 1)/(1.04 − 1)). Simplify geometric sum: sum of 1.04 + 1.04² + … + 1.04^10, first term a=1.04, r=1.04, n=10. Sum = 1.04(1.04^10 − 1)/(0.04). Compute 1.04^10 ≈ 1.480244, so numerator = 1.04 × 0.480244 ≈ 0.49945, divided by 0.04 ≈ 12.4863, multiplied by 2000 gives £24,972.60. Alternatively, use the formula for future value of an annuity due.
这是一个几何级数：第一笔存款计息10年，第二笔计息9年，依此类推。10次存款后的余额为 S = 2000(1.04^10 + 1.04^9 + … + 1.04)。运用几何级数求和：首项 a = 1.04，公比 r = 1.04，项数 n = 10，和 = 1.04×(1.04^10 − 1)/0.04。计算 1.04^10 ≈ 1.480244，分子约 0.49945，除以0.04得12.4863，乘以2000约为24,972.60英镑。也可以使用即时年金终值公式。
5. Case Study 4: Matrix Transformations in Computer Graphics | 案例四：计算机图形学中的矩阵变换
A graphic designer wants to rotate a logo by 90° anticlockwise about the origin. The logo is defined by the coordinates A(3,1), B(5,2) and C(2,4). Find the new coordinates after rotation using a matrix transformation.
一位平面设计师要将一个标志绕原点逆时针旋转90°。标志由坐标 A(3,1)、B(5,2) 和 C(2,4) 定义。试用矩阵变换求出旋转后的新坐标。
The rotation matrix for +90° is: [0 −1; 1 0]. Apply to each column vector: A′ = [0 −1; 1 0][3; 1] = [−1; 3]; B′ = [0 −1; 1 0][5; 2] = [−2; 5]; C′ = [0 −1; 1 0][2; 4] = [−4; 2]. So the new vertices are A′(−1,3), B′(−2,5), C′(−4,2).
逆时针旋转90°的矩阵为：[0 −1; 1 0]。分别作用于列向量：A′ = [0 −1; 1 0][3; 1] = [−1; 3]；B′ = [−2; 5]；C′ = [−4; 2]。因此新顶点坐标为 A′(−1,3)、B′(−2,5)、C′(−4,2)。
To enlarge by scale factor 2 and then rotate, matrix multiplication would combine the operations, remembering order matters.
若先放大2倍再旋转，可将矩阵相乘，注意顺序至关重要。
6. Case Study 5: Trigonometric Navigation Problem | 案例五：三角函数在航海中的应用
A ship leaves port and sails 15 km on a bearing of 040°, then changes course to 120° and sails 20 km. Find the direct distance and bearing from the starting point to the final position.
一艘船离开港口，沿方位角040°航行15公里，然后调整航向至120°并航行20公里。求从起点到终点的直线距离和方位角。
Draw a diagram. Let O be the start. The first leg ends at point A: OA = 15 at 40° from north. The second leg AB = 20 at 120°, so angle between OA and AB: north at A is parallel, so the turn from 40° to 120° is 80°. In triangle OAB, OA=15, AB=20, included angle ∠OAB = 180° − 80° = 100°. Use cosine rule: OB² = 15² + 20² − 2×15×20×cos(100°). cos(100°) ≈ −0.1736. OB² = 225 + 400 − 600×(−0.1736) = 625 + 104.16 = 729.16, OB ≈ 27.0 km. Sine rule for bearing: sin(∠AOB)/20 = sin(100°)/27.0 → sin(∠AOB) = (20×0.9848)/27.0 ≈ 0.7295, angle AOB ≈ 46.8°. The bearing of B from O is 40° + 46.8° = 086.8°.
画图分析。设 O 为起点，第一段结束于 A 点：OA = 15 公里，方向040°。第二段 AB = 20 公里，方向120°，因此从 OA 方向转到 AB 方向的夹角为80°。在三角形 OAB 中，OA=15，AB=20，∠OAB = 180° − 80° = 100°。用余弦定理：OB² = 15² + 20² − 2×15×20×cos100°，cos100° ≈ −0.1736，得 OB² ≈ 729.16，OB ≈ 27.0 km。用正弦定理求方位角：sin(∠AOB)/20 = sin100°/27.0，sin(∠AOB) ≈ 0.7295，∠AOB ≈ 46.8°。终点的方位角为 040° + 46.8° = 086.8°。
7. Case Study 6: Connected Rates of Change in Biology | 案例六：生物学中的关联变化率
The radius of a circular bacterial colony grows at a constant rate of 0.2 mm per hour. Find the rate at which the area is increasing when the radius is 5 mm.
一个圆形细菌菌落的半径以每小时0.2毫米的恒定速率增长。求半径为5毫米时，面积的增加速率。
We know dR/dt = 0.2 mm/h. Area A = πR², so dA/dR = 2πR. Using the chain rule: dA/dt = dA/dR × dR/dt = 2πR × 0.2. At R=5 mm, dA/dt = 2π(5)×0.2 = 2π × 1 = 2π ≈ 6.28 mm² per hour.
已知 dR/dt = 0.2 mm/h。面积 A = πR²，故 dA/dR = 2πR。运用链式法则：dA/dt = dA/dR × dR/dt = 2πR × 0.2。当 R=5 mm 时，dA/dt = 2π×5×0.2 = 2π ≈ 6.28 mm²/h。
Always state the units of the rate clearly; the problem asks for the rate of area increase, so mm²/h is appropriate.
务必清晰注明速率单位；本题要求面积的增加速率，故单位用 mm²/h 是合适的。
8. Case Study 7: Probability Trees and Decision Making | 案例七：概率树与决策分析
A medical test for a disease has a sensitivity of 95% (probability of a positive result if the disease is present) and a specificity of 90% (probability of a negative result if the disease is not present). The disease occurs in 2% of the population. If a randomly selected person tests positive, what is the probability they actually have the disease?
某疾病检测的灵敏度为95%（若患病，检测呈阳性的概率），特异性为90%（若未患病，检测呈阴性的概率）。该疾病在人群中的发病率为2%。若随机一人检测呈阳性，其实际患病的概率是多少？
Use Bayes’ theorem via a tree. Define events: D = has disease, ¬D = no disease; + = positive test. P(D)=0.02, P(¬D)=0.98. P(+|D)=0.95, P(+|¬D)=1−0.90=0.10. P(+) = P(D)×P(+|D) + P(¬D)×P(+|¬D) = 0.02×0.95 + 0.98×0.10 = 0.019 + 0.098 = 0.117. Required P(D|+) = (0.02×0.95)/0.117 = 0.019/0.117 ≈ 0.1624 (16.2%).
利用树状图或贝叶斯定理求解。设 D 为患病，¬D 为未患病；+ 为阳性。P(D)=0.02，P(¬D)=0.98。P(+|D)=0.95，P(+|¬D)=1−0.90=0.10。总体阳性概率 P(+) = 0.02×0.95 + 0.98×0.10 = 0.019 + 0.098 = 0.117。所求 P(D|+) = (0.02×0.95)/0.117 ≈ 0.1624，即16.2%。
This counter‑intuitive result shows why base rate (prevalence) is crucial in interpreting a positive test.
这个有悖直觉的结果表明，在解读阳性检测结果时，基础概率（患病率）至关重要。
9. Strategy for Tackling Case Study Exam Questions | 应对考试案例分析的策略
Read the entire question through once, then annotate key numbers and verbs. Underline or circle the specific mathematical task – for example, ‘show that’, ‘find the maximum’, ‘prove’. Create a clear layout starting with ‘Let … be …’ statements and a list of known formulas. As you work, label intermediate answers so an examiner can follow your reasoning. When you have a final numerical answer, write a short sentence connecting it back to the context.
先通读全题，然后标注关键数字和动词。在具体的数学任务下画线或画圈，例如“证明”、“求最大值”、“求证”。从一开始就用“设……为……”的清晰格式列出已知公式。解题过程中，标记中间结果，以便考官追踪你的推理过程。得出最终数值答案后，写一句简短的话，将答案与具体情境联系起来。
10. Common Pitfalls and How to Avoid Them | 常见陷阱及避免方法
One frequent mistake is forgetting to check inequality boundaries when a variable must be an integer – like chairs in Case Study 1. Another is misinterpreting bearings: always start from north and measure clockwise. In differentiation, students sometimes find the stationary point but forget to confirm it is a maximum. Always perform a second derivative test or examine the sign change. With probability, double‑check that all branches add to 1 and that you have not confused conditional probability with joint probability.
一个常见错误是当变量必须为整数时忘记检验不等式边界——如案例1中的椅子数量。另一个错误是误解方位角：永远从正北开始顺时针测量。在微分应用中，学生有时找到驻点却忘记确认它是最大值，应该进行二阶导数检验或检查符号变化。在概率问题中，务必检查所有分支的概率之和为1，且不要混淆条件概率与联合概率。
11. Practice Tips and Resources | 练习建议与资源
Build a set of index cards, each summarising a methodology: ‘Steps for Linear Programming’, ‘Steps for Related Rates’, ‘Geometric Series Formulae’, and so on. Use past CCEA Further Maths papers to identify how case studies are structured – often part (a) is straightforward modelling, (b) calculation, (c) interpretation. Time yourself writing full solutions under exam conditions. Collaborate with peers to explain a case study aloud; teaching is the best way to deepen understanding.
制作一套索引卡片，每张总结一种方法：“线性规划解题步骤”、“相关变化率步骤”、“几何级数公式”等。利用 CCEA 进阶数学历年真题，分析案例分析的设题结构——通常 (a) 部分为简单建模，(b) 计算，(c) 解释。在限时条件下计时作答。与同学合作，大声讲解案例；教授他人是深化理解的最佳方式。
12. Conclusion | 结语
Case study exercises in Year 11 CCEA Further Mathematics are not just about getting the right number; they train you to think like a mathematician. By breaking down a complex situation into manageable steps and linking pure methods with realistic contexts, you build a skillset that extends far beyond the exam hall. Return to these examples, vary the numbers, and create your own problems. The confidence you gain will be your strongest asset when you tackle the real examination.
CCEA 11年级进阶数学中的案例分析练习不仅仅是算出正确答案，它们旨在训练你像数学家一样思考。通过将复杂情境分解为可操作的步骤，将纯数学方法与现实情境相结合，你所培养的技能将远远超越考场。反复演练这些例子，改变其中的数据，并尝试自己编题。由此获得的自信将是你面对真正考试时最强大的资本。
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