📚 Case Study Practice for Year 8 CAIE Further Mathematics | Year 8 CAIE 进阶数学：案例分析实战演练
In Year 8 CAIE Further Mathematics, applying mathematical concepts to real-world scenarios is not just an exam skill – it is the heart of the subject. The following case studies show how algebra, geometry, calculus, and probability can be combined to model and solve practical problems. Each study includes a clear problem statement, a step-by-step solution, and notes on common pitfalls. Working through these examples will sharpen your analytical thinking and prepare you for higher-level challenges.
在 Year 8 CAIE 进阶数学中，将数学概念应用到现实情境不仅是一项考试技能，更是学科的核心。以下案例分析展示了如何结合代数、几何、微积分和概率来建模并解决实际问题。每个案例都包括清晰的问题陈述、分步解答以及常见误区提示。通过这些练习，你的分析思维将得到锻炼，为更高层次的学习做好准备。
1. Introduction to Case Study Method | 案例分析导论
A case study in mathematics is a structured investigation of a real-life situation that requires you to identify relevant information, formulate a mathematical model, perform computations, and interpret the results. In Further Mathematics, this often involves multiple topics – from quadratic functions to trigonometric ratios and rates of change. The key is to read carefully, define variables, and always check whether your answer makes sense in the original context.
数学中的案例研究是对现实情境的结构化探究，要求你识别相关信息、建立数学模型、进行计算并解释结果。在进阶数学中，这通常涉及多个主题——从二次函数到三角比和变化率。关键在于仔细阅读、定义变量、并始终检查答案在原本语境中是否合理。
Below are eight carefully selected case studies that mirror the type of problems you will encounter in CAIE assessments. They cover maximisation, kinematics, growth models, mixtures, indirect measurement, optimisation with derivatives, game analysis, and financial sequences.
以下是八个精心挑选的案例研究，反映了你在 CAIE 考试中会遇到的题型。它们涵盖最值问题、运动学、增长模型、混合问题、间接测量、导数优化、游戏分析和金融数列。
2. Case Study 1: Maximising Area with a Fixed Perimeter | 案例 1：固定周长下的最大面积
A farmer has 60 metres of fencing and wants to enclose a rectangular plot along a straight riverbank. One side of the rectangle will be the river and does not require fencing. What dimensions give the maximum enclosed area?
一位农民有 60 米长的围栏，想沿着一条笔直的河岸围出一个矩形地块。矩形的一边靠河，不需要围栏。问什么尺寸能让围出的面积最大？
. This is a quadratic with a negative coefficient of x², so its maximum occurs at the vertex.
设垂直于河岸的边长为 x 米，则平行于河岸的边使用剩余的围栏：长度 = 60 − 2x。面积 A = x(60 − 2x) = 60x − 2x²。这是一个 x² 系数为负的二次函数，因此在顶点处取得最大值。
Therefore the dimensions are 15 m (perpendicular) and 30 m (parallel), giving area 450 m².
对于 ax² + bx + c，顶点的 x 坐标为 x = −b/(2a)。这里 a = −2，b = 60，所以 x = −60/(2 × −2) = 15。因此尺寸为 15 米（垂直边）和 30 米（平行边），面积为 450 平方米。
the second derivative or complete the square to confirm the nature of the extreme. Here f'(x) = 60 − 4x; f'(15)=0 and f''=−4 < 0, so it is a maximum. The farmer obtains the greatest area with these measurements.
总是检查二阶导数或配方以确认极值性质。这里 f'(x) = 60 − 4x；f'(15)=0，f''(x)=−4 < 0，所以是极大值。农民用这些尺寸得到了最大面积。
3. Case Study 2: Motion Along a Straight Line | 案例 2：直线运动
metres. Find the velocity and acceleration at time t, the times when the particle is at rest, and the total distance travelled in the first 5 seconds.
一质点沿直线运动，t 秒后它到原点的位移为 s(t) = t³ − 9t² + 24t 米。求任意时刻 t 的速度和加速度，质点静止的时刻，以及前 5 秒内经过的总路程。
3t² − 18t + 24 = 0 → divide by 3: t² − 6t + 8 = 0 → (t−2)(t−4)=0, so t = 2 s and t = 4 s.
速度 v(t) = ds/dt = 3t² − 18t + 24。加速度 a(t) = dv/dt = 6t − 18。质点静止时 v=0：解 3t² − 18t + 24 = 0 → 除以 3：t² − 6t + 8 = 0 → (t−2)(t−4)=0，因此 t = 2 秒和 t = 4 秒。
+48=20; s(4)=64−144+96=16; s(5)=125−225+120=20. Between 0 and 2 s, v>0 (moving forward); 2–4 s v<0 (moving backward); 4–5 s v>0. Distance = |s(2)−s(0)| + |s(4)−s(2)| + |s(5)−s(4)| = |20−0| + |16−20| + |20−16| = 20+4+4 = 28 metres.
求总路程需分析运动方向。s(0)=0，s(2)=20；s(4)=16；s(5)=20。0–2 秒间 v>0（前进）；2–4 秒 v<0（后退）；4–5 秒 v>0。路程 = |s(2)−s(0)| + |s(4)−s(2)| + |s(5)−s(4)| = 20+4+4 = 28 米。
4. Case Study 3: Exponential Growth of Bacteria | 案例 3：细菌的指数增长
A bacterial culture initially contains 500 cells and grows at a rate proportional to its size. After 2 hours, the population has increased to 800. Determine the population after 5 hours and the doubling time.
某种细菌培养液起初含有 500 个细胞，其增长速率与种群大小成正比。2 小时后数量增加到 800。求 5 小时后的数量以及翻倍时间。
00 e²ᵏ → e²ᵏ = 1.6. Take natural logs: 2k = ln 1.6 → k = ½ ln 1.6 ≈ 0.5 × 0.4700 = 0.2350.
模型为 P(t) = P₀ eᵏᵗ，其中 P₀ = 500。t=2 时 P(2)=800，故 800 = 500 e²ᵏ → e²ᵏ = 1.6。取自然对数：2k = ln 1.6 → k = ½ ln 1.6 ≈ 0.2350。
At t=5: P(5) = 500 e⁵ᵏ = 500 e5×0.235 ≈ 500 e1.175 ≈ 500 × 3.238 = 1619 cells. Doubling time T satisfies 2P₀ = P₀ eᵏᵀ → 2 = eᵏᵀ → T = (ln 2)/k ≈ 0.6931/0.2350 ≈ 2.95 hours.
t=5 时：P(5) = 500 e⁵ᵏ ≈ 500 × 3.238 = 1619 个细胞。翻倍时间 T 满足 2P₀ = P₀ eᵏᵀ → 2 = eᵏᵀ → T = (ln 2)/k ≈ 0.6931/0.2350 ≈ 2.95 小时。
Note that in Further Mathematics you might also solve using logₐ or recognise this as a geometric sequence if growth is discrete. Continuous exponential models are preferred for natural growth processes.
请注意，在进阶数学中你也可以用对数的一般形式来解决，或者如果增长是离散的，可将其视作几何数列。对于自然增长过程，连续指数模型更为常用。
5. Case Study 4: Mixing Chemicals – System of Equations | 案例 4：化学品混合——方程组求解
A chemist has two acid solutions: one with 20% concentration and another with 50% concentration. She needs 10 litres of a 35% acid solution. How many litres of each solution must she mix?
一位化学家有两种酸溶液：一种浓度为 20%，另一种浓度为 50%。她需要 10 升 35% 的酸溶液。每种溶液各需多少升？
Let x be the volume of 20% solution, y be the volume of 50% n total volume: x + y = 10. Total acid: 0.20x + 0.50y = 0.35 × 10 = 3.5. Solve the system.
设 x 为 20% 溶液的体积（升），y 为 50% 溶液的体积。则总体积：x + y = 10。酸总量：0.20x + 0.50y = 0.35 × 10 = 3.5。解该方程组。
) = 3.5 → 0.20x + 5 − 0.50x = 3.5 → −0.30x = −1.5 → x = 5. Then y = 5. So mix 5 L of each.
由第一个方程得 y = 10 − x。代入第二个：0.20x + 0.50(10 − x) = 3.5 → 0.20x + 5 − 0.50x = 3.5 → −0.30x = −1.5 → x = 5。于是 y = 5。因此每种混合 5 升。
Always interpret the context: both values are positive and sum to 10, which is realistic. This method can be extended to three or more mixtures, often solved by matrix methods in later studies.
记得联系实际：两个值均为正且总和为 10，符合实际。该方法可推广到三种或更多混合物，后期学习中常用矩阵方法求解。
6. Case Study 5: Calculating Height Using Trigonometry | 案例 5：利用三角学计算高度
From a point A on level ground, the angle of elevation to the top of a tower is 28°. After walking 150 metres directly towards the tower to point B, the angle of elevation becomes 42°. Find the height of the tower.
从水平地面上一点 A 测得塔顶的仰角为 28°。朝着塔的方向直走 150 米到达点 B 后，仰角变为 42°。求塔的高度。
A: tan 28° = h/(d + 150). Equate the two expressions for h: d tan 42° = (d + 150) tan 28°.
设塔高为 h 米，B 到塔底的距离为 d 米。则在 B 点：tan 42° = h/d → h = d tan 42°。在 A 点：tan 28° = h/(d + 150)。令 h 的两个表达式相等：d tan 42° = (d + 150) tan 28°。
17) = 150 × 0.5317 → 0.3687 d ≈ 79.755 → d ≈ 216.35 m. Then h = 216.35 × 0.9004 ≈ 194.8 m.
解出 d：d (tan 42° − tan 28°) = 150 tan 28°。利用 tan 42° ≈ 0.9004，tan 28° ≈ 0.5317，得到 d (0.9004 − 0.5317) = 150 × 0.5317 → 0.3687 d ≈ 79.755 → d ≈ 216.35 米。然后 h = 216.35 × 0.9004 ≈ 194.8 米。
Thus the tower is approximately 195 metres tall. This technique is widely used in surveying and requires careful handling of trigonometric ratios and identities.
因此塔高约为 195 米。此技术在测量中广泛应用，需要熟练处理三角比和恒等式。
7. Case Study 6: Minimising Production Cost Using Derivatives | 案例 6：利用导数最小化生产成本
.02q³ − 3q² + 200q + 5000. Determine the production level that minimises the cost, and find the minimum cost.
一家工厂发现生产 q 件小装置的总成本 C（千美元）由 C(q) = 0.02q³ − 3q² + 200q + 5000 给出。求使成本最小的产量以及最小成本。
To minimise C, differentiate and find critical points: C'(q) = 0.06q² − 6q + 200. Set C' − 6q + 200 = 0 → multiply by 50: 3q² − 300q + 10000 = 0 → divide by 1: q² − 100q + 3333.33… This is messy, so use quadratic formula: q = [6 ± √36 − 4×0.06×200]/(2×0.06) = [6 ± √36 − 48]/0.12. The discriminant is negative! Let's recalculate: 6²=36, 4×0.06×200 = 4×12=48, so 36-48 = -12. No real roots. This means C'(q) is always positive? Check: C'' es sign. So there is an inflection, but no stationary minimum. This is a poor example. Let's revise C(q) to have a minimum: e.g., C(q) = 0.02q³ − 1.8q² + 60q + 2000. Then C'(q)=0.06q² − 3.6q +60 = 0 → divide 0.06: q² − 60q + 1000 = 0 → (q-20)(q-40)? 20×40=800, not 1000. Better use C(q)=2q³ − 21q² + 60q + 100. Then C'=6q² − 42q +60 =0→ divided 6: q² −7q +10=0→ (q-2)(q-5)=0. q=2 or 5. C''=12q-42. At q=2, C''=−18 max; at q=5, C''=18 min. So min at q=5. Let's use that: total cost C(q) = 2q³ − 21q² + 60q + 100 (in $ rs. This avoids discriminant problems. I will rewrite the case to be mathematically sound.
求 C 的最小值，首先求导并找临界点：C'(q) = 6q² − 42q + 60。令 C'(q)=0：6q² − 42q + 60 = 0 → 除以 6：q² − 7q + 10 = 0 → (q−2)(q−5)=0。所以 q=2 或 q=5。用二阶导检验：C''(q) = 12q − 42，在 q=2 时 C''=−18 < 0（极大值），在 q=5 时 C''=18 > 0（极小值）。因此产量 q=5 时成本最小。最小成本 C(5)= 2×125 − 21×25 + 60×5 + 100 = 250 − 525 + 300 + 100 = 125 美元。
Thus the factory should produce 5 units to achieve the minimum cost of $125. This example shows how differentiation helps businesses optimise operations.
因此工厂应生产 5 件以达到最低成本 125 美元。该例子展示了微分如何帮助企业优化运营。
8. Case Study 7: Probability in a Carnival Game | 案例 7：游乐园游戏中的概率
In a carnival game, a player draws two cards from a standard deck of 52 without replacement. If both cards are aces, the player wins a big prize. If exactly one card is an ace, the player wins a small prize. Find the probability of winning each prize, and the expected number of big prize winners if 650 people play the game.
在一个游乐园游戏中，玩家从一副标准的 52 张扑克牌中无放回地抽取两张。如果两张都是 A，玩家赢得大奖；如果恰好一张是 A，赢得小奖。求每种奖的获奖概率，以及如果 650 人参与游戏，期望的大奖得主人数。
ability of exactly one ace: pick one ace from 4: C(4,1)=4; pick one non-ace from 48: C(48,1)=48; number of ways = 4×48=192. P(1 ace) = 192/1326 = 32/221 ≈ 0.1448.
两张 A 的概率：从 4 张 A 中选 2 张的方式：C(4,2)=6。总选牌方式：C(52,2)=1326。P(两张 A) = 6/1326 = 1/221 ≈ 0.00452。恰好一张 A 的概率：选一张 A：C(4,1)=4；选一张非 A：C(48,1)=48；方式数 = 4×48=192。P(一张 A) = 192/1326 = 32/221 ≈ 0.1448。
× (32/221) ≈ 94.1. Such calculations help the carnival set ticket prices to ensure profit.
650 人中大奖的期望人数 = 650 × (1/221) ≈ 2.94，即约 3 人获奖。小奖期望人数 ≈ 94.1。此类计算帮助游乐园设定票价以保证盈利。
9. Case Study 8: Savings Plan – Arithmetic and Geometric Sequences | 案例 8：储蓄计划——等差数列与等比数列
Liam wants to save money in two different ways. In Plan A, he deposits $50 the first month and increases the deposit by $5 each month (arithmetic). In Plan B, he deposits $50 the first month and increases the deposit by 2% each month (geometric). Compare the total saved after 12 months for both plans.
利亚姆想用两种方式存钱。方案 A：第一个月存 50 美元，之后每月存款额增加 5 美元（等差数列）。方案 B：第一个月存 50 美元，之后每月存款额增加 2%（等比数列）。比较 12 个月后两种方案的总存款额。
Last term l = a + (n−1)d = 50 + 11×5 = 105. Sum S₁₂ = n/2 (a + l) = 6 × (50+105) = 6×155 = $930.
方案 A：首项 a=50，公差 d=5，n=12。末项 l = a + (n−1)d = 50 + 11×5 = 105。总和 S₁₂ = n/2 (a + l) = 6 × (50+105) = 6×155 = 930 美元。
= 50 (1.02¹² − 1)/0.02. Compute 1.02¹² ≈ 1.2682, so S₁₂ ≈ 50 (0.2682)/0.02 = 50 × 13.41 = $670.50.
方案 B：首项 50，公比 r=1.02。等比数列求和 S₁₂ = a (rⁿ − 1)/(r − 1) = 50 (1.02¹² − 1)/0.02。计算 1.02¹² ≈ 1.2682，因此 S₁₂ ≈ 50 (0.2682)/0.02 = 50 × 13.41 = 670.50 美元。
Arithmetic plan saves more in this short term (930 vs 670.50). However, if the plan ran for many years, the geometric growth could overtake. Financial modelling often requires comparing such sequences.
短期来看，等差数列方案存得更多（930 对比 670.50）。但如果计划持续多年，等比增长可能会反超。金融建模常需要比较此类数列。
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