📚 Year 10 CAIE Additional Mathematics Case Study Practical Exercises | CAIE 进阶数学案例分析实战演练
Case study problems form an essential part of the CAIE IGCSE Additional Mathematics syllabus. They challenge you to translate real-life scenarios into mathematical models, often combining algebra, calculus, and trigonometry. Mastering these problems requires a systematic approach: understand the problem, formulate equations, solve using appropriate techniques, and interpret the results. This article presents a series of practical exercises, guiding you through four complete case studies with detailed step‑by‑step reasoning, all aligned with Year 10 learning objectives.
案例分析问题是 CAIE IGCSE 进阶数学大纲的重要组成部分。这类问题要求你将现实场景转化为数学模型，通常综合运用代数、微积分和三角学。要掌握它们需要系统的方法：理解问题、建立方程、用适当的技术求解并解释结果。本文通过四个完整的案例进行实战演练，逐步给出详细推理过程，全部紧扣 Year 10 的学习目标。
1. Understanding Case Study Problems | 理解案例分析问题
In Additional Mathematics, case study problems typically appear as word problems with a real‑world context. Your first task is to read the entire problem carefully, identifying the known quantities, the unknowns, and any constraints. Highlight key information, such as fixed perimeters, given rates, or initial values, so that you can later build a correct mathematical model.
在进阶数学中，案例分析题通常以现实背景的文字题形式出现。你的首要任务是通读全题，仔细识别已知量、未知量以及所有约束条件。圈出关键信息，比如固定的周长、给定的速率或初始值，以便后续建立正确的数学模型。
2. Problem Formulation: Translating Words into Mathematics | 问题建模：将文字转化为数学
Once you understand the problem, the next step is to express it mathematically. Define appropriate variables, write down any equations relating them, and construct a function that needs to be optimised or analysed. By using the constraints to reduce the number of variables, you can often express the quantity of interest as a function of a single variable, ready for calculus techniques.
理解问题后，下一步是用数学语言表达出来。定义合适的变量，写出它们之间的关系式，并构造函数以便进行优化或分析。借助约束条件消去多余变量，你往往能将目标量表示为单一变量的函数，为微积分方法做好准备。
3. Case Study 1: Maximizing Area of a Rectangular Enclosure | 案例1：矩形围场面积最大化
A farmer has 100 metres of fencing and wishes to create a rectangular enclosure against a long straight wall. The wall forms one side of the enclosure, so only three sides require fencing. The goal is to determine the dimensions that maximise the enclosed area.
一位农民有 100 米围栏，想靠着一面很长的直墙围一个矩形场地。墙壁作为场地的一边，因此只需为另外三边设置围栏。目标是求出使围成面积最大化的尺寸。
Let the side perpendicular to the wall be x metres. The side parallel to the wall then uses the remaining fencing, so its length is (100 − 2x) metres. The area A (in square metres) can be expressed as A = x(100 − 2x) = 100x − 2x².
设垂直于墙的边长为 x 米，那么平行于墙的边长就用剩余的围栏，其长度为 (100 − 2x) 米。面积 A（平方米）可表示为 A = x(100 − 2x) = 100x − 2x²。
4. Solving Case 1 with Differentiation | 用微分法求解案例1
To find the maximum area, differentiate A with respect to x: dA/dx = 100 − 4x. Set the derivative equal to zero to locate stationary points: 100 − 4x = 0 → x = 25. The second derivative is d²A/dx² = −4, which is negative, confirming that the stationary point yields a maximum.
为求最大面积，对 x 求导：dA/dx = 100 − 4x。令导数为零以寻找驻点：100 − 4x = 0 → x = 25。二阶导数为 d²A/dx² = −4，为负值，确认该驻点对应最大值。
Thus, the optimal perpendicular side is 25 m. The parallel side is 100 − 2×25 = 50 m. Substituting back gives the maximum area A = 25 × 50 = 1250 m².
因此，最优的垂直边长为 25 米，平行边长为 100 − 2×25 = 50 米。代回得到最大面积 A = 25 × 50 = 1250 平方米。
5. Interpreting Results and Verifying | 结果解释与验证
It is always wise to check the endpoints of the feasible domain. Since both x and the parallel side must be positive, we require x > 0 and 100 − 2x > 0, giving 0 < x < 50. At x = 0, A = 0; as x → 50, A → 0. The stationary point at x = 25 therefore gives the global maximum. The answer is sensible because the enclosure uses half the fence for the parallel side, a classic result.
检验可行域的端点总是明智的。由于 x 和平行边都必须为正，我们有 x > 0 且 100 − 2x > 0，即 0 < x < 50。当 x = 0 时 A = 0；当 x → 50 时 A → 0。因此 x = 25 处的驻点对应全局最大值。这一结果合乎常理——围栏的一半用于平行边，是经典的优化结论。
6. Case Study 2: Kinematics of a Particle | 案例2：质点运动学
A particle moves along a straight line. Its displacement s metres from an origin at time t seconds is given by s = 2t³ − 15t² + 36t. We are asked to analyse its velocity, acceleration, and to find when it is at rest, its maximum displacement, and the total distance travelled in the first five seconds.
一质点沿直线运动，它在时刻 t 秒时相对于原点的位移 s 米由 s = 2t³ − 15t² + 36t 给出。我们需要分析其速度、加速度，求质点静止的时刻、最大位移以及前五秒内经过的总路程。
7. Applying Calculus to Displacement, Velocity, Acceleration | 应用微积分于位移、速度、加速度
Velocity v is the first derivative of displacement with respect to time: v = ds/dt = 6t² − 30t + 36. Acceleration a is the derivative of velocity: a = dv/dt = 12t − 30. For instance, at t = 1 s, v = 6 − 30 + 36 = 12 m/s, and a = 12 − 30 = −18 m/s², indicating the particle is slowing down in the positive direction.
速度 v 是位移对时间的一阶导数：v = ds/dt = 6t² − 30t + 36。加速度 a 是速度的导数：a = dv/dt = 12t − 30。例如，在 t = 1 秒时，v = 6 − 30 + 36 = 12 m/s，a = 12 − 30 = −18 m/s²，表明质点正在正方向上减速。
To find when the particle is instantaneously at rest, set v = 0: 6t² − 30t + 36 = 0 ⇒ t² − 5t + 6 = 0 ⇒ (t − 2)(t − 3) = 0, giving t = 2 s and t = 3 s.
要求质点瞬时静止的时刻，令 v = 0：6t² − 30t + 36 = 0 ⇒ t² − 5t + 6 = 0 ⇒ (t − 2)(t − 3) = 0，得 t = 2 s 和 t = 3 s。
8. Finding Maximum Displacement and Total Distance | 求最大位移与总路程
We examine the displacement at the rest times and at the boundaries. At t = 0, s = 0. At t = 2, s = 2(8) − 15(4) + 72 = 16 − 60 + 72 = 28 m. At t = 3, s = 2(27) − 15(9) + 108 = 54 − 135 + 108 = 27 m. The sign of v changes from positive to negative at t = 2, so s has a local maximum of 28 m. At t = 3, v changes from negative to positive, giving a local minimum of 27 m. To find the total distance travelled from t = 0 to t = 5, we first need s(5) = 2(125) − 15(25) + 180 = 250 − 375 + 180 = 55 m. The path is divided into segments: |28 − 0| + |27 − 28| + |55 − 27| = 28 + 1 + 28 = 57 m.
我们检查静止时刻和边界处的位移。t = 0 时 s = 0；t = 2 时，s = 2(8) − 15(4) + 72 = 16 − 60 + 72 = 28 m；t = 3 时，s = 2(27) − 15(9) + 108 = 54 − 135 + 108 = 27 m。在 t = 2 处速度由正变负，因此 s 有局部最大值 28 m；在 t = 3 处速度由负变正，s 有局部最小值 27 m。要求 t = 0 到 t = 5 的总路程，我们首先计算 s(5) = 2(125) − 15(25) + 180 = 250 − 375 + 180 = 55 m。路径分为若干段：|28 − 0| + |27 − 28| + |55 − 27| = 28 + 1 + 28 = 57 m。
9. Case Study 3: Exponential Growth of a Bacterial Culture | 案例3：细菌培养的指数增长
A bacterial culture initially contains 2000 bacteria and grows at a rate proportional to its population. After exactly
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