CCEA Pre-U Further Maths: Case Study Practice | 案例分析实战演练

📚 CCEA Pre-U Further Maths: Case Study Practice | 案例分析实战演练

The Pre-U Further Mathematics specification from CCEA features a distinctive Case Study component that challenges students to apply a broad range of mathematical ideas to real‑world scenarios. Unlike tightly‑structured examination questions, the Case Study demands independent thinking, modelling skills and the ability to communicate mathematical reasoning clearly. In this article, we walk through four complete case studies, modelling everything from infectious diseases to investment portfolios, and provide bilingual commentary to help you master the technique.

CCEA的 Pre‑U 进阶数学考试中有一个独特的案例分析部分，它要求学生将广泛的数学工具应用于真实情境。与结构固定的常规试题不同，案例分析需要独立思考、建模能力和清晰的推理表达。本文将通过四个完整的案例——从传染病传播到投资组合——以中英双语详细演示解题过程，帮助你彻底掌握这一题型的技巧。

1. Overview of CCEA Pre-U Case Studies | CCEA Pre-U 案例研究概述

In the CCEA Pre-U Further Mathematics paper, the Case Study typically presents a multi‑paragraph scenario followed by a series of linked questions. The scenario may be drawn from mechanics, statistics, operational research or pure mathematics, often blending two or more areas. Marks are awarded for both the accuracy of the solution and the clarity of the mathematical argument, so thorough annotation is essential.

在CCEA Pre‑U进阶数学试卷中，案例分析通常给出一段多自然段的情境，随后提出一系列相互关联的问题。情境可能来自力学、统计学、运筹学或纯粹数学，且常常融合多个领域。评分既看解答的准确性，也看推理过程的清晰程度，因此详尽的注释至关重要。

Examiners frequently report that students lose marks by jumping straight into calculations without first defining variables, stating assumptions, and outlining the modelling approach. A structured plan – understand the problem, formulate a model, solve, interpret and validate – is the key to high scores.

考官常指出，学生常因未先定义变量、陈述假设并概述建模思路就直接计算而失分。一个结构化的流程——理解问题、建立模型、求解、解释并验证——是获取高分的关键。

2. Mathematical Modelling Framework | 数学建模框架

Every case study can be approached using the following five‑stage cycle:

每一个案例分析都可以按照以下五步循环来处理：

Define variables and parameters with units. | 定义变量与参数并注明单位。
Construct a mathematical model (equations, inequalities, constraints). | 构建数学模型（方程、不等式、约束条件）。
Solve using appropriate analytic or numerical techniques. | 求解，运用合适的解析方法或数值方法。
Interpret the solution in the original context and check whether it makes sense. | 解释，把解答放回原情境中并检验是否合理。
Validate by considering limitations, sensitivity to assumptions, and possible refinements. | 验证，考虑局限性、假设的敏感性以及可能的改进方向。

This framework will be used throughout the case studies that follow, and examiners expect to see evidence of each stage in your written response.

这一框架将贯穿后续所有案例，考官期望在你的书面回答中看到每一个步骤的证据。

3. Case Study 1: Epidemic Spread (Differential Equations) | 案例一：疫情传播（微分方程）

A town of 50 000 inhabitants experiences a novel virus outbreak. Public health officials model the spread using a logistic differential equation:

dP/dt = kP (1 − P/M)

where P is the number of infected individuals at time t (in days), M = 50000 is the carrying capacity (the whole population), and k is a constant to be determined. On day 10, there are 1200 cases and on day 20 there are 5800 cases. The task is to determine k, predict the number of cases after 30 days, and estimate when 40% of the population will have been infected.

某城镇有 50 000 居民，爆发了一种新型病毒。公共卫生官员使用逻辑斯谛微分方程来模拟传播过程：

dP/dt = kP (1 − P/M)

其中 P 是时刻 t（天）时的感染人数，M = 50000 是承载容量（即总人口），k 为待定常数。第 10 天有 1200 例，第 20 天有 5800 例。要求确定 k，预测第 30 天的感染人数，并估算何时 40% 的人口将被感染。

4. Case Study 1: Solution Walkthrough | 案例一：解题过程详解

Step 1 – Define variables. Let P(t) be the number of infected individuals at time t (days). The maximum population is M = 50 000.

第一步 – 定义变量。 令 P(t) 为 t（天）时刻的感染人数。最大人口数为 M = 50 000。

Step 2 – Model. The logistic equation is separable. We write

dP/[P(1 − P/M)] = k dt

第二步 – 建模。 逻辑斯谛方程是可分离的。写作

dP/[P(1 − P/M)] = k dt

Step 3 – Solve. Using partial fractions,

1/[P(1 − P/M)] = 1/P + 1/(M − P)

Integrating gives

ln P − ln(M − P) = kt + C

Thus

P/(M − P) = A eᵏᵗ, where A = eᶜ.

At t = 0, assume P(0) = P₀, then A = P₀/(M − P₀). However P₀ is unknown. We will determine k and P₀ using the given data points.

第三步 – 求解。 利用部分分式，

1/[P(1 − P/M)] = 1/P + 1/(M − P)

积分得

ln P − ln(M − P) = kt + C

因此

P/(M − P) = A eᵏᵗ，其中 A = eᶜ。

设 t = 0 时 P(0) = P₀，则 A = P₀/(M − P₀)。但 P₀ 未知，我们利用已知数据点来确定 k 和 P₀。

We have two data points: (t₁=10, P₁=1200) and (t₂=20, P₂=5800). Substituting into the solved form:

1200/(50000 − 1200) = A e¹⁰ᵏ → 1200/48800 = A e¹⁰ᵏ → 0.02459 ≈ A e¹⁰ᵏ

5800/(50000 − 5800) = A e²⁰ᵏ → 5800/44200 = A e²⁰ᵏ → 0.13122 ≈ A e²⁰ᵏ

已知两个数据点：(t₁=10, P₁=1200) 和 (t₂=20, P₂=5800)。代入求解式：

Dividing the second equation by the first eliminates A:

0.13122 / 0.02459 = e²⁰ᵏ / e¹⁰ᵏ = e¹⁰ᵏ → 5.335 = e¹⁰ᵏ

Thus 10k = ln 5.335 → k ≈ 0.1676 day⁻¹.

Substitute back: A = 0.02459 / e¹·⁶⁷⁶ ≈ 0.02459 / 5.335 ≈ 0.004609.

Now P₀/(50000 − P₀) = A => P₀ = 50000A / (1+A) ≈ 230. So initially about 230 people were infected.

两式相除以消去 A：

0.13122 / 0.02459 = e²⁰ᵏ / e¹⁰ᵏ = e¹⁰ᵏ → 5.335 = e¹⁰ᵏ

因此 10k = ln 5.335 → k ≈ 0.1676 天⁻¹。

代回：A = 0.02459 / e¹·⁶⁷⁶ ≈ 0.02459 / 5.335 ≈ 0.004609。

于是 P₀/(50000 − P₀) = A ⇒ P₀ = 50000A / (1+A) ≈ 230。所以最初约有 230 人感染。

Prediction for t = 30 days:

P(30)/(50000 − P(30)) = 0.004609 × e⁰·¹⁶⁷⁶ˣ³⁰ = 0.004609 × e⁵·⁰²⁸ ≈ 0.004609 × 152.4 ≈ 0.7025

Solving gives P(30) ≈ 50000 × 0.7025 / 1.7025 ≈ 20630 cases.

预测 t = 30 天：

P(30)/(50000 − P(30)) = 0.004609 × e⁰·¹⁶⁷⁶ˣ³⁰ = 0.004609 × e⁵·⁰²⁸ ≈ 0.004609 × 152.4 ≈ 0.7025

解得 P(30) ≈ 50000 × 0.7025 / 1.7025 ≈ 20630 例。

Time to reach 40% infected (P = 20000):

20000/30000 = A eᵏᵗ → 0.6667 = 0.004609 e⁰·¹⁶⁷⁶ᵗ → e⁰·¹⁶⁷⁶ᵗ = 144.6 → 0.1676 t = ln 144.6 ≈ 4.974 → t ≈ 29.7 days.

达到 40% 感染所需时间（P = 20000）：

20000/30000 = A eᵏᵗ → 0.6667 = 0.004609 e⁰·¹⁶⁷⁶ᵗ → e⁰·¹⁶⁷⁶ᵗ = 144.6 → 0.1676 t = ln 144.6 ≈ 4.974 → t ≈ 29.7 天。

Interpretation: The model suggests a rapid rise, with nearly half the town infected by day 30. In reality, public health interventions would change k, so the model is only valid for the early, unimpeded phase.

解释： 模型显示感染快速上升，到第 30 天将近一半的镇民被感染。现实中，公共卫生干预会改变 k 值，因此该模型仅适用于早期未受抑制的阶段。

5. Case Study 2: Structural Forces (Vectors & Mechanics) | 案例二：结构受力分析（向量与力学）

A heavy uniform beam AB of length 10 metres and weight 800 N rests in equilibrium with the end A on rough horizontal ground and the end B against a smooth vertical wall. The beam makes an angle of 60° with the horizontal. A load of 600 N is placed at a point 3 m from B. Determine the reaction forces at A and B, and the minimum coefficient of friction at A to prevent slipping.

一根长 10 米、重 800 N 的均匀梁 AB 处于平衡状态，A 端置于粗糙水平地面上，B 端靠在光滑竖直墙上。梁与水平面成 60° 角。距 B 端 3 米处放有 600 N 的负载。求 A 端和 B 端的反力，以及防止 A 端滑动的最小摩擦系数。

6. Case Study 2: Solving Equilibrium | 案例二：求解平衡条件

Step 1 – Define variables. Let R_Ax, R_Ay be the horizontal and vertical reactions at A, and R_B be the normal reaction at B (wall is smooth, so only horizontal). The beam’s weight 800 N acts at its centre, 5 m from each end. The 600 N load acts 3 m from B, thus 7 m from A.

第一步 – 定义变量。 设 R_Ax, R_Ay 为 A 端的水平和竖直反力，R_B 为 B 端的法向反力（墙光滑，只有水平力）。梁自重 800 N 作用在中心，距两端各 5 m。600 N 负载作用在距 B 端 3 m 处，即距 A 端 7 m。

Step 2 – Resolve forces horizontally and vertically.

Horizontally: R_Ax − R_B = 0 → R_Ax = R_B.

Vertically: R_Ay − 800 − 600 = 0 → R_Ay = 1400 N.

第二步 – 水平和竖直方向分解。

水平方向：R_Ax − R_B = 0 → R_Ax = R_B。

竖直方向：R_Ay − 800 − 600 = 0 → R_Ay = 1400 N。

Step 3 – Take moments about A (to eliminate A reactions). The perpendicular distance of each force from A must be considered. Place A at the origin, with the beam along a line at 60°. The horizontal, vertical geometry is easier using components. Better: use the fact that moment = force × perpendicular distance along the horizontal/vertical plane. If we resolve forces into horizontal and vertical components, the moment about A is sum of (force × lever arm). The lever arm for a vertical force is the horizontal distance from A, and for a horizontal force it is the vertical distance from A.

第三步 – 对 A 点取矩（消除 A 的反力）。 需考虑每个力到 A 的垂直距离。将 A 设为原点，梁沿 60° 方向。利用水平和垂直分量计算更方便：力矩 = 力 × 到 A 的垂直距离（水平或竖直方向）。垂直力的力臂是到 A 的水平距离，水平力的力臂是到 A 的竖直距离。

Beam length 10 m, angle 60° → horizontal projection of beam length = 10 cos60° = 5 m, vertical projection = 10 sin60° = 5√3 ≈ 8.66 m. Centre of beam is at midpoint: horizontal distance from A = 5 cos60° = 2.5 m, vertical distance = 5 sin60° ≈ 4.33 m. The 600 N load is 7 m from A along the beam: horizontal distance from A = 7 cos60° = 3.5 m, vertical = 7 sin60° ≈ 6.06 m. The reaction at B, R_B, is horizontal, so its line of action is at B: horizontal distance from A = 10 cos60° = 5 m, vertical from A = 10 sin60° = 8.66 m. The moment due to R_B about A is R_B × vertical distance (since horizontal force) = R_B × 8.66, tending to rotate clockwise (negative). Taking anticlockwise as positive, the weight and load produce clockwise moments (negative). So summing moments about A:

R_B × 8.66 − 800 × 2.5 (horizontal distance for vertical weight) − 600 × 3.5 = 0 ? Wait, weight is vertical, so its moment arm is horizontal distance from A: 800 N at 2.5 m → moment = 800 × 2.5 (clockwise, negative). 600 N at 3.5 m → 600 × 3.5 (clockwise, negative). R_B is horizontal, so its moment arm is vertical distance from A: 8.66 m, and it creates anticlockwise moment (positive). Equation:

8.66 R_B − 800×2.5 − 600×3.5 = 0 → 8.66 R_B = 2000 + 2100 = 4100 → R_B ≈ 473.5 N.

Thus R_Ax = R_B ≈ 473.5 N.

梁长 10 m，夹角 60° → 梁的水平投影 = 10 cos60° = 5 m，垂直投影 = 10 sin60° = 5√3 ≈ 8.66 m。梁的中心在中点：到 A 的水平距离 = 5 cos60° = 2.5 m，垂直距离 = 5 sin60° ≈ 4.33 m。600 N 负载位于距 A 沿梁 7 m 处：水平距离 = 7 cos60° = 3.5 m，垂直 = 7 sin60° ≈ 6.06 m。B 端反力 R_B 为水平方向，作用点在 B：到 A 的水平 = 5 m，垂直 = 8.66 m。R_B 对 A 的力矩为 R_B × 垂直距离（因为是水平力）= R_B × 8.66，趋向逆时针（正方向）。重量和负载产生顺时针力矩（负）。取逆时针为正，对 A 取矩：

8.66 R_B − 800×2.5 − 600×3.5 = 0 → 8.66 R_B = 2000 + 2100 = 4100 → R_B ≈ 473.5 N。

因此 R_Ax = R_B ≈ 473.5 N。

Step 4 – Minimum friction coefficient. At A, the horizontal friction force is R_Ax, and the normal reaction is R_Ay = 1400 N. For limiting equilibrium, friction = μ R_Ay, so μ_min = R_Ax / R_Ay = 473.5 / 1400 ≈ 0.338.

第四步 – 最小摩擦系数。 A 端水平摩擦力为 R_Ax，法向反力为 R_Ay = 1400 N。极限平衡时，摩擦力 = μ R_Ay，故 μ_min = R_Ax / R_Ay = 473.5 / 1400 ≈ 0.338。

Interpretation: If the coefficient of friction is less than 0.34, the beam will slip. The result seems plausible, and the assumptions (smooth wall, uniform beam) are typical.

解释： 若摩擦系数小于 0.34，梁将发生滑动。结果合理，所用假设（光滑墙、均匀梁）均为典型情况。

7. Case Study 3: Investment Portfolio (Linear Programming) | 案例三：投资组合（线性规划）

An investor has £60 000 to allocate between three types of investment: low‑risk bonds, medium‑risk stocks and high‑risk equities. Expected annual returns are 3%, 6% and 10% respectively. Risk ratings (on a scale 1–10) are 1, 4 and 8. Regulations require that at least 20% of the total amount must be in bonds, no more than 50% in equities, and the average risk rating of the portfolio must not exceed 5. The objective is to maximise annual return. Formulate and solve this linear programming problem.

某投资者有 60 000 英镑可分配于三种投资：低风险债券、中风险股票和高风险股权。预期年收益分别为 3%、6% 和 10%。风险评级（1–10 分制）分别为 1、4 和 8。规定要求至少 20% 的总资金须投入债券，股权投入不得超过 50%，且投资组合的平均风险评级不得超过 5。目标为最大化年收益。请建立并求解此线性规划问题。

8. Case Study 3: Graphical & Simplex Methods | 案例三：图解法与单纯形法

Step 1 – Define decision variables. Let x₁, x₂, x₃ be the amounts (in £) invested in bonds, stocks and equities respectively. All ≥ 0.

第一步 – 定义决策变量。 令 x₁、x₂、x₃ 分别为投资于债券、股票和股权的金额（英镑）。均 ≥ 0。

Step 2 – Constraints.

Total investment: x₁ + x₂ + x₃ = 60 000

Minimum bonds: x₁ ≥ 0.2 × 60 000 = 12 000

Maximum equities: x₃ ≤ 0.5 × 60 000 = 30 000

Average risk: (1x₁ + 4x₂ + 8x₃) / (x₁ + x₂ + x₃) ≤ 5. Since total = 60 000, this simplifies to x₁ + 4x₂ + 8x₃ ≤ 300 000 (multiply both sides by 60 000).

第二步 – 约束条件。

总投资额：x₁ + x₂ + x₃ = 60 000

最低债券比例：x₁ ≥ 12 000

最高股权比例：x₃ ≤ 30 000

平均风险：(x₁ + 4x₂ + 8x₃) / 60 000 ≤ 5，两边乘 60 000 得 x₁ + 4x₂ + 8x₃ ≤ 300 000。

Step 3 – Objective. Maximise total return R = 0.03x₁ + 0.06x₂ + 0.10x₃.

第三步 – 目标函数。 最大化总收益 R = 0.03x₁ + 0.06x₂ + 0.10x₃。

To reduce to two variables, use the equality constraint: x₂ = 60 000 − x₁ − x₃. Substitute into the inequality constraints and objective:

x₁ ≥ 12 000; x₃ ≤ 30 000; and x₂ ≥ 0 → x₁ + x₃ ≤ 60 000.

Risk constraint: x₁ + 4(60 000 − x₁ − x₃) + 8x₃ ≤ 300 000 → −3x₁ + 4x₃ ≤ 60 000 (since x₁ + 240 000 − 4x₁ − 4x₃ + 8x₃ = −3x₁ + 4x₃ + 240 000 ≤ 300 000, so −3x₁ + 4x₃ ≤ 60 000).

Objective becomes: R = 0.03x₁ + 0.06(60 000 − x₁ − x₃) + 0.10x₃ = 3600 − 0.03x₁ + 0.04x₃. So we need to maximise R = 3600 + 0.04x₃ − 0.03x₁, subject to:

x₁ ≥ 12 000, x₃ ≤ 30 000, x₁ + x₃ ≤ 60 000, −3x₁ + 4x₃ ≤ 60 000, x₁, x₃ ≥ 0.

利用等式约束消元：x₂ = 60 000 − x₁ − x₃。代入不等式和目标函数：

x₁ ≥ 12 000；x₃ ≤ 30 000；且 x₂ ≥ 0 → x₁ + x₃ ≤ 60 000。

风险约束：x₁ + 4(60 000 − x₁ − x₃) + 8x₃ ≤ 300 000 → −3x₁ + 4x₃ ≤ 60 000。

目标函数变为：R = 0.03x₁ + 0.06(60 000 − x₁ − x₃) + 0.10x₃ = 3600 − 0.03x₁ + 0.04x₃。因此需 最大化 R = 3600 + 0.04x₃ − 0.03x₁，满足：

x₁ ≥ 12 000，x₃ ≤ 30 000，x₁ + x₃ ≤ 60 000，−3x₁ + 4x₃ ≤ 60 000，x₁, x₃ ≥ 0。

Step 4 – Graphical solution. Plot the feasible region in the (x₁, x₃) plane. The line −3x₁ + 4x₃ = 60 000 passes through (0, 15000) and (−20000,0) – only the portion with x₁ ≥ 0, x₃ ≥ 0 matters. The line x₁ + x₃

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CCEA Pre-U Further Maths: Case Study Practice | 案例分析实战演练