GCSE CIE Further Mathematics: Case Study Practice | GCSE CIE 进阶数学：案例分析实战演练

📚 GCSE CIE Further Mathematics: Case Study Practice | GCSE CIE 进阶数学：案例分析实战演练

Case study practice is one of the most effective ways to solidify your understanding of the CIE IGCSE Further Mathematics syllabus. By working through realistic, multi-step problems that connect topics such as calculus, matrices, vectors and probability, you learn to apply techniques in context and develop the problem-solving fluency required for top marks. This article presents a single extended case study based on a fictional technology firm, TechNova, taking you from initial modelling to final decision-making using a wide range of Further Mathematics skills.

案例分析实战是巩固 CIE IGCSE 进阶数学课程最有效的方法之一。通过逐步解决连接微积分、矩阵、向量和概率等多个主题的真实情境问题，你可以在实际背景中应用技巧，培养获得高分所必需的解题流畅度。本文以一个虚构的科技公司 TechNova 为背景，呈现一个完整的扩展案例分析，带你从初始建模到最终决策，运用多种进阶数学技能。

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1. Case Overview | 案例概述

TechNova produces two electronic devices: smartwatches (model A) and wireless earphones (model B). The company wants to maximise profit, allocate production resources efficiently, forecast seasonal sales, and design cost‑effective packaging. We will model these aspects using functions, solve equations, differentiate, integrate, use matrices and vectors, and assess risk with probability.

TechNova 生产两种电子设备：智能手表（型号 A）和无线耳机（型号 B）。公司希望最大化利润、高效分配生产资源、预测季节性销售、设计经济包装。我们将运用函数建模、解方程、微分、积分、矩阵、向量以及用概率评估风险等方法处理这些方面。

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2. Revenue, Cost and Profit Modelling | 收入、成本与利润建模

Let x be the number of smartwatches produced per week (in hundreds). The selling price per unit decreases linearly with quantity: revenue function R(x) = 200x – 2x² (in £1000). The total cost function is C(x) = 50x + 150 (in £1000). Therefore, the weekly profit (in £1000) is modelled by:

设 x 为每周生产的智能手表数量（以百台计）。单台售价随产量线性下降：收入函数为 R(x) = 200x – 2x²（单位：千英镑）。总成本函数为 C(x) = 50x + 150（单位：千英镑）。因此，周利润（千英镑）可用以下函数表示：

P(x) = R(x) – C(x) = –2x² + 150x – 150

We will use this quadratic profit function to find the optimum production level and perform break‑even analysis.

我们将利用这个二次利润函数求出最优产量并进行盈亏平衡分析。

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3. Differentiation for Profit Maximisation | 微分法求利润最大化

To maximise P(x), we differentiate and set the derivative to zero:

为最大化 P(x)，对其求导并令导数为零：

P'(x) = –4x + 150 = 0 ⇒ x = 37.5

This gives a critical point at x = 37.5 (i.e. 3750 smartwatches). To confirm it is a maximum, we check the second derivative:

这给出临界点 x = 37.5（即 3750 只手表）。为确认其为最大值，我们检查二阶导数：

P''(x) = –4 < 0 ⇒ maximum

Thus, producing 3750 smartwatches per week yields the maximum profit. The maximum profit value is P(37.5) = –2(37.5)² + 150(37.5) – 150 = 2662.5 (£1000), i.e. £2,662,500.

因此，每周生产 3750 只智能手表可获得最大利润。代入得最大利润值 P(37.5) = –2(37.5)² + 150(37.5) – 150 = 2662.5（千英镑），即 2,662,500 英镑。

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4. Break‑even Analysis with Quadratics | 二次方程盈亏平衡分析

Break‑even occurs when profit is zero: P(x) = 0. We solve:

盈亏平衡发生在利润为零时：P(x) = 0。解方程：

–2x² + 150x – 150 = 0 ⇔ 2x² – 150x + 150 = 0

Using the quadratic formula x = [150 ± √150² – 4×2×150] / (2×2):

运用求根公式 x = [150 ± √150² – 4×2×150] / (2×2)：

Discriminant Δ = 22500 – 1200 = 21300

x = (150 ± √21300) / 4 ≈ (150 ± 145.95) / 4

This gives x ≈ 1.01 and x ≈ 74.0. Therefore, the firm breaks even when producing about 101 or 7400 smartwatches per week. Production between these levels generates a profit.

解得 x ≈ 1.01 和 x ≈ 74.0。因此，当每周生产约 101 只或 7400 只智能手表时达到盈亏平衡。产量在这两个值之间时产生利润。

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5. Resource Allocation Using Matrices | 矩阵在资源分配中的应用

Now consider both products: let y be the number of earphone units (in hundreds) produced per week. Each smartwatch requires 2 labour hours and 3 units of raw material; each earphone takes 1.5 labour hours and 2 units of material. Available resources are 120 labour hours and 180 material units per week. This system can be written in matrix form:

现在考虑两种产品：设 y 为每周生产的耳机数量（以百台计）。每只手表需 2 个工时和 3 单位原材料；每副耳机需 1.5 个工时和 2 单位原材料。每周可用资源为 120 个工时和 180 单位原材料。该系统可用矩阵表示为：

⎡ 2 1.5 ⎤ ⎡ x ⎤ ⎡ 120 ⎤
⎣ 3 2 ⎦ ⎣ y ⎦ = ⎣ 180 ⎦

Let A = ⎡ 2 1.5 ⎤, X = ⎡ x ⎤, B = ⎡ 120 ⎤. To find X, we compute the inverse of A:

记 A = ⎡ 2 1.5 ⎤，X = ⎡ x ⎤，B = ⎡ 120 ⎤。为求 X，先计算 A 的逆矩阵：

det(A) = 2×2 – 1.5×3 = 4 – 4.5 = –0.5

A⁻¹ = 1/(–0.5) ⎡ 2 –1.5 ⎤ = –2 ⎡ 2 –1.5 ⎤ = ⎡ –4 3 ⎤
                   ⎣ –3 2 ⎦      ⎣ –3 2 ⎦   ⎣ 6 –4 ⎦

Then X = A⁻¹ B = ⎡ –4 3 ⎤ ⎡ 120 ⎤ = ⎡ –480 + 540 ⎤ = ⎡ 60 ⎤
                   ⎣ 6 –4 ⎦ ⎣ 180 ⎦   ⎣ 720 – 720 ⎦  ⎣ 0 ⎦

Thus x = 60, y = 0. This tells us that under these resource constraints, only smartwatches can be produced, at 6000 units per week. The firm must consider relaxing material constraints to build earphones profitably.

因此 x = 60，y = 0。这表明在当前资源约束下，只能生产智能手表，每周 6000 只。公司必须考虑放宽原材料限制才能盈利地生产耳机。

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6. Probability and Risk Assessment | 概率与风险评估

Market research indicates that demand for earphones next quarter follows a discrete probability distribution:

市场调研显示下季度耳机需求服从以下离散概率分布：

Demand (hundreds)	Probability
20	0.2
30	0.5
40	0.3

(hundreds), i.e. 3100 units. To quantify risk, we compute variance and standard deviation:

期望需求 E(D) = 20×0.2 + 30×0.5 + 40×0.3 = 4 + 15 + 12 = 31（百台），即 3100 台。为量化风险，我们计算方差和标准差：

Var(D) = (20–31)²×0.2 + (30–31)²×0.5 + (40–31)²×0.3
= 121×0.2 + 1×0.5 + 81×0.3 = 24.2 + 0.5 + 24.3 = 49

σ = √49 = 7

The standard deviation of 700 units indicates considerable variability. TechNova can use this to set safety stock levels.

标准差为 7 (即 700 台)，表明存在较大波动。TechNova 可利用此信息设定安全库存水平。

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7. Geometric Design: Packaging Optimisation | 几何设计：包装优化

TechNova plans a cylindrical box for the earphones. The volume must be 1000 cm³. Let radius r cm and height h cm. The volume constraint is:

TechNova 计划为耳机设计圆柱形包装盒。容积须为 1000 cm³。设底面半径 r cm，高 h cm。体积约束为：

V = πr²h = 1000 ⇒ h = 1000/(πr²)

The total surface area A (to minimise material cost) is:

总表面积 A（需最小化材料成本）为：

A = 2πr² + 2πrh = 2πr² + 2000/r

Differentiate and find the stationary point:

求导并找驻点：

dA/dr = 4πr – 2000/r² = 0 ⇒ 4πr³ = 2000 ⇒ r³ = 500/π

Hence r = ∛(500/π) ≈ 5.42 cm. Check minimum using second derivative:

因此 r = ∛(500/π) ≈ 5.42 cm。用二阶导数验证极小值：

d²A/dr² = 4π + 4000/r³ > 0 ⇒ minimum

Then h ≈ 1000/(π × 5.42²) ≈ 10.84 cm. The optimal can has radius about 5.4 cm and height about 10.8 cm.

随后 h ≈ 1000/(π × 5.42²) ≈ 10.84 cm。最优包装盒半径约 5.4 cm，高约 10.8 cm。

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8. Integration for Accumulated Change | 积分法求累积变化

The marginal cost function for earphone production is estimated as MC(y) = 0.3y² – 2y + 15 (in £ per hundred units). To find the total cost increase when output rises from 0 to 20 hundred units, we integrate:

耳机生产的边际成本函数估计为 MC(y) = 0.3y² – 2y + 15（单位：英镑/百台）。求产量从 0 增加到 20（百台）时总成本的增加量，我们进行积分：

ΔC = ∫₀²⁰ (0.3y² – 2y + 15) dy

Compute:

计算：

= [0.1y³ – y² + 15y]₀²⁰
= (0.1×8000 – 400 + 300) – 0
= 800 – 400 + 300 = 700

Thus, producing the first 2000 earphones costs an additional £700 (in the given units, this is £700). This integral represents the area under the marginal cost curve.

因此，生产首批 2000 副耳机的额外成本为 700 英镑（按给定单位）。该积分表示边际成本曲线下的面积。

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9. Trigonometric Functions in Seasonal Sales | 三角函数在季节性销售中的应用

Quarterly sales of smartwatches follow a sinusoidal pattern. The sales S(t) hundreds of units in quarter t are modelled by:

智能手表的季度销售量呈正弦规律。第 t 季度销售量 S(t)（百台）的模型为：

S(t) = 500 + 200 sin(πt/2)

To find the quarter with maximum sales, set the sine term to its maximum value 1:

为找到销售最高的季度，令正弦项取最大值 1：

sin(πt/2) = 1 ⇒ πt/2 = π/2 + 2kπ ⇒ t = 1 + 4k

Thus, sales peak in quarters 1, 5, 9, … (i.e. every first quarter). The maximum sales volume is 500 + 200 = 700 hundreds, or 70,000 units.

因此，销量在每第 1 季度达到峰值（即 t=1, 5, 9…），峰值为 500 + 200 = 700 百台，即 70,000 台。

If the firm wants to know when sales first drop to 400 hundreds, solve:

若公司想知道销量何时首次降至 400 百台，解方程：

500 + 200 sin(πt/2) = 400 ⇒ sin(πt/2) = –0.5

The principal solution for sin(θ) = –0.5 is θ = –π/6 or 7π/6. Using the positive angle, πt/2 = 7π/6 ⇒ t = 7/3 ≈ 2.33, meaning partway through quarter 2. This cyclical insight helps in inventory planning.

主值解 sin(θ)=–0.5 得 θ=–π/6 或 7π/6。取正角，πt/2 = 7π/6 ⇒ t = 7/3 ≈ 2.33，即在第 2 季度中段。这一周期性规律有助于库存规划。

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10. Vectors and Motion Analysis | 向量与运动分析

A delivery drone leaves the warehouse at point A(2, 3) and travels to store B(8, 7) (coordinates in km). The displacement vector is:

一架送货无人机从位于点 A(2, 3) 的仓库飞往商店 B(8, 7)（坐标单位：公里）。位移向量为：

AB = (8–2, 7–3) = (6, 4)

The straight‑line distance is |AB| = √6² + 4² = √52 = 2√13 ≈ 7.21 km. Suppose wind has velocity vector w = ( –1, 2 ) km/h, and the drone’s intended velocity relative to air is v = (15, 10) km/h. The resultant ground velocity is:

直线距离为 |AB| = √6² + 4² = √52 = 2√13 ≈ 7.21 km。假设风速向量为 w = ( –1, 2 ) km/h，无人机相对于空气的预定速度为 v = (15, 10) km/h。则合地面速度为：

v_ground = v + w = (14, 12)

The drone’s actual speed is |v_ground| = √14² + 12² = √196 + 144 = √340 ≈ 18.44 km/h. The angle between the drone's heading and the wind vector can be found via the dot product:

无人机实际速率为 |v_ground| = √14² + 12² = √340 ≈ 18.44 km/h。无人机航向与风向的夹角可通过点积求得：

v · w = 15×(–1) + 10×2 = –15 + 20 = 5

|v| = √15² + 10² = √325, |w| = √((–1)² + 2²) = √5

cos θ = 5 / (√325 × √5) = 5 / √1625 ≈ 5/40.31 ≈ 0.124 ⇒ θ ≈ 82.9°

The drone will need to adjust its heading to compensate for the wind, an application of vector addition directly relevant to the CIE syllabus.

无人机需调整航向以抵消风力影响，这一向量加法应用与 CIE 课程紧密相关。

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11. Synthesis and Decision Making | 综合与决策

Bringing the findings together: the profit‑maximising smartwatch output is 3750 units, while break‑even occurs at 101 and 7400 units. Matrix resource analysis shows earphone production is currently unviable without extra materials. Earphone demand is expected around 3100 units with a standard deviation of 700, suggesting moderate risk. Optimal packaging uses a radius of 5.4 cm, saving material. Seasonal peaks in Q1 guide marketing spend, and the delivery drone must account for wind drift.

综合各项结论：利润最大化下的智能手表产量为 3750 台，盈亏平衡点为 101 台和 7400 台。矩阵资源分析显示，在没有额外材料的情况下耳机生产尚不可行。耳机需求预期为 3100 台，标准差 700，风险适中。最优包装半径为 5.4 cm，节省材料。第一季度销量高峰指引营销投入，而送货无人机需考虑风偏。

These connected steps illustrate how CIE Further Mathematics techniques do not exist in isolation. The skills of modelling, calculus, algebra, vectors and probability must be woven together to solve real‑world business problems, exactly as exam questions increasingly expect.

这些环环相扣的步骤表明，CIE 进阶数学的技巧并不是孤立的。建模、微积分、代数、向量和概率等技能必须融合在一起，用以解决现实世界的商业问题，这也正是考试题目越发注重的能力要求。

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12. Practice Reflections and Further Steps | 练习反思与进阶步骤

After working through this case study, review each technique's clarity: can you differentiate a quadratic confidently? Do you remember the quadratic formula and discriminant? Are you able to set up and solve a matrix equation using the inverse? Practise similar problems with varying numbers to strengthen your fluency. The CIE exam often presents unstructured problems, so being comfortable translating a written scenario into mathematical language is key.

完成本案例分析后，回顾每项技术的清晰度：你能自信地对二次函数求导吗？还记得求根公式与判别式吗？你能建立并用逆矩阵求解矩阵方程吗？用不同数字练习类似问题以增强熟练度。CIE 考试常呈现非结构化题目，因此习惯于将文字情境转化为数学语言至关重要。

on to find consumer surplus. This will stretch your skills and build deeper understanding.

作为额外挑战，可调整案例：将利润函数改为三次函数，引入第三种产品并使用 3×3 矩阵，或用积分求消费者剩余。这将拓展你的技能并建立更深的理解。

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