Case Study in Action: Applying Further Maths to Real Problems | 进阶数学案例实战：将知识应用于实际问题

📚 Case Study in Action: Applying Further Maths to Real Problems | 进阶数学案例实战：将知识应用于实际问题

In AQA Level 2 Certificate in Further Mathematics, you are often expected to apply pure mathematical techniques to unfamiliar, real-world scenarios. This article guides you through a detailed case study that combines optimisation, matrix transformations, trigonometric modelling, and sequence predictions – all typical of higher-level GCSE further maths questions. By working through each stage, you will sharpen both your problem-solving strategy and your exam readiness.

在 AQA 进阶数学证书考试中，你经常需要将纯数学技巧应用到不熟悉的真实场景中。本文将通过一个详细的案例研究，带你综合运用优化、矩阵变换、三角函数建模和数列预测——这些都是进阶数学的典型考点。逐步完成每个环节，你将提高问题解决策略，并为考试做好更充分的准备。

1. Setting Up the Case Study | 案例背景设定

A design agency has been asked to create a new product range for a client. The project includes three interlinked mathematical tasks: designing an open-top box with maximum volume from a fixed rectangular sheet, animating a logo using matrix transformations, and forecasting sales growth over twelve months. Each task draws on a different area of the Further Maths syllabus, making this an ideal integrated revision exercise.

一家设计机构接到客户要求，设计一个全新产品系列。该项目包含三个相互关联的数学任务：使用固定尺寸的矩形板材设计一个容积最大的无盖盒子；利用矩阵变换为标志制作动画效果；以及预测未来十二个月的销售增长。每项任务都涉及进阶数学大纲的不同领域，这使其成为绝佳的综合复习练习。

2. Optimising a Box Volume Using Calculus | 使用微积分优化盒子的体积

The packaging design starts with a rectangular sheet of card measuring 30 cm by 20 cm. A square of side x cm is cut from each corner, and the flaps are folded up to form an open box. The volume V in cubic centimetres depends on x.

包装设计从一张 30 cm × 20 cm 的矩形卡纸开始。从每个角剪去边长为 x cm 的正方形，再将四周折起制成无盖盒。体积 V（单位为立方厘米）取决于 x。

V = x(30 − 2x)(20 − 2x)

Expanding the brackets gives a cubic function: V(x) = 4x³ − 100x² + 600x. To find the maximum volume, we differentiate and set the derivative to zero.

展开括号得到三次函数：V(x) = 4x³ − 100x² + 600x。为求最大体积，我们对其求导并令导数为零。

V'(x) = 12x² − 200x + 600

Divide through by 4: 3x² − 50x + 150 = 0. Solving the quadratic using the formula gives the critical values.

两边除以 4：3x² − 50x + 150 = 0。使用求根公式解此二次方程得到临界值。

x = [50 ± √(2500 − 1800)] / (2 × 3) = [50 ± √700] / 6

So x ≈ (50 − 26.46)/6 ≈ 3.92 (the other root exceeds the sheet size). Always check the domain: 0 < x < 10 to keep dimensions positive.

即 x ≈ (50 − 26.46)/6 ≈ 3.92（另一个根超出纸张尺寸）。务必检查定义域：0 < x < 10 以确保尺寸为正。

3. Applying Matrix Transformations to a Logo | 对标志应用矩阵变换

The company logo is a triangle with vertices A(2,1), B(4,1) and C(3,3). The design brief requires a 45° rotation about the origin. The rotation matrix is given by:

公司标志是一个三角形，顶点为 A(2,1)、B(4,1) 和 C(3,3)。设计要求绕原点旋转 45°。旋转矩阵如下：

cos45°	−sin45°
sin45°	cos45°

Since cos45° = sin45° = √2/2, the matrix simplifies to:

因为 cos45° = sin45° = √2/2，矩阵简化为：

√2/2	−√2/2
√2/2	√2/2

Multiply this matrix by each position vector. For instance, for A(2,1) the image A’ is calculated as:

将此矩阵乘以每个位置向量。例如，对于 A(2,1)，其像 A’ 计算如下：

x’ = 2×(√2/2) + 1×(−√2/2) = √2/2 ≈ 0.71

y’ = 2×(√2/2) + 1×(√2/2) = 3√2/2 ≈ 2.12

Repeat for all vertices to obtain the rotated logo. In Further Maths exams, leaving answers in surd form is preferred unless a decimal approximation is requested.

对所有顶点重复此步骤以获得旋转后的标志。在进阶数学考试中，除非要求小数近似，否则最好保留根号形式。

4. Using Trigonometric Functions to Model Oscillation | 使用三角函数建模振荡

A mechanical part in the product oscillates, and its angular displacement θ (in degrees) after t seconds is modelled by:

产品中的一个机械部件会摆动，其 t 秒后的角位移 θ（以度为单位）由下式建模：

θ(t) = 30 sin(πt/2) + 15

The amplitude is 30°, the equilibrium position is raised by 15°, and the angular frequency is π/2 rad/s. The period T is found from 2π/ω:

振幅为 30°，平衡位置抬高了 15°，角频率为 π/2 rad/s。周期 T 可由 2π/ω 求得：

T = 2π ÷ (π/2) = 4 seconds

The maximum displacement is 15° + 30° = 45°, and the minimum is 15° − 30° = −15°. Understanding how to extract these parameters from a trigonometric model is a key skill.

最大位移为 15° + 30° = 45°，最小为 15° − 30° = −15°。理解如何从三角模型中提取这些参数是一项关键技能。

5. Predicting Sales Growth with Sequences and Binomial Expansion | 数列与二项式展开预测销售增长

Sales are predicted to grow by 5% each month. Starting with 1000 units in month 1, the sequence of monthly sales forms a geometric progression:

预计销售量每月增长 5%。从第一个月 1000 件开始，每月销售量形成一个等比数列：

U_n = 1000 × 1.05^(n−1)

To estimate month 12 sales without a calculator, we use the binomial expansion of (1 + 0.05)^11. Expanding up to the x³ term:

为在不使用计算器的情况下估算第 12 个月的销售量，我们使用 (1 + 0.05)^11 的二项式展开，并展开至 x³ 项：

(1 + x)^n ≈ 1 + nx + [n(n−1)/2]x² + [n(n−1)(n−2)/6]x³

With n = 11, x = 0.05:

1 + 11×0.05 = 1.55
+ (11×10)/2 × 0.0025 = 55 × 0.0025 = 0.1375
+ (11×10×9)/6 × 0.000125 = 165 × 0.000125 = 0.020625

Sum ≈ 1.708125, so U_12 ≈ 1708 units. The exact value from a calculator is 1000×1.05^11 ≈ 1710.34, showing the approximation is impressively close.

和约为 1.708125，因此 U_12 ≈ 1708 件。使用计算器得到的精确值为 1000×1.05^11 ≈ 1710.34，可见近似值非常接近。

6. Checking Stationary Points with Second Derivative | 用二阶导数检验驻点

Returning to the volume problem, we must confirm that x ≈ 3.92 yields a maximum. The second derivative is:

回到体积问题，我们必须确认 x ≈ 3.92 对应的是最大值。二阶导数为：

V”(x) = 24x − 200

At x = 3.92, V”(3.92) = 24×3.92 − 200 = 94.08 − 200 = −105.92, which is negative. This confirms a local maximum. The optimal side of the cut-out square is therefore 3.92 cm, giving a maximum volume.

在 x = 3.92 处，V”(3.92) = 24×3.92 − 200 = 94.08 − 200 = −105.92，结果为负。这确认了局部最大值。因此剪去正方形的最佳边长为 3.92 cm，从而得到最大体积。

V_max ≈ 3.92 × (30−7.84) × (20−7.84) ≈ 3.92 × 22.16 × 12.16 ≈ 1056 cm³

7. Matrix Multiplication and Inverse for Composite Transformations | 矩阵乘法与逆矩阵用于复合变换

Next, the design requires the logo to be first rotated by 45° and then scaled by a factor of 2. Scaling matrix S and rotation matrix R are:

接下来，设计要求标志先旋转 45°，再放大 2 倍。缩放矩阵 S 和旋转矩阵 R 为：

2	0
0	2

and R as before. The composite transformation is given by the product M = S × R. Matrix multiplication is not commutative, but here the order matches the sequence (rotate then scale), so we multiply S by R.

和之前的 R。复合变换由乘积 M = S × R 给出。矩阵乘法不可交换，但这里的顺序与操作序列匹配（先旋转后缩放），因此我们用 S 乘以 R。

M =

2	0
0	2

√2/2	−√2/2
√2/2	√2/2

√2	−√2
√2	√2

To reverse the transformation, we need the inverse matrix M⁻¹. Since det(M) = (√2)(√2) − (−√2)(√2) = 2 + 2 = 4, the inverse is:

为逆向变换，我们需要逆矩阵 M⁻¹。由于 det(M) = (√2)(√2) − (−√2)(√2) = 2 + 2 = 4，其逆为：

M⁻¹ = (1/4) ×

√2	√2
−√2	√2

8. Solving Trigonometric Equations in Context | 在上下文中解三角方程

When does the oscillating part reach an angle of 20° for the first time within one period (0 ≤ t ≤ 4)? We set θ(t) = 20 and solve.

摆动部件在一个周期内（0 ≤ t ≤ 4）何时首次达到 20° 角？我们设 θ(t) = 20 并求解。

30 sin(πt/2) + 15 = 20 → sin(πt/2) = 5/30 = 1/6

Therefore, πt/2 = arcsin(1/6) or π − arcsin(1/6). Using a calculator, arcsin(1/6) ≈ 0.1674 rad. The first positive solution is:

因此，πt/2 = arcsin(1/6) 或 π − arcsin(1/6)。使用计算器 arcsin(1/6) ≈ 0.1674 rad。第一个正解为：

t = 2 × 0.1674 / π ≈ 0.107 seconds

The second solution gives t ≈ 2 × (π − 0.1674) / π ≈ 1.89 s. In the exam, always check the range and list all possible solutions within it.

第二个解给出 t ≈ 2 × (π − 0.1674) / π ≈ 1.89 秒。考试中，务必检查定义域并列出其内所有可能的解。

9. Analysing Accuracy and Limitations | 分析准确性与局限性

Each mathematical model carries assumptions. The volume optimisation assumes zero cardboard thickness and perfect folds. The trigonometric oscillation ignores damping and friction. The sales forecast assumes a constant growth rate without market fluctuations. Recognising these limitations demonstrates deeper understanding and can earn marks in evaluation questions.

每个数学模型都带有假设条件。体积优化假设纸板厚度为零且折叠完美。三角函数振荡忽略阻尼和摩擦。销售预测假定增长率恒定，无市场波动。认识到这些局限性展现了更深入的理解，并能在评估题中得分。

Moreover, the binomial approximation is excellent for small x and moderate n, but errors grow if n is large or x is not small. Always discuss the domain of validity when using expansions.

此外，当 x 较小且 n 适中时二项式近似效果极佳，但若 n 很大或 x 并非很小，误差就会增大。使用展开时务必讨论其有效范围。

10. Summary of Key Skills and Exam Tips | 关键技能总结与考试技巧

This case study has integrated the following AQA Further Maths skills: differentiation and optimisation, matrix transformations and composites, solving trigonometric equations, geometric sequences and binomial expansion, and the use of the second derivative test. To succeed in exam problem-solving:

本案例综合演练了以下 AQA 进阶数学技能：微分与优化、矩阵变换及复合变换、解三角方程、等比数列与二项式展开、以及二阶导数检验。要在考试中成功解题：

Break the problem into manageable mathematical steps.
Clearly state any formulas before substituting numbers.
Keep exact values (surd form) until the final answer if possible.
Always verify that your solution fits the context (e.g., positive lengths).
Present matrix multiplications neatly and check the determinant when finding inverses.
In trigonometric problems, sketch a graph or use the CAST diagram to find all solutions.

将问题分解为可操作的数学步骤。
在代入数值前清晰写出所用公式。
尽可能保留精确值（根号形式）直到最后答案。
始终验证解是否符合实际背景（例如长度必须为正）。
矩阵乘法要书写整齐，求逆时检查行列式。
在三角问题中，画草图或使用 CAST 图找出所有解。

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Case Study in Action: Applying Further Maths to Real Problems | 进阶数学案例实战：将知识应用于实际问题