📚 GCSE AQA Further Maths: Case Study Practice | GCSE AQA 进阶数学：案例分析实战演练
In the AQA Level 2 Certificate in Further Mathematics, the highest marks are reserved for questions that force you to combine several topics inside one multi-step context. Such questions rarely say "differentiate this function"; instead they describe a real situation and expect you to build the model, choose the technique, justify the answer and interpret it in context. This article works through four extended case studies - optimisation, kinematics, matrix transformations and proof - and breaks each one into clear, examinable steps with paired English and Chinese explanations.
在 AQA 进阶数学二级证书考试中，最高分往往留给那些要求你在同一个多步骤情境中综合运用多个知识点的题目。这类题很少直接说 "对这个函数求导"，而是描述一个真实情境，要求你自己建立模型、选择方法、给出理由，并把结果放回情境中解释。本文通过四个完整案例——最优化、运动学、矩阵变换与证明——逐步拆解解题过程，并配有中英文对照讲解。
1. Why Case Study Questions Matter | 为什么案例分析题如此重要
Case study questions test transfer, not recall. The examiner assumes you already know how to differentiate a cubic, form a determinant or complete an algebraic proof; what is being assessed is whether you can recognise which of those tools the situation demands, and whether you can chain them together without losing accuracy along the way. A single slip in the model at step one usually destroys every later mark, so careful modelling is worth as much as fluent algebra.
案例分析题考查的是迁移能力，而不是记忆能力。考官默认你已经会求三次函数的导数、会计算行列式、会完成代数证明；真正被考查的，是你能否看出这个情境需要哪一种工具，以及能否把它们连贯地串起来而不在中途出错。第一步建模一旦出错，后面几乎所有分数都会丢失，因此严谨建模与熟练运算同等重要。
The four cases below cover the four strands that appear most often in AQA Further Maths case study questions:
下面四个案例覆盖了 AQA 进阶数学案例分析题中最常出现的四条主线：
- Calculus applied to a geometric model (optimisation). | 微积分应用于几何模型（最优化）。
- Calculus applied to motion (kinematics). | 微积分应用于运动（运动学）。
- Matrix algebra applied to transformations and simultaneous equations. | 矩阵代数应用于变换与联立方程。
- Algebraic proof with general integers. | 使用一般整数的代数证明。
Each case follows the same four-stage discipline: model, differentiate or manipulate, justify, interpret.
每个案例都遵循同样的四步纪律：建模、求导或变形、论证、解释。
2. Case Study 1: Modelling the Open Box | 案例一：无盖盒子的建模
A square piece of cardboard measures 30 cm by 30 cm. Small squares of side length x cm are cut from each corner, and the sides are folded up to form an open box. Our goal is to find the value of x that maximises the volume of the box.
一张正方形纸板，边长为 30 厘米。从每个角剪去边长为 x 厘米的小正方形，然后将四边折起，制成一个无盖盒子。目标是求使盒子容积最大的 x 值。
After cutting, the base is a square of side (30 − 2x) cm and the height of the box is x cm, so the volume V in cm³ is:
剪切之后，底面是边长为 (30 − 2x) 厘米的正方形，盒子的高为 x 厘米，因此容积 V（单位 cm³）为：
V = x(30 − 2x)²
area and at x > 15 the cardboard cannot be folded into a box at all. Recording this domain is not optional: it is worth a mark on its own and it decides which stationary point you eventually accept.
只有当底面边长为正时，模型才有物理意义，即 0 < x < 15。当 x = 15 时底面积为零，x > 15 时纸板根本无法折成盒子。写出这个定义域不是可选项：它本身通常占一分，而且决定了你最终接受哪个驻点。
Expanding the bracket gives a cubic, and writing it in descending powers makes differentiation easier:
展开括号得到一个三次函数，按降幂排列便于求导：
V = x(900 − 120x + 4x²) = 4x³ − 120x² + 900x
This is the function we now maximise.
这就是我们接下来要求最大值的函数。
3. Case Study 1: Differentiation and Critical Points | 案例一：求导与临界点
Differentiating term by term with the power rule gives the first derivative:
用幂法则逐项求导，得到一阶导数：
dV/dx = 12x² − 240x + 900
Taking out the common factor 12 leaves a quadratic that factorises neatly:
提取公因式 12 后，剩下的二次式可以整齐地分解因式：
dV/dx = 12(x² − 20x + 75) = 12(x − 5)(x − 15)
e x = 15 gives a volume of zero and is a boundary case rather than a maximum of the model.
令 dV/dx = 0 得 x = 5 或 x = 15。两者都是这个三次函数的驻点，但只有 x = 5 严格落在定义域 0 < x < 15 之内；x = 15 对应的容积为零，属于边界情形，而不是模型的最大值。
A quick numerical check confirms that the cubic really does peak near x = 5:
做一个简单的数值检查，可以确认该三次函数确实在 x = 5 附近取到峰值：
| x (cm) | Base side (30 − 2x) | V = x(30 − 2x)² |
| 4 | 22 | 1936 |
| 5 | 20 | 2000 |
| 6 | 18 | 1944 |
ise to 2000 at x = 5 and fall again, which is exactly the behaviour a maximum should show.
数值先升到 x = 5 时的 2000，然后回落，这正是极大值应有的表现。
4. Case Study 1: Second Derivative Test and Conclusion | 案例一：二阶导数检验与结论
To prove rigorously that x = 5 gives a maximum rather than a minimum, differentiate again:
要严格证明 x = 5 给出的是极大值而非极小值，再求一次导：
d²V/dx² = 24x − 240
ves 24(15) − 240 = +120, a minimum, which agrees with the observation that the volume collapses to zero at the edge of the domain.
代入 x = 5，得 24(5) − 240 = −120，为负，说明曲线开口向下，该驻点是极大值点。代入 x = 15，得 24(15) − 240 = +120，为极小值，这与"容积在定义域边缘塌缩为零"的观察一致。
The maximum volume is then found by substituting back into the original model, not into the derivative:
接着把 x 代回原始模型（而不是代回导数式）求最大容积：
V = 5 × (30 − 10)² = 5 × 400 = 2000 cm³
sion in context, namely that a 5 cm square should be cut from each corner to give a maximum volume of 2000 cm³.
一个完整的答案要写清三点：x = 5 厘米这个值、推理过程（dV/dx = 0 且 d²V/dx² < 0）、以及结合情境的结论，即每个角应剪去边长 5 厘米的正方形，最大容积为 2000 cm³。
5. Case Study 2: Kinematics - Velocity and Acceleration | 案例二：运动学——速度与加速度
Velocity is the derivative of displacement and acceleration is the derivative of velocity.
一个质点沿直线运动，在 t 秒时距离固定点 O 的位移为 s 米，关系式为 s = t³ − 6t² + 9t + 2，其中 0 ≤ t ≤ 4。速度是位移的导数，加速度是速度的导数。
v = ds/dt = 3t² − 12t + 9 = 3(t − 1)(t − 3)
a = dv/dt = 6t − 12
g a = 0 gives t = 2, where the velocity reaches its minimum value of 3(4) − 24 + 9 = −3 m/s.
令 v = 0 得 t = 1 和 t = 3，因此质点在这两个时刻瞬时静止；在这段时间之间它沿负方向运动。令 a = 0 得 t = 2，此时速度取最小值 3(4) − 24 + 9 = −3 m/s。
