📚 Case Study Practice in AQA Further Maths | AQA 进阶数学：案例分析实战演练
Real-world problems often require you to take a mathematical model, interpret unfamiliar data, and apply multiple techniques from your AQA Level 2 Certificate in Further Mathematics. The following case studies blend pure, mechanics, statistics and discrete elements, giving you the chance to reason, justify and communicate mathematically – and building the skills assessed in the written papers.
现实世界的问题往往要求你使用数学模型、解读陌生数据，并综合运用 AQA Level 2 进阶数学的多项技术。下列案例融合了纯数学、力学、统计和离散元素，让你有机会进行数学推理、证明和交流，从而培养笔试中所考察的核心能力。
1. Optimising a Packaging Design | 包装设计的优化
A company wants to manufacture an open-top box from a rectangular sheet of cardboard measuring 30 cm by 20 cm. Small squares of side length x cm are cut from each corner, and the sides are folded up. The volume V (cm³) of the box is given by V = x(30 − 2x)(20 − 2x).
一家公司要用一块 30 cm × 20 cm 的长方形纸板制作无盖盒。从每个角切去边长为 x cm 的小正方形，再将四边折起。盒子的体积 V (cm³) 为 V = x(30 − 2x)(20 − 2x)。
The aim is to find the value of x that maximises V. Expand the expression: V = 4x³ − 100x² + 600x. Differentiate: dV/dx = 12x² − 200x + 600. Set dV/dx = 0 and solve 12x² − 200x + 600 = 0. Dividing by 4 gives 3x² − 50x + 150 = 0. Using the quadratic formula:
x = [50 ± √(2500 − 1800)] / 6 = [50 ± √700] / 6 ≈ [50 ± 26.46] / 6
目标是找出使 V 最大化的 x 值。展开表达式：V = 4x³ − 100x² + 600x。求导：dV/dx = 12x² − 200x + 600。令 dV/dx = 0，解 12x² − 200x + 600 = 0。除以 4 得 3x² − 50x + 150 = 0，使用求根公式。
The critical values are x ≈ 12.74 cm (reject, because 20 − 2x would be negative) and x ≈ 3.93 cm. To confirm it is a maximum, use the second derivative d²V/dx² = 24x − 200. At x ≈ 3.93, d²V/dx² < 0, so the volume is maximised. The optimum cut-out size is approximately 3.93 cm.
临界值为 x ≈ 12.74 cm（舍去，因为 20 − 2x 为负）和 x ≈ 3.93 cm。利用二阶导数 d²V/dx² = 24x − 200 验证其为最大值。当 x ≈ 3.93 时，d²V/dx² < 0，所以体积最大。最佳切割边长约为 3.93 cm。
From a case-study perspective, you also need to consider practical constraints: the material cost, the aesthetic proportion, and the fact that x must lie between 0 and 10. Always check the domain of your variable in modelling questions.
从案例分析的角度看，你还需要考虑实际约束：材料成本、外观比例，以及 x 必须在 0 与 10 之间。在建模题中务必检查变量的定义域。
2. Motion of a Particle with Variable Acceleration | 变加速运动的物体
A particle moves along a straight line so that its displacement s (m) from a fixed point at time t (s) is s = t³ − 9t² + 24t. This model describes a short journey of a robotic arm. The case study requires you to interpret when the particle changes direction and to calculate the total distance travelled between t = 0 and t = 5.
一质点沿直线运动，其相对于固定点的位移 s (m) 与时间 t (s) 的关系为 s = t³ − 9t² + 24t。此模型描述了一条机械臂的短程运动。该案例要求你判断质点何时改变运动方向，并计算在 t = 0 到 t = 5 之间经过的总路程。
Velocity v = ds/dt = 3t² − 18t + 24. Factorise: v = 3(t² − 6t + 8) = 3(t − 2)(t − 4). The particle is at rest when v = 0, i.e. at t = 2 and t = 4. By testing signs, v > 0 for t < 2, v < 0 for 2 < t < 4, and v > 0 for t > 4. So the particle changes direction at t = 2 and t = 4.
速度 v = ds/dt = 3t² − 18t + 24。因式分解：v = 3(t² − 6t + 8) = 3(t − 2)(t − 4)。质点静止时 v = 0，即 t = 2 和 t = 4。通过符号检验可知，t < 2 时 v > 0，2 < t < 4 时 v < 0，t > 4 时 v > 0。因此质点在 t = 2 和 t = 4 改变方向。
Find displacements: s(0) = 0, s(2) = 8 − 36 + 48 = 20, s(4) = 64 − 144 + 96 = 16, s(5) = 125 − 225 + 120 = 20. Total distance = |20 − 0| + |16 − 20| + |20 − 16| = 20 + 4 + 4 = 28 m.
计算各位移：s(0) = 0，s(2) = 8 − 36 + 48 = 20，s(4) = 64 − 144 + 96 = 16，s(5) = 125 − 225 + 120 = 20。总路程 = |20 − 0| + |16 − 20| + |20 − 16| = 20 + 4 + 4 = 28 m。
In a case-study report, you would explain that the robot arm advances, retreats slightly, then advances again – and the total wear on the track relates to the distance, not just the net displacement.
在案例研究报告中，你需要解释机械臂先前进、再略微后退、然后再前进——轨道总磨损与总路程有关，而不仅仅是净位移。
3. Matrix Transformations in Computer Graphics | 计算机图形中的矩阵变换
A logo design is stored as a set of coordinates: A(1, 2), B(3, 2), C(3, 1). The graphic designer wants to rotate the logo by 90° anticlockwise about the origin and then enlarge it by a scale factor of 2. The rotation matrix R and enlargement matrix E are:
一个标志设计以坐标集存储：A(1, 2), B(3, 2), C(3, 1)。图形设计师想将标志绕原点逆时针旋转 90°，再以比例因子 2 放大。旋转矩阵 R 和放大矩阵 E 分别为：
R = [0 −1; 1 0], E = [2 0; 0 2]
The combined transformation matrix is M = E R (applying rotation first, then enlargement): M = [2 0; 0 2] × [0 −1; 1 0] = [0 −2; 2 0].
组合变换矩阵 M = E R（先旋转后放大）：M = [2 0; 0 2] × [0 −1; 1 0] = [0 −2; 2 0]。
Apply M to each coordinate (write position vectors as columns): A’ = [0 −2; 2 0] × [1; 2] = [−4; 2] → (−4, 2); B’ = [0 −2; 2 0] × [3; 2] = [−4; 6]; C’ = [0 −2; 2 0] × [3; 1] = [−2; 6].
将 M 作用于每个坐标（将位置向量写成列向量）：A’ = [0 −2; 2 0] × [1; 2] = [−4; 2] → (−4, 2)；B’ = [0 −2; 2 0] × [3; 2] = [−4; 6]；C’ = [0 −2; 2 0] × [3; 1] = [−2; 6]。
You can see that the order of transformations matters. If enlargement had been applied first, the result would have been different: R E gives [0 −1; 1 0] × [2 0; 0 2] = [0 −2; 2 0] – coincidentally the same in this special case, but generally matrices do not commute. Always justify the order in your case study.
可以看到变换顺序至关重要。假如先放大，结果将不同：R E = [0 −1; 1 0] × [2 0; 0 2] = [0 −2; 2 0]——本特例中巧合相同，但一般而言矩阵乘法不可交换。在案例研究中务必论证顺序。
4. Modelling Seasonal Temperatures with Trigonometry | 用三角学模拟季节性温度
A meteorological office models the average monthly temperature T (°C) in a city using T = 14 + 6 sin(30t)°, where t is the month number (1 for January, 12 for December). The angle 30t is in degrees.
某气象局用公式 T = 14 + 6 sin(30t)° 模拟某城市月平均气温 T (°C)，其中 t 为月份编号（1 表示一月，12 表示十二月）。角度 30t 以度为单位。
Find the maximum temperature and the month in which it occurs. The sine function reaches its maximum of 1 when 30t = 90° + 360°k. For t between 1 and 12, 30t = 90 gives t = 3 (March). Max T = 14 + 6 × 1 = 20 °C. Minimum occurs when sin(30t) = −1: 30t = 270° → t = 9 (September), giving T = 8 °C.
求最高气温及其出现的月份。正弦函数在 30t = 90° + 360°k 时取得最大值 1。对于 1 ≤ t ≤ 12，30t = 90 得 t = 3（三月）。最高气温 = 14 + 6 × 1 = 20 °C。最低气温出现在 sin(30t) = −1 时：30t = 270° → t = 9（九月），气温为 8 °C。
Now consider a revised model T = 14 + 6 sin(30t − 60)°. Describe the transformation: this represents a phase shift of +60° in the angle, i.e. a translation of 2 months to the right. The peak now occurs when 30t − 60 = 90 → 30t = 150 → t = 5 (May). The model can be adapted to fit different climatic patterns.
现考虑修正模型 T = 14 + 6 sin(30t − 60)°。描述其变换：这表示角度提前 60°，即图像向右平移 2 个月。峰值现在出现在 30t − 60 = 90 → 30t = 150 → t = 5（五月）。该模型可调整以适应不同气候模式。
5. Using Functions to Model Profit and Breakeven | 利用函数模拟利润与盈亏平衡
A small business learns that its daily cost £C and revenue £R for producing x units of a gadget are C(x) = 0.2x² + 4x + 50 and R(x) = 10x − 0.1x². The profit function P(x) = R(x) − C(x).
某小企业了解到，生产 x 件小器具的每日成本 £C 和收入 £R 分别为 C(x) = 0.2x² + 4x + 50 和 R(x) = 10x − 0.1x²。利润函数 P(x) = R(x) − C(x)。
Simplify: P(x) = (10x − 0.1x²) − (0.2x² + 4x + 50) = −0.3x² + 6x − 50. To find breakeven points, solve P(x) = 0: −0.3x² + 6x − 50 = 0. Multiply by −10: 3x² − 60x + 500 = 0, or divide by 1: x = [60 ± √(3600 − 600)] / (0.6)?? Let’s use original: x = [−6 ± √(36 − 4(−0.3)(−50))] / (2(−0.3)) = [−6 ± √(36 − 60)] / (−0.6). Discriminant = −24, no real roots – so the business never breaks even under this model. The maximum profit is found by completing the square or differentiating: P‘(x) = −0.6x + 6 = 0 → x = 10, giving P(10) = −0.3(100) + 60 − 50 = −30 + 10 = −20. So maximum profit is −£20, meaning a minimum loss of £20 at x = 10. The case study advises changing the pricing or cost structure.
化简：P(x) = (10x − 0.1x²) − (0.2x² + 4x + 50) = −0.3x² + 6x − 50。求盈亏平衡点，解 P(x) = 0：−0.3x² + 6x − 50 = 0。判别式 = 36 − 60 = −24，无实根——在此模型下企业永无盈亏平衡。最大利润通过配方法或求导得到：P‘(x) = −0.6x + 6 = 0 → x = 10，得 P(10) = −30 + 10 = −20。因此最大“利润”为 −£20，即最低亏损 £20。该案例建议改变定价或成本结构。
6. Sequences and Loan Repayment Modelling | 数列与贷款偿还模型
A loan of £2000 is to be repaid in monthly instalments of £150, with interest charged at 1% per month on the outstanding balance. Let uₙ be the amount owed after n months. The recurrence relation is uₙ₊₁ = 1.01uₙ − 150, with u₀ = 2000.
一笔 £2000 的贷款按月偿还 £150，未偿还余额按月利率 1% 计息。设 uₙ 为 n 个月后的欠款额。递推关系为 uₙ₊₁ = 1.01uₙ − 150，u₀ = 2000。
Find a closed form. First, find the fixed point u* such that u* = 1.01u* − 150 → 0.01u* = 150 → u* = 15000. The general solution can be written as uₙ = u* + ( u₀ − u* ) × 1.01ⁿ = 15000 + (2000 − 15000) × 1.01ⁿ = 15000 − 13000 × 1.01ⁿ. This formula allows the company to predict when the loan will be paid off: set uₙ = 0 and solve 15000 − 13000 × 1.01ⁿ = 0 → 1.01ⁿ = 15000/13000 ≈ 1.1538. Taking logs: n = ln(1.1538) / ln(1.01) ≈ 0.1431 / 0.00995 ≈ 14.4. So after 15 months the loan will be fully repaid (with the 15th payment being slightly less than £150).
寻找闭式。首先求不动点 u* 满足 u* = 1.01u* − 150 → 0.01u* = 150 → u* = 15000。通解可写为 uₙ = u* + ( u₀ − u* ) × 1.01ⁿ = 15000 − 13000 × 1.01ⁿ。该公式使企业能够预测还清贷款的时间：设 uₙ = 0，解 15000 − 13000 × 1.01ⁿ = 0 → 1.01ⁿ ≈ 1.1538。取对数：n ≈ 14.4。因此 15 个月后贷款将全部还清（第 15 期还款略低于 £150）。
This case study highlights the power of sequences and logarithms in financial planning, directly linking to the AQA Further Maths requirement to use iterative and algebraic methods.
本案例突出了数列与对数在财务规划中的威力，直接对接 AQA 进阶数学中运用迭代和代数方法的要求。
7. Probability and Expected Value in a Game of Chance | 机会游戏中的概率与期望值
A fairground game costs £2 to play. A participant rolls two fair dice. If the sum is 7, they win £10; if the sum is 11 or 12, they win £5; otherwise, they win nothing. Let the random variable X be the net gain (winnings minus the cost to play).
一款游乐场游戏每次收费 £2。参与者掷两个公平骰子。若和为 7，赢得 £10；若和为 11 或 12，赢得 £5；否则无奖金。设随机变量 X 为净收益（赢款减去玩游戏的费用）。
First, find the probabilities: P(sum = 7) = 6/36 = 1/6; P(sum = 11 or 12) = (2+1)/36 = 3/36 = 1/12; P(other) = 27/36 = 3/4. Now construct the net gain: If sum=7, net gain = £10 − £2 = £8; sum=11/12 → £5 − £2 = £3; else → −£2.
计算概率：P(和为 7) = 6/36 = 1/6；P(和为 11 或 12) = (2+1)/36 = 3/36 = 1/12；P(其他) = 27/36 = 3/4。构造净收益：和为 7 时净收益 = £10 − £2 = £8；和为 11/12 → £5 − £2 = £3；其他 → −£2。
The expected net gain E(X) = 8×(1/6) + 3×(1/12) + (−2)×(3/4) = 8/6 + 3/12 − 6/4 = 4/3 + 1/4 − 3/2. Convert to twelfths: 16/12 + 3/12 − 18/12 = 1/12 ≈ £0.0833. The positive expected value means the organiser will, on average, lose about 8 pence per game – so from the player’s perspective the game is slightly favourable. A case study might recommend adjusting the prize or cost to ensure profitability.
期望净收益 E(X) = 8×(1/6) + 3×(1/12) + (−2)×(3/4) = 8/6 + 3/12 − 6/4 = 4/3 + 1/4 − 3/2。通分为十二分之：16/12 + 3/12 − 18/12 = 1/12 ≈ £0.0833。正的期望值意味着主办方平均每局亏损约 8 便士——因此从玩家角度看游戏略微有利。案例研究可能建议调整奖金或费用以确保盈利。
8. Geometric Reasoning with Coordinate Geometry | 用坐标几何进行几何推理
An engineer needs to check whether a fabricated bracket forms a right angle. Three points on the bracket are given as A(1, −3), B(4, y) and C(6, 5). The angle ABC must be 90° for safety specifications. Find the possible value(s) of y.
工程师需要检验一个制造支架是否形成直角。支架上三点为 A(1, −3), B(4, y) 和 C(6, 5)。安全规范要求 ∠ABC 必须为 90°。求 y 的可能值。
Vectors: BA = (1 − 4, −3 − y) = (−3, −3 − y); BC = (6 − 4, 5 − y) = (2, 5 − y). For perpendicular, dot product = 0: (−3)(2) + (−3 − y)(5 − y) = 0 → −6 + (−3 − y)(5 − y) = 0. Expand: (−3 − y)(5 − y) = −15 + 3y − 5y + y² = y² − 2y − 15. So equation: −6 + y² − 2y − 15 = 0 → y² − 2y − 21 = 0. Solve: y = [2 ± √(4 + 84)] / 2 = [2 ± √88] / 2 = 1 ± √22. Thus two possible positions for B exist, both satisfying the right-angle condition.
向量：BA = (1 − 4, −3 − y) = (−3, −3 − y)；BC = (6 − 4, 5 − y) = (2, 5 − y)。垂直条件点积为 0：(−3)(2) + (−3 − y)(5 − y) = 0 → −6 + (−3 − y)(5 − y) = 0。展开得 y² − 2y − 15，方程式化为 y² − 2y − 21 = 0。求解：y = [2 ± √88] / 2 = 1 ± √22。因此存在两个可能的 B 点位置，均满足直角条件。
This case study demonstrates that further mathematics techniques (vectors, dot product, algebraic manipulation) can directly verify manufacturing tolerances, merging pure mathematics with engineering contexts.
本案例表明，进阶数学技术（向量、点积、代数处理）能直接验证制造公差，将纯数学与工程情境融为一体。
9. Statistical Modelling: Correlation and Regression | 统计建模：相关与回归
A biologist measures the wing length L (mm) and body mass M (g) of 5 bird specimens. The data are: (12, 18), (14, 20), (16, 26), (18, 30), (20, 36). The case study asks you to model the relationship using a linear regression line of M on L and to predict the mass for a wing length of 17 mm.
一位生物学家测量了 5 只鸟标本的翅长 L (mm) 与体重 M (g)。数据为：(12, 18), (14, 20), (16, 26), (18, 30), (20, 36)。本案例要求你用 M 对 L 的线性回归线建模，并预测翅长 17 mm 时的体重。
Calculate means: L̄ = (12+14+16+18+20)/5 = 16; M̄ = (18+20+26+30+36)/5 = 26. Sum of products: Σ(Lᵢ − L̄)(Mᵢ − M̄) = (−4)(−8) + (−2)(−6) + (0)(0) + (2)(4) + (4)(10) = 32 + 12 + 0 + 8 + 40 = 92. Sum of squares: Σ(Lᵢ − L̄)² = 16 + 4 + 0 + 4 + 16 = 40. Slope b = 92 / 40 = 2.3. Intercept a = M̄ − b L̄ = 26 − 2.3×16 = 26 − 36.8 = −10.8. Regression line: M = −10.8 + 2.3L.
计算均值：L̄ = 16；M̄ = 26。乘积和：Σ(Lᵢ − L
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