📚 Year 9 AQA Further Maths: Case Study Practice Drills | Year 9 AQA 进阶数学：案例分析实战演练
In Further Maths, applying concepts to real-world situations is essential for deep understanding and exam success. This article presents a series of case study drills that mirror the style and challenge of AQA questions. Each case study integrates multiple topics—algebra, geometry, statistics and more—requiring you to select the right methods and communicate reasoning clearly. Work through these examples to sharpen your problem-solving skills and build confidence.
在进阶数学中，将概念应用于实际情境对于深入理解和考试成功至关重要。本文呈现一系列案例分析演练，模拟 AQA 试题的风格与挑战。每个案例融合了多个主题——代数、几何、统计等——要求你选择合适的解题方法并清晰地表达推理过程。通过这些例题的训练，你将提升问题解决能力并建立信心。
1. Case Study 1 – Maximising Profit with Quadratics | 用二次函数求利润最大化
A small business finds its weekly profit £ ant to know the maximum possible profit and how many units they should produce to achieve it.
一家小型企业发现其每周销售 x 件的利润 £P 由 P(x) = -2x² + 120x - 500 模拟。董事会想知道可能获得的最大利润以及应生产多少件才能实现。
We recognise this as a quadratic function with a negative coefficient of x², so its graph is a ∩-shaped parabola and the vertex gives the maximum point.
我们识别出这是一个二次函数，x² 系数为负，因此其图像是 ∩ 形抛物线，顶点给出最大值点。
vertex is found by x = -b/(2a). Here a = -2, b = 120, so x = -120/(2×-2) = -120/(-4) = 30.
顶点的 x 坐标由 x = -b/(2a) 求得。这里 a = -2, b = 120，所以 x = -120/(2×-2) = -120/(-4) = 30。
to P(x): P(30) = -2(30)² + 120×30 - 500 = -2×900 + 3600 - 500 = -1800 + 3600 - 500 = 1300.
将 x = 30 代入 P(x) 求出最大利润：P(30) = -2(30)² + 120×30 - 500 = -2×900 + 3600 - 500 = -1800 + 3600 - 500 = 1300。
Hence, the maximum weekly profit is £1300, achieved by producing and selling 30 units.
因此，最大周利润为 £1300，通过生产和销售 30 件产品达到。
(solving P(x)=0), which occur at x=5 and x=55, indicating the viable range of production.
在更广的背景下，公司还可以利用该模型看到在盈亏平衡点（解 P(x)=0）利润为零，这出现在 x=5 和 x=55 处，表明可行的生产范围。
2. Case Study 2 – Population Growth Using Exponential Functions | 用指数函数模拟人口增长
A town's population t years after 2024 is given by Q(t) = 5000 × 1.03t. The council needs to predict the population in 2034 and determine how long it takes for the population to double.
某城镇在 2024 年后 t 年的人口由 Q(t) = 5000 × 1.03t 给出。市政局需要预测 2034 年的人口，并确定人口翻倍所需的时间。
In 2034, t = 10. We compute Q(10) = 5000 × 1.0310. Using a calculator, 1.0310 Q(10) ≈ 5000 × 1.343916 = 6719.58. Rounding to the nearest person gives about 6720 people.
在 2034 年，t = 10。我们计算 Q(10) = 5000 × 1.0310。用计算器得 1.0310 ≈ 1.343916，所以 Q(10) ≈ 5000 × 1.343916 = 6719.58。四舍五入到整数约为 6720 人。
To find the doubling time, we need to solve 5000 × 1.03t = 10000, which simplifies to 1.03t = 2.
为了求翻倍时间，我们需要解 5000 × 1.03t = 10000，简化得 1.03t = 2。
gs, t ≈ 0.6931 / 0.02956 ≈ 23.45 years. So the population will double in about 23.5 years.
取对数：t = log(2) / log(1.03)。用自然对数得 t ≈ 0.6931 / 0.02956 ≈ 23.45 年。因此人口约在 23.5 年后翻倍。
The model assumes a constant growth rate; in reality, growth might slow due to limited resources, but it gives a useful short-term projection.
该模型假设增长率恒定；现实中增长可能因资源有限而减缓，但它提供了有用的短期预测。
3. Case Study 3 – Surveying Height with Trigonometry | 用三角学测量高度
A surveyor stands 50 metres away from a tall building. She measures the angle of elevation to the top as 32°. Her eye-level is 1.6 m above the ground. Find the height of the building.
一位测量员站在离高楼 50 米远的地方。她测得楼顶的仰角为 32°。她的视线高度离地面 1.6 米。求楼的高度。
r eye-level to the top. In the right-angled triangle, tan(32°) = opposite/adjacent = h/50.
设 h 为从她视线高度到楼顶的垂直距离。在直角三角形中，tan(32°) = 对边/邻边 = h/50。
So h = 50 × tan(32°). Using tan(32°) ≈ 0.6249 gives h ≈ 50 × 0.6249 = 31.245 m.
因此 h = 50 × tan(32°)。tan(32°) ≈ 0.6249，得 h ≈ 50 × 0.6249 = 31.245 米。
e total building height = h + eye-level = 31.245 + 1.6 = 32.845 m, which rounds to 32.8 m.
则大楼总高度 = h + 视线高度 = 31.245 + 1.6 = 32.845 米，四舍五入为 32.8 米。
If the sun is behind the building and casts a shadow of length 62 m when the angle of elevation of the sun is 28°, we could verify the height using similar triangles, reinforcing the reliability of the method.
如果太阳在建筑物后方，且在太阳仰角为 28° 时投下 62 米长的影子，我们可以通过相似三角形验证高度，从而强化该方法的可靠性。
4. Case Study 4 – Savings Plan with Arithmetic Sequences | 用等差数列规划储蓄
A student saves money every week, starting with £5 in the first week, £7 in the second, £9 in the third, and so on, forming an arithmetic sequence. She wants to know the amount saved in week 20 and the total saved after 20 weeks.
一名学生每周存钱，第一周 £5，第二周 £7，第三周 £9，依此类推，构成等差数列。她想知道第 20 周的存款额以及 20 周后的总储蓄。
s first term a = 5 and common difference d = 2. The n-th term is given by uₙ = a + (n-1)d.
该数列的首项 a = 5，公差 d = 2。第 n 项由 uₙ = a + (n-1)d 给出。
For week 20, u₂₀ = 5 + 19×2 = 5 + 38 = 43. So she saves £43 in week 20.
第 20 周 u₂₀ = 5 + 19×2 = 5 + 38 = 43。因此她在第 20 周存 £43。
of an arithmetic series is Sₙ = n/2 [2a + (n-1)d] = n/2 (a + l), where l is the last term.
等差数列前 n 项和为 Sₙ = n/2 [2a + (n-1)d] = n/2 (a + l)，其中 l 为末项。
So S₂₀ = 20/2 (5 + 43) = 10 × 48 = 480. The total saved after 20 weeks is £480.
因此 S₂₀ = 20/2 (5 + 43) = 10 × 48 = 480。20 周后总储蓄 £480。
Comparing with a regular weekly savings plan of £25, the arithmetic plan starts lower but grows, demonstrating how incremental increases accelerate long-term savings.
与每周固定 £25 的储蓄计划相比，等差数列计划起初较低但逐渐增长，展示了增量增长如何加速长期储蓄。
5. Case Study 5 – Navigation Using Vectors | 用向量进行导航
A drone flies from point A to B with displacement vector (200i + 150j) metres, where i is east and j is north. It then flies from B to C with vector (-100i + 200j) metres. Find the straight-line distance from A to C and the bearing from A.
一架无人机从 A 点飞往 B 点，位移向量为 (200i + 150j) 米，其中 i 指向东，j 指向北。然后它从 B 飞往 C，向量为 (-100i + 200j) 米。求 A 到 C 的直线距离以及从 A 出发的方位角。
isplacement from A to C is the vector sum: AC = (200 - 100)i + (150 + 200)j = 100i + 350j.
从 A 到 C 的总位移为向量和：AC = (200 - 100)i + (150 + 200)j = 100i + 350j。
Distance AC = |AC| = √100² + 350² = √10000 + 122500 = √132500 = 50√53 ≈ 364.0 metres.
距离 AC = |AC| = √100² + 350² = √10000 + 122500 = √132500 = 50√53 ≈ 364.0 米。
h is given by tan θ = 100/350, so θ = arctan(2/7) ≈ 15.9°. Thus the bearing is about 016°.
方位角从正北顺时针测量。东偏北的角度 θ 满足 tan θ = 100/350，因此 θ = arctan(2/7) ≈ 15.9°。故方位角约为 016°。
Navigators often use this vector method to combine legs of a journey. If wind or current vectors were added, the drone's resultant path would be the vector sum of its own velocity and the wind velocity.
导航员常用这种向量方法结合各段航程。如果加入风或水流向量，无人机的实际轨迹将是自身速度与风速的向量和。
6. Case Study 6 – Analysing Sporting Performance with Statistics | 用统计分析运动成绩
A basketball player's points per game over a ten-match sample are: 18, 21, 15, 30, 22, 19, 25, 28, 17, 24. The coach wants to find the mean, median, interquartile range, and standard deviation to assess consistency.
一位篮球运动员在十场比赛中的每场得分样本为：18, 21, 15, 30, 22, 19, 25, 28, 17, 24。教练想要求出平均值、中位数、四分位距和标准差以评估其稳定性。
2, 24, 25, 28, 30. The mean = sum/10 = (15+17+18+19+21+22+24+25+28+30)/10 = 219/10 = 21.9.
首先排序数据：15, 17, 18, 19, 21, 22, 24, 25, 28, 30。平均值 = 总和/10 = (15+17+18+19+21+22+24+25+28+30)/10 = 219/10 = 21.9。
er quartile Q3 is median of upper half (22,24,25,28,30) = 25. IQR = Q3 - Q1 = 25 - 18 = 7.
中位数是第 5 和第 6 个值的平均：(21+22)/2 = 21.5。下四分位数 Q1 是下半部分（15,17,18,19,21）的中位数 = 18；上四分位数 Q3 是上半部分（22,24,25,28,30）的中位数 = 25。四分位距 IQR = 25 - 18 = 7。
41, (21-21.9)²=0.81, (22-21.9)²=0.01, (24-21.9)²=4.41, (25-21.9)²=9.61, (28-21.9)²=37.21, (30-21.9)²=65.61. Sum = 213.9. σ = √213.9/10 ≈ √21.39 ≈ 4.62.
标准差 σ = √[Σ(x - 平均值)²/n]。计算平方偏差： .81, (22-21.9)²=0.01, (24-21.9)²=4.41, (25-21.9)²=9.61, (28-21.9)²=37.21, (30-21.9)²=65.61。总和 = 213.9。σ = √213.9/10 ≈ √21.39 ≈ 4.62。
The IQR of 7 indicates moderate spread, while the standard deviation of 4.62 suggests scores typically vary by about 4.6 points from the mean, which is reasonable for consistent performance.
四分位距 7 表明中等分散程度，而标准差 4.62 表示得分通常偏离平均值约 4.6 分，对于稳定的表现而言是合理的。
7. Case Study 7 – Traffic Flow and Simultaneous Equations | 交通流量与联立方程
x, y, z are flows on three branches. Determine the flows and check whether the system is consistent.
三条道路交汇于一个路口。交通流量（每小时车辆数）满足：x + y = 500，y + z = 600，x - z = -100，其中 x, y, z 为三条分支的流量。确定流量并检查系统是否一致。
100 - x = -100 → -100 = -100, which is always true. This means the equations are dependent—there are infinitely many solutions.
由第一个方程：y = 500 - x。代入第二个：(500 - x) + z = 600 → z = 100 + x。代入第三个：x - (100 + x) = -100 → x - 100 - x = -100 → -100 = -100，恒成立。这意味着方程是相依的——存在无穷多解。
traffic bounds), y = 500 - x, z = 100 + x. For example, if x = 200, then y = 300, z = 300.
我们将通解表示为：x 可取任意实数（在合理的交通范围内），y = 500 - x，z = 100 + x。例如，若 x = 200，则 y = 300，z = 300。
In practice, additional information like road capacity would restrict the possible values of x. This case study illustrates how simultaneous equations model network flows.
实际上，诸如道路容量等额外信息会限制 x 的可能取值。本案例说明了联立方程如何模拟网络流量。
8. Case Study 8 – Packaging Optimization with Calculus (Graphical Differentiation) | 用微积分（图解微分）优化包装设计
A company designs a box with a square base of side x cm and height h cm. The volume must be 1000 cm³. The material for the base costs twice as much per cm² as the sides. Express the cost in terms of x only and find the dimensions that minimize the cost.
一家公司设计一个底面为正方形、边长为 x cm、高为 h cm 的盒子。体积必须为 1000 cm³。底面材料每平方厘米的成本是侧面的两倍。将成本表示为仅含 x 的表达式，并找出使成本最小的尺寸。
se cost = 2c. Total cost C = 2c × x² + c × 4xh = 2c x² + 4c x (1000/x²) = 2c x² + 4000c/x.
体积 V = x²h = 1000 → h = 1000/x²。底面积 = x²，侧面积 = 4xh。设侧面每平方厘米成本为 c，则底成本 = 2c。总成本 C = 2c × x² + c × 4xh = 2c x² + 4c x (1000/x²) = 2c x² + 4000c/x。
x² + 4000/x. Using differentiation (though Year 9 may use graphical or numeric methods), f'(x) = 4x - 4000/x². Set f'(x) = 0: 4x = 4000/x² → x³ = 1000 → x = 10 cm.
为最小化 C，可忽略常数 c，最小化 f(x) = 2x² + 4000/x。利用微分（虽然 Year 9 可能使用图解或数值方法），f'(x) = 4x - 4000/x²。令 f'(x) = 0：4x = 4000/x² → x³ = 1000 → x = 10 cm。
box is a cube 10×10×10 cm. The minimum cost is C = 2c·100 + 4000c/10 = 200c + 400c = 600c.
则 h = 1000/10² = 10 cm。因此最优盒子是一个 10×10×10 厘米的立方体。最低成本为 C = 2c·100 + 4000c/10 = 200c + 400c = 600c。
This demonstrates the practical power of algebraic manipulation and calculus in reducing production expenses.
这展示了代数操作和微积分在降低生产成本方面的实际作用。
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