📚 Case Study Practice in Advanced Mathematics | 剑桥Year 9进阶数学：案例分析实战演练
Case studies form an essential part of the Cambridge Year 9 Advanced Mathematics curriculum, bridging theoretical knowledge with real-world applications. In this article, we explore ten carefully designed case studies that challenge students to apply concepts such as quadratic functions, calculus, logarithms, sequences, and probability in practical situations. Each case study is structured to develop analytical thinking, problem-solving skills, and mathematical communication – key competencies for success in further study and examinations. By working through these examples, learners will gain confidence in modelling and interpreting data, making informed decisions based on mathematical reasoning.
案例分析是剑桥Year 9进阶数学课程的重要组成部分，它将理论知识与现实应用紧密相连。本文精选十个精心设计的案例，要求学生将二次函数、微积分、对数、数列和概率等概念灵活运用于实际情境中。每个案例都旨在培养分析思维、问题解决能力和数学交流技巧——这些是进一步学习和考试成功的关键能力。通过演练这些实例，学习者将增强建模和解释数据的信心，学会基于数学推理做出明智决策。
1. Introduction to Case Study Analysis | 案例分析导论
Approaching a case study requires a structured method. First, read the problem carefully and identify the variables involved. Next, translate the description into mathematical expressions or equations. Then, apply the appropriate techniques to solve the problem. Finally, interpret your answer in the context of the original scenario, checking for reasonableness. This framework – understand, represent, solve, and interpret – underpins all the examples that follow.
处理案例分析需要有条理的方法。首先，仔细阅读问题并找出涉及的变量。接着，将文字描述转化为数学表达式或方程。然后，运用适当的方法求解。最后，在原始情境中解释答案，并检查其合理性。这个“理解—表示—求解—解释”的框架贯穿了以下所有实例。
2. Case 1: Maximising Area – Quadratic Optimisation | 案例一：面积最大化——二次函数优化
A farmer has 60 metres of fencing to enclose a rectangular garden against an existing wall. One side of the rectangle is the wall, so only three sides require fencing. Let the width perpendicular to the wall be x metres. Then the length parallel to the wall is (60 − 2x) metres. The area A of the garden is given by A = x(60 − 2x). This simplifies to A = −2x² + 60x. The function is a downward-opening quadratic, so its maximum occurs at the vertex. The x-coordinate of the vertex is x = −b/(2a) = −60/(2 × (−2)) = 15. The maximum area is A = 15 × (60 − 30) = 450 m².
一位农民用60米长的栅栏沿着已有的墙壁围一个矩形花园。矩形的一边是墙，因此只需围三边。设垂直于墙的宽度为x米。那么平行于墙的长度为(60 − 2x)米。花园面积A = x(60 − 2x)，化简得A = −2x² + 60x。这是一个开口向下的二次函数，最大值出现在顶点。顶点的x坐标为x = −b/(2a) = −60/(2 × (−2)) = 15。最大面积为A = 15 × (60 − 30) = 450平方米。
Thus, the optimal dimensions are width 15 m and length 30 m. This case highlights how quadratic models can be used to solve optimisation problems without calculus, although taking the derivative of A with respect to x would give the same result.
因此，最优尺寸为宽15米、长30米。这个案例展示了如何在没有微积分的情况下使用二次模型解决优化问题，尽管对A关于x求导也能得到相同结果。
3. Case 2: Investing Money – Compound Interest and Logarithms | 案例二：投资理财——复利与对数
Emma invests £2000 in a savings account that pays 4.5% interest compounded annually. She wants to know how many years it will take for her investment to double. The amount after t years is given by A = P(1 + r)ᵗ, where P = 2000, r = 0.045. We need A = 4000, so 2000(1.045)ᵗ = 4000. Dividing both sides by 2000 gives (1.045)ᵗ = 2. Taking logarithms of both sides: t ln(1.045) = ln 2. Evaluating: t = ln 2 / ln(1.045) ≈ 0.6931 / 0.0440 ≈ 15.75 years. So it will take slightly less than 16 years to double.
艾玛将2000英镑存入一个年利率为4.5%且每年复利的储蓄账户。她想知道需要多少年投资才能翻倍。t年后的金额为A = P(1 + r)ᵗ，其中P = 2000，r = 0.045。要求A = 4000，因此2000(1.045)ᵗ = 4000。两边除以2000得(1.045)ᵗ = 2。两边取对数：t ln(1.045) = ln 2。计算得t = ln 2 / ln(1.045) ≈ 0.6931 / 0.0440 ≈ 15.75年。因此大约不到16年即可翻倍。
Logarithms are essential for solving exponential equations. In this case, natural logarithms were used, but logarithms to any base would work as long as consistency is maintained. The analysis also introduces the ‘Rule of 72’, a mental shortcut: 72/4.5 = 16 years, which approximates the exact result.
对数对于求解指数方程至关重要。此处使用自然对数，但只要保持一致，任何底数的对数均可。该分析还介绍了“72法则”，这是一种心算捷径：72/4.5 = 16年，与精确结果相近。
4. Case 3: Modeling Motion – Kinematics and Calculus | 案例三：运动建模——运动学与微积分
A particle moves along a straight line. Its velocity v (in m/s) after t seconds is given by v(t) = 6t² − 4t + 3. Find the acceleration at t = 2 s and the displacement over the first 3 seconds. Acceleration is the derivative of velocity: a(t) = dv/dt = 12t − 4. At t = 2, a(2) = 12×2 − 4 = 20 m/s². Displacement is the integral of velocity from t = 0 to t = 3: s = ∫₀³ (6t² − 4t + 3) dt = [2t³ − 2t² + 3t]₀³ = (2×27 − 2×9 + 9) − 0 = 54 − 18 + 9 = 45 m.
一个质点沿直线运动。t秒后的速度v（单位m/s）为v(t) = 6t² − 4t + 3。求t = 2 s时的加速度以及前三秒的位移。加速度是速度的导数：a(t) = dv/dt = 12t − 4。当t = 2时，a(2) = 12×2 − 4 = 20 m/s²。位移是速度从t = 0到t = 3的积分：s = ∫₀³ (6t² − 4t + 3) dt = [2t³ − 2t² + 3t]₀³ = (2×27 − 2×9 + 9) − 0 = 54 − 18 + 9 = 45 m。
The application of differentiation and integration in kinematics is a core skill. Notice that the units of acceleration are metres per second squared, and displacement is given in metres. Always check that derived physical quantities have appropriate units.
在运动学中应用微分和积分是一项核心技能。注意加速度的单位是米每二次方秒，位移的单位是米。务必检查推导出的物理量是否具有合适的单位。
5. Case 4: Designing a Box – Surface Area and Volume | 案例四：盒子设计——表面积与体积
A manufacturer needs to design an open rectangular box with a square base and a volume of 500 cm³. The base is of side length x cm. Express the height h in terms of x, then find the surface area A as a function of x. Since volume V = x²h = 500, we have h = 500/x². The box has an open top, so its surface area is the base plus four sides: A = x² + 4xh = x² + 4x(500/x²) = x² + 2000/x. To minimise the material used, one would differentiate A with respect to x and set it to zero: dA/dx = 2x − 2000/x² = 0 ⇒ 2x = 2000/x² ⇒ 2x³ = 2000 ⇒ x³ = 1000 ⇒ x = 10 cm. So the optimal base side is 10 cm, with height 500/100 = 5 cm.
一家制造商需要设计一个开口矩形盒子，其底面为正方形且体积为500 cm³。设底面边长为x cm。用x表示高h，然后求表面积A关于x的函数。因为体积V = x²h = 500，所以h = 500/x²。盒子无盖，因此表面积为底面积加四个侧面：A = x² + 4xh = x² + 4x(500/x²) = x² + 2000/x。为最小化材料用量，可对A关于x求导并令其为零：dA/dx = 2x − 2000/x² = 0 ⇒ 2x = 2000/x² ⇒ 2x³ = 2000 ⇒ x³ = 1000 ⇒ x = 10 cm。因此最优底面边长为10 cm，高为500/100 = 5 cm。
This optimisation exercise demonstrates the importance of expressing a quantity in terms of a single variable before applying calculus. The resulting dimensions yield the minimum surface area of 100 + 200 = 300 cm².
这个优化练习展示了在运用微积分之前将量表示为单一变量的重要性。最终尺寸给出的最小表面积为100 + 200 = 300 cm²。
6. Case 5: Population Growth – Exponential Models | 案例五：人口增长——指数模型
A population of bacteria doubles every 3 hours. Initially, there are 500 bacteria. Derive an exponential model N(t) = N₀ × 2^(t/3) and predict the population after 10 hours. Also, when will the population reach 10,000? Using N₀ = 500, N(10) = 500 × 2^(10/3) = 500 × 2^(3.333…) ≈ 500 × 10.079 = 5,040 bacteria. To find t when N = 10,000, we set 500 × 2^(t/3) = 10,000 ⇒ 2^(t/3) = 20. Taking logs: (t/3) ln 2 = ln 20 ⇒ t = 3 ln 20 / ln 2 ≈ 3 × 2.9957 / 0.6931 ≈ 12.97 hours. So after about 13 hours the population surpasses 10,000.
某种细菌每3小时数量翻倍。最初有500个细菌。推导指数模型N(t) = N₀ × 2^(t/3)，并预测10小时后的数量。同时，何时数量会达到10,000？利用N₀ = 500，N(10) = 500 × 2^(10/3) = 500 × 2^(3.333…) ≈ 500 × 10.079 = 5,040个细菌。要求t使N = 10,000，设500 × 2^(t/3) = 10,000 ⇒ 2^(t/3) = 20。取对数：(t/3) ln 2 = ln 20 ⇒ t = 3 ln 20 / ln 2 ≈ 3 × 2.9957 / 0.6931 ≈ 12.97小时。因此大约13小时后数量超过10,000。
Exponential growth models are widely used in biology, finance, and physics. The key step is identifying the doubling time or growth rate and constructing the function correctly.
指数增长模型广泛应用于生物学、金融和物理学。关键步骤是识别倍增时间或增长率，并正确构造函数。
7. Case 6: Geometry and Trigonometry – Measuring Heights | 案例六：几何与三角学——测量高度
From a point 30 metres away from the base of a tree, the angle of elevation to the top is 28°. Use trigonometry to find the height of the tree. Let h be the height. Then tan 28° = h/30, so h = 30 tan 28° ≈ 30 × 0.5317 = 15.95 m. If the observer’s eye level is 1.6 m above the ground, the actual tree height becomes 15.95 + 1.6 = 17.55 m. This case illustrates the practical use of the tangent ratio in surveying and field measurements.
离树底部30米处，测得树顶的仰角为28°。使用三角学求树高。设高度为h。则tan 28° = h/30，因此h = 30 tan 28° ≈ 30 × 0.5317 = 15.95 m。若观测者眼睛离地高度为1.6 m，则树的实际高度为15.95 + 1.6 = 17.55 m。此案例展示了正切比在测量和实地勘测中的实际应用。
8. Case 7: Data Analysis – Linear Regression and Prediction | 案例七：数据分析——线性回归与预测
The table below shows the test scores of students against hours of study:
| Hours (x) | Score (y) |
|---|---|
| 2 | 45 |
| 3 | 52 |
| 5 | 65 |
| 7 | 78 |
Calculate the least-squares regression line of y on x and estimate the score for 6 hours of study. First, compute means: x̄ = (2+3+5+7)/4 = 4.25, ȳ = (45+52+65+78)/4 = 60. The sums: Σ(x – x̄)(y – ȳ) = (2-4.25)(45-60)+(3-4.25)(52-60)+(5-4.25)(65-60)+(7-4.25)(78-60) = (−2.25)(−15)+(−1.25)(−8)+(0.75)(5)+(2.75)(18) = 33.75+10+3.75+49.5 = 97; Σ(x – x̄)² = (−2.25)²+(−1.25)²+(0.75)²+(2.75)² = 5.0625+1.5625+0.5625+7.5625 = 14.75. Slope b = 97/14.75 ≈ 6.576. Intercept a = ȳ – b x̄ = 60 – 6.576×4.25 ≈ 60 – 27.948 = 32.052. The line is y = 6.58x + 32.05 (approx). For x = 6, predicted score = 6.58×6 + 32.05 ≈ 39.48 + 32.05 = 71.53 ≈ 72.
下表显示学生学习时间与考试成绩的对比数据。（表格见上）计算y关于x的最小二乘回归线，并估算学习6小时的分数。首先计算均值：x̄ = 4.25，ȳ = 60。求和：Σ(x – x̄)(y – ȳ) = 97；Σ(x – x̄)² = 14.75。斜率b = 97/14.75 ≈ 6.576。截距a = 60 – 6.576×4.25 ≈ 32.052。回归线为y = 6.58x + 32.05（近似）。对x = 6，预测分数约为72。
Linear regression provides a model for making predictions within the range of observed data. However, extrapolating far beyond this range could be unreliable.
线性回归为在观测数据范围内进行预测提供了一个模型。然而，远超此范围的外推可能不可靠。
9. Case 8: Breaking Even – Cost and Revenue Analysis | 案例八：盈亏平衡——成本与收入分析
A company produces a gadget with fixed costs of £5000 and variable cost of £12 per unit. The selling price is £20 per unit. Determine the break-even point – the number of units where total cost equals total revenue. Let x be the number of units. Total cost C = 5000 + 12x. Total revenue R = 20x. Setting C = R: 5000 + 12x = 20x ⇒ 5000 = 8x ⇒ x = 625 units. For any production above 625 units, the company makes a profit. If only 500 units are produced, the loss is C − R = (5000 + 12×500) − 20×500 = 5000 + 6000 − 10000 = £1000. This simple linear model is a fundamental tool in business.
一家公司生产一款小设备，固定成本为5000英镑，每单位可变成本为12英镑。售价为每单位20英镑。确定盈亏平衡点——即总成本等于总收入的产量。设x为产品数量。总成本C = 5000 + 12x。总收入R = 20x。令C = R：5000 + 12x = 20x ⇒ 5000 = 8x ⇒ x = 625单位。任何超过625单位的产量，公司均可盈利。若仅生产500单位，则亏损为C − R = £1000。这一简单的线性模型是商业中的基本工具。
10. Case 9: Using Sequences – Loan Repayment Plan | 案例九：数列应用——贷款偿还计划
A loan of £3000 is to be repaid in monthly instalments. Each month, £250 is paid, but interest of 1.5% of the outstanding balance is added before the payment. Model the outstanding balance as a sequence. Let uₙ be the balance after n months, with u₀ = 3000. The recurrence relation is uₙ₊₁ = 1.015 uₙ − 250. Find the balance after 6 months and determine when the loan will be paid off. Compute iteratively: u₁ = 1.015×3000 − 250 = 3045 − 250 = 2795; u₂ = 1.015×2795 − 250 = 2836.925 − 250 = 2586.925; u₃ ≈ 1.015×2586.925 − 250 = 2624.73 − 250 = 2374.73; u₄ = 1.015×2374.73 − 250 = 2410.35 − 250 = 2160.35; u₅ = 1.015×2160.35 − 250 = 2192.76 − 250 = 1942.76; u₆ = 1.015×1942.76 − 250 = 1971.90 − 250 = 1721.90. The sequence decreases slowly. To find payoff month, continue until balance ≤ 0. You can also solve using the formula for geometric and arithmetic progression combined, but iteration gives roughly month 15 or 16.
一笔3000英镑的贷款需按月分期偿还。每月支付250英镑，但在支付前会加上未偿余额1.5%的利息。将未偿余额建模为一个数列。设uₙ为n个月后的余额，u₀ = 3000。递推关系为uₙ₊₁ = 1.015 uₙ − 250。求6个月后的余额，并确定何时还清贷款。迭代计算：u₁ = 2795；u₂ ≈ 2586.93；u₃ ≈ 2374.73；u₄ ≈ 2160.35；u₅ ≈ 1942.76；u₆ ≈ 1721.90。余额缓慢减少。继续迭代直至余额≤0，大约在第15或16个月还清。
Sequences and recurrence relations are powerful for modelling financial scenarios. This case encourages exploring both iterative methods and closed-form solutions.
数列与递推关系对于金融场景建模非常有用。这个案例鼓励学生探索迭代法和封闭式解法。
11. Case 10: Probability in Games – Expected Value | 案例十：游戏中的概率——期望值
A game involves rolling a fair six-sided die. The prize money depends on the roll: rolling a 1 or 2 wins £5, a 3 or 4 wins £10, a 5 wins £15, and a 6 wins £30. It costs £12 to play. Should a player expect to profit in the long run? The probabilities are: P(£5) = 2/6 = 1/3, P(£10) = 1/3, P(£15) = 1/6, P(£30) = 1/6. The expected winnings E = 5×(1/3) + 10×(1/3) + 15×(1/6) + 30×(1/6) = 5/3 + 10/3 + 15/6 + 30/6 = 15/3 + 45/6 = 5 + 7.5 = £12.50. The expected profit is £12.50 − £12 = £0.50 per game. So in the long run, the player gains 50 pence per game on average, making it a favourable game.
一个游戏需要投掷一枚均匀的六面骰子。奖金取决于结果：掷出1或2赢5英镑，3或4赢10英镑，5赢15英镑，6赢30英镑。每次玩花费12英镑。玩家长期来看预期能盈利吗？概率为：P(5£) = 2/6 = 1/3，P(10£) = 1/3，P(15£) = 1/6，P(30£) = 1/6。预期奖金E = 5×(1/3) + 10×(1/3) + 15×(1/6) + 30×(1/6) = 12.50英镑。预期利润为12.50 − 12 = 0.50英镑每局。因此长期来看，玩家平均每局赚50便士，这是一个有利的游戏。
Expected value calculations are fundamental in decision making under uncertainty. They reveal that short-term outcomes may vary, but the average result over many trials converges to the expected value.
期望值计算是不确定情况下决策的基础。它揭示了短期结果可能各异，但大量试验后的平均结果会收敛于期望值。
12. Conclusion: Tips for Tackling Case Studies | 总结：案例分析应对技巧
Success in case study analysis comes from regular practice and a systematic approach. Always define variables clearly, draw diagrams where possible, and verify your solutions by substituting back into original conditions. When using formulas, ensure you understand the meaning of each parameter rather than just plugging numbers. Finally, communicate your reasoning step by step in writing – clear communication of mathematical ideas is as important as obtaining the correct answer. The ten examples above represent typical applications you may encounter. Work through them carefully, then attempt variations to build confidence.
要成功完成案例分析，需依靠定期练习和系统方法。始终清晰地定义变量，尽可能画出示意图，并通过代入原始条件验证解答。使用公式时，确保理解每个参数的含义，而非仅仅代数字。最后，逐步写出推理过程——清晰地表达数学思想与得到正确答案同样重要。以上十个实例代表了可能遇到的典型应用。请仔细推敲，然后尝试变式练习，以增强信心。
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