Case Study Practice in Year 10 CAIE Maths | 剑桥IGCSE数学案例分析实战演练

📚 Case Study Practice in Year 10 CAIE Maths | 剑桥IGCSE数学案例分析实战演练

Case study questions in the CAIE IGCSE Mathematics syllabus require you to apply mathematical concepts to real-world scenarios. They test not only your calculation skills but also your ability to interpret information, form equations, and justify conclusions. This article walks you through several typical case studies, showing you how to break down problems, model them mathematically, and present solutions clearly.

在 CAIE IGCSE 数学大纲中，案例分析题要求你将数学概念应用于真实世界情境。它们不仅考查计算能力，更考查你解读信息、建立方程和论证结论的能力。本文将通过多个典型案例分析，带你一步步拆解问题、建立数学模型并清晰呈现解答。

1. Understanding Case Study Questions | 理解案例分析题

Case study questions are often lengthy and contain tables, graphs, or descriptive paragraphs. The first step is to read the entire question carefully, highlighting key numerical data and the exact requirements. Identify what you are being asked to find — is it a maximum area, a total cost, a probability, or an interpretation of a trend?

案例分析题通常篇幅较长，包含表格、图表或描述性段落。第一步是仔细阅读整个题目，标出关键数值和具体要求。明确你需要求解什么——是最大面积、总成本、概率，还是对趋势的解读？

Organise the given information logically. For algebraic problems, define your variables first. For graphical problems, note the axes scales and units. Always check whether intermediate answers need rounding or if exact values are required.

将所给信息有逻辑地整理好。对于代数问题，先定义变量。对于图像问题，注意坐标轴尺度和单位。始终检查中间答案是否需要四捨五入，还是需要精确值。

2. Case Study 1: Optimising a Rectangular Enclosure | 案例研究1：优化矩形围栏

A farmer has 100 m of fencing and wants to enclose a rectangular area against a straight river, using the river as one side. Find the dimensions that maximise the enclosed area.

一位农场主有 100 米长的围栏，他想靠着一笔直的河流围出一块矩形区域，利用河流作为一边。求使面积最大化的尺寸。

Step 1: Let the side perpendicular to the river be x metres, and the side parallel to the river be y metres. Since the river forms one side, the total fencing used is 2x + y = 100.

步骤1：设垂直河流的边为 x 米，平行河流的边为 y 米。因为河流作为一边，所用围栏总长为 2x + y = 100。

Step 2: Express y = 100 – 2x. The area A = x * y = x(100 – 2x) = 100x – 2x². This is a quadratic function. To maximise, find the vertex.

步骤2：表达 y = 100 – 2x。面积 A = x * y = x(100 – 2x) = 100x – 2x²。这是一个二次函数。为求最大值，找到顶点。

For quadratic ax² + bx + c, the vertex occurs at x = -b/(2a). Here a = -2, b = 100, so x = -100/(2 * -2) = 25.

对于二次函数 ax² + bx + c，顶点在 x = -b/(2a)。此处 a = -2，b = 100，因此 x = -100/(2 * -2) = 25。

Then y = 100 – 2*25 = 50. Maximum area = 25 * 50 = 1250 m². Thus, the optimal dimensions are 25 m by 50 m.

于是 y = 100 – 2*25 = 50。最大面积 = 25 * 50 = 1250 m²。所以最佳尺寸为 25 米乘 50 米。

A(x) = 100x – 2x², x = 25, A_{max} = 1250 m²

In your solution, always confirm that the second derivative is negative or use the shape of the quadratic to prove maximum. Also, check domain: x > 0 and y > 0, so 0 < x < 50, which is satisfied.

解题时，需通过二阶导数为负或二次函数开口向下确认是最大值。也要检查定义域：x > 0 且 y > 0，因此 0 < x < 50，满足。

3. Case Study 2: Compound Interest and Savings Plan | 案例研究2：复利与储蓄计划

An investor deposits $5000 in an account offering 4% annual interest compounded quarterly. How much will be in the account after 3 years? Compare this with simple interest at the same rate.

一位投资者将 5000 美元存入一个年利率 4%、按季复利的账户。3 年后账户内会有多少钱？请与同利率下的单利进行比较。

Compound interest formula: A = P(1 + r/n)^(nt), where P = 5000, r = 0.04, n = 4, t = 3.

复利公式：A = P(1 + r/n)^(nt)，其中 P = 5000，r = 0.04，n = 4，t = 3。

Calculate inside: (1 + 0.04/4) = (1 + 0.01) = 1.01. Exponent: 4*3 = 12. So A = 5000 * (1.01)¹².

计算： (1 + 0.04/4) = (1 + 0.01) = 1.01。指数：4*3 = 12。则 A = 5000 * (1.01)¹²。

Compute (1.01)¹² ≈ 1.126825. Then A ≈ 5000 * 1.126825 = $5,634.13 (to the nearest cent).

计算 (1.01)¹² ≈ 1.126825。那么 A ≈ 5000 * 1.126825 = 5,634.13 美元（精确到分）。

Simple interest: I = P * r * t = 5000 * 0.04 * 3 = 600. Total = 5000 + 600 = $5,600. The compound interest yields an extra $34.13 over 3 years.

单利：I = P * r * t = 5000 * 0.04 * 3 = 600。总额 = 5000 + 600 = 5,600 美元。复利在 3 年内多收益 34.13 美元。

Year	Simple Balance	Compound Balance
0	$5,000	$5,000
1	$5,200	$5,203.02
2	$5,400	$5,414.78
3	$5,600	$5,634.13

Clearly, compound interest grows faster because interest is earned on previously accumulated interest. In case study answers, show all substitution steps and final rounding as instructed.

显然，复利增长更快，因为利息本身也会产生利息。在案例分析答案中，要展示所有代入步骤，并按要求进行最终保留。

4. Case Study 3: Interpreting Distance-Time Graphs | 案例研究3：解释距离-时间图

A cyclist travels from town A to town B. The graph shows distance from A in km over time in hours. The journey has three stages: steady speed for 1.5 hours covering 30 km; a 30-minute rest; then a slower speed for the remaining 20 km over 2 hours. Analyse the motion.

一名骑车人从 A 镇前往 B 镇。图像展示了离 A 镇的距离（公里）与时间（小时）的关系。旅程分三段：匀速骑行 1.5 小时，前进 30 公里；休息 30 分钟；然后以较慢的速度骑行剩余的 20 公里，用时 2 小时。请分析该运动。

Plot key points: at t=0, distance=0; t=1.5, d=30; t=2.0, d=30 (rest); t=4.0, d=50. Calculate speeds: first part speed = 30/1.5 = 20 km/h. Second part speed = 20/2 = 10 km/h. The rest shows a horizontal line segment, indicating zero velocity.

标出关键点：t=0 时，距离=0；t=1.5，d=30；t=2.0，d=30（休息）；t=4.0，d=50。计算速度：第一部分速度 = 30/1.5 = 20 km/h。第二部分速度 = 20/2 = 10 km/h。休息段为水平线段，表示速度为零。

A common exam question asks for the average speed for the whole journey: total distance = 50 km, total time = 4 h, average speed = 50/4 = 12.5 km/h. Note this is different from the average of the two speeds (15 km/h).

常见考题是求全程的平均速度：总距离 = 50 km，总时间 = 4 h，平均速度 = 50/4 = 12.5 km/h。注意这不同于两个速度的平均值（15 km/h）。

Interpreting graphs requires careful reading of axis scales and units. When describing motion, use phrases like ‘constant speed’, ‘stationary’, ‘acceleration’ if applicable.

解读图像需要仔细阅读坐标轴刻度和单位。描述运动时，要用“匀速”、“静止”等短语，如涉及加速也要指出。

5. Case Study 4: Statistical Data Analysis | 案例研究4：统计数据分析

A school recorded the test scores of 40 students (out of 50) for a maths assessment. The data is grouped: 0-10 (3 students), 11-20 (5), 21-30 (9), 31-40 (15), 41-50 (8). Estimate the mean score and find the modal class. Discuss whether the median lies in the 31-40 interval.

某学校记录了 40 名学生一次数学评估的考试分数（满分 50）。数据分组如下：0-10（3 人），11-20（5 人），21-30（9 人），31-40（15 人），41-50（8 人）。估算平均分并找出众数所在的组。讨论中位数是否落在 31-40 区间。

To estimate the mean for grouped data, use midpoints: 5, 15.5, 25.5, 35.5, 45.5. Multiply each by frequency: 5*3 = 15, 15.5*5 = 77.5, 25.5*9 = 229.5, 35.5*15 = 532.5, 45.5*8 = 364. Sum of fx = 1218.5. Mean ≈ 1218.5 / 40 = 30.4625.

估算分组数据的平均值，使用组中值：5, 15.5, 25.5, 35.5, 45.5。分别乘频数：5*3=15，15.5*5=77.5，25.5*9=229.5，35.5*15=532.5，45.5*8=364。fx 总和=1218.5。均值≈1218.5/40=30.4625。

The modal class is the one with the highest frequency: 31-40. For median position: (40+1)/2 = 20.5th value. Cumulative frequencies: 3, 8, 17, 32, 40. The 20.5th value lies in the 31-40 interval, so yes, the median is in that class.

众数所在组是频数最高的区间：31-40。中位数位置：(40+1)/2 = 第 20.5 个值。累积频数：3, 8, 17, 32, 40。第 20.5 个值落在 31-40 区间，因此中位数在该组内。

Always check whether data boundaries are inclusive. In this case, intervals are distinct, but in some cases boundaries like 10.5 are needed for continuous data. Provide reasons backed by calculations.

始终检查数据边界是否包含在内。本例中各区间互斥，但若是连续数据，需要边界如 10.5。用计算支撑你的推理。

6. Case Study 5: Volume and Surface Area of a Cylinder | 案例研究5：圆柱体的体积与表面积

A manufacturer designs a cylindrical can to hold 500 cm³ of soup. The material for the base costs twice as much per cm² as the material for the curved side. Find the dimensions (radius r and height h) that minimise the material cost. Give answers correct to 3 significant figures.

某制造商设计了一个圆柱形罐头来装 500 cm³ 的汤。底部材料每平方厘米的成本是侧面材料成本的两倍。找出使材料成本最小的尺寸（半径 r 和高 h）。答案精确到 3 位有效数字。

Volume: V = πr²h = 500, so h = 500 / (πr²). Cost: Let cost per cm² of side material be c. Then side cost = c * curved area = c * 2πrh. Base cost = 2c * πr² (only one base, the top is presumably included? Assume both top and base cost double? Need clarity: typically both top and base would cost twice if stated ‘base material’ but might mean both circular ends. In many problems, ‘material for the ends’ costs double. For this case, assume both top and bottom are made of the expensive material. So base + top area = 2πr², cost = 2c * 2πr²? Actually, careful: ‘base’ could mean the circular parts. I’ll assume both circular faces cost double. So total cost C = (side cost) c * 2πrh + (ends cost) 2c * 2πr² = 2cπrh + 4cπr².

体积：V = πr²h = 500，故 h = 500 / (πr²)。成本：设侧面材料每 cm² 成本为 c。侧面成本 = c * 曲面面积 = c * 2πrh。两底面成本（假设均使用昂贵材料） = 2c * 2πr² = 4cπr²。总成本 C = 2cπrh + 4cπr²。

Substitute h: C = 2cπr(500/(πr²)) + 4cπr² = 1000c/r + 4cπr². To minimise, differentiate with respect to r: dC/dr = -1000c/r² + 8cπr = 0 => 8cπr = 1000c/r² => 8πr³ = 1000 => r³ = 1000/(8π) = 125/π => r = (125/π)^(1/3). Calculate r ≈ (39.7887…)^(1/3)? Wait, 125/π = 125/3.14159… ≈ 39.7887. Cube root ≈ 3.41 cm (since 3.41³ = 39.6). More precisely: 3.41³=39.6, 3.42³≈40.0, so r ≈ 3.41 cm. Then h = 500/(π*(3.41)²) ≈ 500/(π*11.6281) ≈ 500/36.53 ≈ 13.7 cm. Check second derivative positive to confirm minimum. The optimal can is roughly radius 3.41 cm and height 13.7 cm.

代入 h：C = 2cπr(500/(πr²)) + 4cπr² = 1000c/r + 4cπr²。求导以最小化：dC/dr = -1000c/r² + 8cπr = 0 => 8cπr = 1000c/r² => 8πr³ = 1000 => r³ = 1000/(8π) = 125/π => r = (125/π)^(1/3)。计算 r ≈ 3.41 cm。然后 h = 500/(π*(3.41)²) ≈ 500/36.53 ≈ 13.7 cm。验证二阶导数为正以确认最小值。最优罐头约为半径 3.41 cm 高 13.7 cm。

r = (125/π)^(1/3) ≈ 3.41 cm, h ≈ 13.7 cm

Always express final answers to the required significant figures and include units. Show each derivative step clearly, cancelling common factors like c.

最终答案始终按要求保留有效数字并带单位。清晰展示每个求导步骤，消去公因子如 c。

7. Case Study 6: Linear Programming in Context | 案例研究6：线性规划实际应用

A furniture company makes tables and chairs. Each table requires 4 units of wood and 2 hours of labour; each chair requires 1 unit of wood and 3 hours of labour. There are 60 units of wood and 48 hours of labour available per day. Tables yield a profit of $50 each, chairs $30 each. Formulate and solve graphically to maximise profit.

一家家具公司生产桌子和椅子。每张桌子需 4 单位木材和 2 小时人工；每把椅子需 1 单位木材和 3 小时人工。每天可用木材 60 单位，人工 48 小时。桌子每张利润 50 美元，椅子每把 30 美元。列出不等式并图解求最大利润。

Let x = number of tables, y = number of chairs. Constraints: wood: 4x + y ≤ 60; labour: 2x + 3y ≤ 48; x ≥ 0, y ≥ 0. Profit P = 50x + 30y. Graph the feasible region and find vertices. Intersection of 4x+y=60 and 2x+3y=48: solve simultaneously. From first, y = 60 – 4x. Substitute: 2x + 3(60-4x) = 48 => 2x + 180 – 12x = 48 => -10x = -132 => x = 13.2, y = 60 – 52.8 = 7.2. Vertices: (0,0), (0,16) [from labour], (15,0) [from wood], and (13.2,7.2). Since x and y must be integers (whole tables/chairs), test nearby integer points: (13,7) and (14,7) etc., but check feasibility. (13,7): 4*13+7=59 ≤ 60, 2*13+3*7=26+21=47 ≤ 48, profit = 50*13+30*7 = 650+210=860. (14,6): 4*14+6=62 > 60, not feasible. (12,8): wood 4*12+8=56 OK, labour 24+24=48 OK, profit=600+240=840. So maximum integer profit is $860 for 13 tables and 7 chairs. Always state integer constraints if applicable.

设 x = 桌子数量，y = 椅子数量。约束条件：木材：4x + y ≤ 60；人工：2x + 3y ≤ 48；x ≥ 0, y ≥ 0。利润 P = 50x + 30y。画出可行域并求顶点。求 4x+y=60 与 2x+3y=48 交点：解方程组得 x = 13.2, y = 7.2。顶点：(0,0)，(0,16)，(15,0)，(13.2,7.2)。因为 x, y 必须是整数，测试附近整数点 (13,7) 利润 860，其他可行整数点利润更小。最大整数利润为 13 张桌子和 7 把椅子，共 860 美元。

Explain why the vertex method works and always check integer answers when discrete items are produced. Show the profit calculation clearly.

解释为何顶点法有效，并在产出为离散物品时务必检查整数解。清晰展示利润计算。

8. Common Pitfalls and How to Avoid Them | 常见错误与避免方法

Misreading units is a frequent mistake — for example, using metres instead of centimetres when substituting into volume formulas. Always highlight units in the question and convert if necessary.

单位读错是常见错误——例如，在代入体积公式时误用米而不是厘米。始终在题目中高亮单位，必要时进行转换。

Another pitfall is forgetting to check the domain of the variable in optimisation problems, leading to mathematically correct but practically meaningless answers (e.g., negative lengths).

另一个陷阱是在优化问题中忘记检查变量的定义域，导致数学正确但实际无意义的答案（如负长度）。

In statistics, using the wrong class midpoint (especially when classes are given as 0-9, 10-19…) causes inaccurate mean estimates. Ensure you confirm the boundaries.

在统计中，使用了错误的组中值（尤其当分组为 0-9, 10-19 时）会导致均值估算不准。务必确认边界。

When using graphs, read scales carefully; a common error is counting grid squares wrongly. Double-check with axis labels.

使用图像时，仔细读刻度；常见错误是数错格子。结合坐标轴标签复核。

9. Exam Tips for Case Studies | 案例分析题应试技巧

Start by writing down what you are given and what you need to find. Use clear variable definitions. If the question has multiple parts, answers from earlier parts often feed into later ones, so check your work.

先把已知量和求解目标写下来。使用清晰的变量定义。若题目有多部分，前部分的答案往往会用于后续部分，因此要核对。

Show all working even when using a calculator; marks are allocated for method. For graph drawing, use a sharp pencil, label axes, and plot points neatly.

即使使用计算器也要展示所有步骤；方法是得分的依据。画图时用尖铅笔，标明坐标轴，整洁地描点。

If a question asks for interpretation, write in full sentences referring to the context, not just numbers. For example, ‘The maximum area of 1250 m² occurs when the width is 25 m,’ not just ‘1250’.

如果题目要求解释，用完整句子结合情境表述，不要只写数字。例如，“最大面积 1250 m² 出现在宽度为 25 m 时”，而不是仅仅“1250”。

10. Practice Problems | 练习题

Here are three short exercises to reinforce your skills:

以下是三道短练习题，巩固你的技能：

Problem A: A rectangular box with a square base and open top has volume 32 cm³. Find the base side length that minimizes the surface area.
提示：设底边长为 x，高为 h，V = x²h = 32，表面积 A = x² + 4xh，用一次导数求解。
Problem B: The number of bacteria in a culture is given by N = 200 × 2^(t/3) after t hours. How long until the population reaches 12 800?
提示：解指数方程，12 800 = 200 × 2^(t/3)。
Problem C: A data set has mean 45 and standard deviation 6. If every value is increased by 20%, find the new mean and standard deviation.
提示：均值乘以 1.2，标准差也乘以 1.2。

Try these before looking at solutions online. The process of modelling and interpreting is what makes case study practice so valuable for the CAIE exams.

在查阅答案前先自己尝试。建模和解读的过程正是案例分析练习对 CAIE 考试如此有价值的原因。

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Case Study Practice in Year 10 CAIE Maths | 剑桥IGCSE数学案例分析实战演练