📚 Case Study Practice for CCEA Year 12 Mathematics | CCEA Year 12 数学案例分析实战演练
In this article, we walk through a realistic case study designed for CCEA Year 12 Mathematics. You will apply concepts from algebra, calculus, optimisation, and statistics to solve a packaging design problem, then interpret data on material strength. This step‑by‑step guide highlights the analytical reasoning expected at AS level, helping you develop confidence for both written exams and coursework‑style tasks.
本文带领你完成一个为 CCEA Year 12 数学设计的真实案例分析。你将运用代数、微积分、最优化和统计知识解决包装设计问题，并解读材料强度数据。这份逐级指导突出 AS 阶段所需的分析推理能力，帮助你为笔试和课程式任务建立信心。
1. Case Introduction – The Canned Food Company | 案例介绍 – 罐头食品公司
A local food producer plans to launch a new soup line in cylindrical cans. Each can must hold exactly 500 cm³ of soup. The company has two priorities: minimise the cost of metal sheeting used for the can, and ensure the lid can withstand a stacking pressure of at least 30 MPa in a warehouse environment. Your task is to recommend the optimal radius and height, and to analyse test data for the proposed metal alloy.
一家本地食品生产商计划推出圆柱形罐头的新汤品系列。每个罐头必须恰好容纳 500 cm³ 的汤。公司有两个重点：尽可能降低罐头所用金属板材的成本，并确保罐盖在仓库环境中能承受至少 30 MPa 的堆叠压力。你的任务是推荐最优的半径和高度，并分析待选金属合金的测试数据。
2. Problem 1 – Setting Up the Optimisation | 问题 1 – 建立最优化模型
Let the can have radius r cm and height h cm. The volume constraint gives V = πr²h = 500. The total surface area A of the cylinder, which directly relates to material usage, is A = 2πr² + 2πrh. Our first goal is to express A in terms of a single variable and find the dimensions that minimise A, before considering separate costs for ends and sides.
设罐头半径为 r cm，高为 h cm。体积约束条件为 V = πr²h = 500。直接关系材料用量的圆柱体总表面积 A 为 A = 2πr² + 2πrh。我们的首要目标是将 A 表示为单一变量的函数，并找出使 A 取最小值的尺寸，之后再考虑罐盖和罐身的不同材质成本。
3. Mathematical Modelling – Surface Area as a Function of r | 数学建模 – 表面积关于 r 的函数
From the volume equation, express h in terms of r: h = 500 / (πr²). Substitute into A:
由体积方程将 h 表示为 r 的函数：h = 500 / (πr²)。代入 A 式：
A(r) = 2πr² + 2πr × (500/(πr²)) = 2πr² + 1000/r
This rational function is differentiable for r > 0 and forms the basis of the optimisation. Notice that as r → 0⁺, 1000/r → ∞; as r → ∞, 2πr² → ∞. Therefore a minimum exists for some finite positive r.
该有理函数在 r > 0 时可导，是我们进行最优化的基础。注意到当 r → 0⁺ 时，1000/r → ∞；当 r → ∞ 时，2πr² → ∞。因此最小值必然存在于某个有限的正 r 处。
4. Calculus – Differentiation to Find the Critical Point | 微积分 – 求导寻找临界点
Differentiate A(r) with respect to r:
对 r 求 A(r) 的导数：
dA/dr = 4πr − 1000/r²
Set dA/dr = 0:
令 dA/dr = 0：
4πr − 1000/r² = 0 ⇒ 4πr³ = 1000 ⇒ r³ = 250/π
Take the cube root: r = ³√(250/π). Using π ≈ 3.1416, compute 250/π ≈ 79.5775, then r ≈ 4.301 cm. The corresponding height is h = 500/(π × 4.301²) ≈ 500/(π × 18.498) ≈ 8.602 cm.
取立方根：r = ³√(250/π)。取 π ≈ 3.1416，计算 250/π ≈ 79.5775，则 r ≈ 4.301 cm。对应的高 h = 500/(π × 4.301²) ≈ 500/(π × 18.498) ≈ 8.602 cm。
5. Confirming the Minimum – Second Derivative Test | 验证最小值 – 二阶导数检验
Compute the second derivative: d²A/dr² = 4π + 2000/r³. At r = ³√(250/π), r³ = 250/π, so:
计算二阶导数：d²A/dr² = 4π + 2000/r³。当 r = ³√(250/π) 时，r³ = 250/π，所以：
d²A/dr² = 4π + 2000/(250/π) = 4π + 8π = 12π > 0
Since the second derivative is positive, the critical point indeed gives a global minimum for the unconstrained surface area. This is a standard result for a cylinder of fixed volume: h = 2r.
由于二阶导数为正，该临界点确为无约束表面积下的全局最小值。这是固定体积圆柱体的标准结果：h = 2r。
6. Real-World Constraints – Introducing Material Costs | 实际约束 – 引入材料成本
The can ends (top and bottom) are made from a thicker steel costing £0.05 per cm², while the lateral side uses a thinner steel at £0.03 per cm². The total cost C (in pence) is:
罐盖和罐底使用较厚钢材，成本为每平方厘米 0.05 英镑；罐身使用较薄钢材，成本为每平方厘米 0.03 英镑。总成本 C（单位：便士，1英镑=100便士）为：
C = 0.05 × 2πr² + 0.03 × 2πrh = 0.10πr² + 0.06πrh
Substitute h = 500/(πr²):
代入 h = 500/(πr²)：
C(r) = 0.10πr² + 0.06πr × (500/(πr²)) = 0.10πr² + 30/r
Now differentiate to minimise cost. dC/dr = 0.20πr − 30/r². Set to zero: 0.20πr³ = 30 ⇒ r³ = 150/π.
接着求导以最小化成本。dC/dr = 0.20πr − 30/r²。令其为零：0.20πr³ = 30 ⇒ r³ = 150/π。
7. Cost-Optimal Dimensions and Sensitivity | 成本最优尺寸与灵敏度
Using π ≈ 3.1416, 150/π ≈ 47.748, so r ≈ ³√47.748 ≈ 3.634 cm. The corresponding height is h = 500/(π × 3.634²) ≈ 500/(π × 13.207) ≈ 12.06 cm. The minimum cost is C ≈ 0.10π(3.634)² + 30/3.634 ≈ 0.10π×13.21 + 8.255 ≈ 4.149 + 8.255 = 12.404 pence (about £0.124). This is lower than the purely area‑minimising design (which would cost about 13.2 pence if calculated with weighted materials).
取 π ≈ 3.1416，150/π ≈ 47.748，r ≈ ³√47.748 ≈ 3.634 cm。对应的高 h = 500/(π × 3.634²) ≈ 500/(π × 13.207) ≈ 12.06 cm。最低成本 C ≈ 0.10π(3.634)² + 30/3.634 ≈ 0.10π×13.21 + 8.255 ≈ 4.149 + 8.255 = 12.404 便士（约 0.124 英镑）。该成本低于纯粹最小表面积设计（若计入加权材料约需 13.2 便士）。
Notice how the different unit costs shift the optimal shape away from h = 2r towards a taller can with a smaller radius, demonstrating the importance of including economic factors.
请注意不同的单位成本如何将最优形状从 h = 2r 推向半径较小、罐身更高的方案，这表明纳入经济因素的重要性。
8. Statistical Data Analysis – Material Strength Test | 统计数据分析 – 材料强度测试
The supplier provides tensile strength data (in MPa) from 15 samples of the proposed alloy:
供应商提供了待选合金 15 个样品的抗拉强度数据（单位：MPa）：
32, 28, 35, 30, 33, 29, 31, 34, 30, 32, 27, 36, 33, 31, 29
We need to verify whether the alloy consistently exceeds the 30 MPa threshold. Organise the data into a stem-and-leaf diagram to visualise the distribution.
我们需要验证该合金是否稳定超过 30 MPa 的门槛。将数据整理成茎叶图以观察分布。
- Stem 2: 7, 8, 9, 9 → four values
- Stem 3: 0, 0, 1, 1, 2, 2, 3, 3, 4, 5, 6 → eleven values
The diagram reveals a slight negative skew, with most values clustered around 30–33 MPa.
该图显示轻微负偏态，多数数值集中在 30–33 MPa 附近。
9. Descriptive Statistics – Mean, Median and Standard Deviation | 描述性统计 – 均值、中位数和标准差
Calculate the key statistics:
计算主要统计量：
Mean x̄ = (32+28+35+30+33+29+31+34+30+32+27+36+33+31+29) / 15 = 470 / 15 = 31.33 MPa.
均值 x̄ = (32+28+35+30+33+29+31+34+30+32+27+36+33+31+29) / 15 = 470 / 15 = 31.33 MPa。
Arrange data in order: 27, 28, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 35, 36. Median = 31 MPa.
数据排序：27, 28, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 35, 36。中位数 = 31 MPa。
Sample standard deviation s = √[ Σ(x − x̄)² / (n−1) ]. Compute squared deviations: (0.67²=0.45, …) sum ≈ 114.67; s ≈ √(114.67/14) ≈ √8.191 ≈ 2.863 MPa.
样本标准差 s = √[ Σ(x − x̄)² / (n−1) ]。计算离差平方和：约等于 114.67；s ≈ √(114.67/14) ≈ √8.191 ≈ 2.863 MPa。
These figures show that while the mean exceeds 30 MPa, the roughly 2.86 MPa spread means some individual samples fall below the target. The company should consider whether the occasional weakness is acceptable or if tighter quality control is needed.
这些数据表明，尽管均值超过 30 MPa，约 2.86 MPa 的离散度意味着个别样品低于目标值。公司需考虑偶发强度不足是否可接受，或是否需要更严格的质量控制。
10. Visual Representation – Histogram and Box Plot | 可视化 – 直方图和箱线图
To present findings to management, create appropriate diagrams. For a histogram, group the data into classes: 27–29, 30–32, 33–36. Frequencies: 4, 7, 4 respectively. The modal class is 30–32 MPa. For a box plot, the five‑number summary is: min = 27, Q1 = 29.5, median = 31, Q3 = 33, max = 36. The interquartile range is 3.5 MPa, indicating moderate variability without outliers.
为了向管理层汇报，需要绘制适当的图表。直方图可将数据分组：27–29、30–32、33–36，频数分别为 4、7、4。众数组为 30–32 MPa。箱线图的五数概括为：最小值 = 27，Q1 = 29.5，中位数 = 31，Q3 = 33，最大值 = 36。四分位距为 3.5 MPa，显示中度变异且无异常值。
| Class Interval (MPa) | Frequency |
|---|---|
| 27 – 29 | 4 |
| 30 – 32 | 7 |
| 33 – 36 | 4 |
Both graphs confirm that the majority of test results lie above 30 MPa, though the lower tail warrants attention.
两幅图表都证实大多数测试结果高于 30 MPa，但下尾部分值得关注。
11. Integrating the Findings – Final Recommendation | 综合结论 – 最终建议
Based on the cost model, the recommended can dimensions are r = 3.63 cm and h = 12.1 cm, giving a material cost of about 12.4 pence per can. The strength analysis shows the alloy meets the average target but exhibits a standard deviation of 2.86 MPa. To reduce the risk of failure below 30 MPa, we recommend either increasing the sheet thickness slightly (which would raise cost) or requesting the supplier to tighten the manufacturing tolerance so that the mean strength rises above 32 MPa.
依据成本模型，推荐的罐头尺寸为 r = 3.63 cm，h = 12.1 cm，每个罐头材料成本约 12.4 便士。强度分析显示该合金达到平均指标，但标准差为 2.86 MPa。为降低低于 30 MPa 的失效风险，我们建议略微增加板材厚度（会提高成本），或要求供应商收紧制造公差，将平均强度提升至 32 MPa 以上。
Present your reasoning using clear mathematical language and supported by the statistical evidence, exactly as required in CCEA examination case‑study questions.
正如 CCEA 考试案例分析题所要求的那样，用清晰的数学语言陈述推理，并以统计证据作为支撑。
12. Common Pitfalls and Exam Tips | 常见错误与应试技巧
- Forgetting to convert units: All lengths in cm, volume in cm³, cost per cm². Keep units consistent throughout.
- 忽略单位换算：所有长度用 cm，体积用 cm³，成本用每 cm²。全程保持单位一致。
- Differentiating 1/r incorrectly: d/dr (1/r) = −1/r². Take care with the negative sign when setting the derivative to zero.
- 对 1/r 求导错误：d/dr (1/r) = −1/r²。设导数为零时注意负号。
- Confusing sample and population standard deviation: Use divisor (n−1) for a sample. In this case n=15, so divide by 14.
- 混淆样本标准差与总体标准差：样本使用除数 (n−1)。本例 n=15，故除以 14。
- Overlooking the second derivative test: Always confirm the nature of the stationary point. A positive second derivative guarantees a minimum.
- 忽略二阶导数检验：永远要确认驻点性质。二阶导数为正则保证最小值。
- Misinterpreting statistical diagrams: The box plot alone does not show mean; pair it with the histogram to comment on shape and spread.
- 误读统计图表：箱线图本身不显示均值；需结合直方图评价形态与分散程度。
13. Conclusion – Building Real-World Problem Solving | 结语 – 构建真实世界的问题解决能力
This case study demonstrates how CCEA Year 12 Mathematics pulls together pure and applied strands. From calculus to descriptive statistics, you must interpret results in context and communicate your reasoning effectively. Practise similar scenarios where constraints change—for example, a limited total cost budget, or a different shaped container—to strengthen your modelling skills.
本案例展示了 CCEA Year 12 数学如何融合纯数学与应用分支。从微积分到描述性统计，你必须结合实际情境解释结果，并有效地传达推理过程。练习类似情景，例如改变约束条件——受限制的总成本预算，或不同形状的容器——以增强你的建模能力。
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