📚 Case Study Practice: Optimising Can Design and Quality Control | 案例分析实战：优化罐体设计与质量控制
Case studies in CCEA Year 13 Mathematics challenge you to bring together pure mathematical techniques and statistical methods to solve real-world problems. This article guides you through a complete worked case study - from optimising the shape of a drinks can to monitoring the filling process with hypothesis tests. By following every step, you will see how calculus, algebra, probability and inferential statistics combine to support practical business decisions.
CCEA 13 年级数学的案例分析要求你将纯数学技巧与统计方法融合，以解决现实问题。本文带你一步步完成一个完整的实例演练——从优化饮料罐的形状，到用假设检验监控灌装过程。通过追踪每一步，你将看到微积分、代数、概率与推断统计如何共同为实际商业决策提供支持。
1. The Business Scenario | 商业情景
A soft drinks company produces cylindrical aluminium cans that must hold exactly 330 ml of liquid. The management team wants to minimise the amount of metal used per can to cut material costs, while ensuring every can meets the capacity requirement. Meanwhile, the quality control department regularly checks the filling process to see whether the average volume matches the declared 330 ml and whether the proportion of leaking cans remains within acceptable limits - currently set at no more than 5%.
一家软饮料公司生产必须精确容纳 330 毫升液体的圆柱形铝罐。管理层希望尽量减少每个罐子的金属用量以削减材料成本，同时确保每个罐子都满足容量要求。与此同时，质量控制部门定期检查灌装过程，确认平均灌装量是否与标示的 330 毫升一致，以及泄漏罐子的比例是否保持在可接受的限值内——目前设定为不超过 5%。
The scenario therefore contains two distinct mathematical strands that must eventually be reported as one coherent business recommendation. The first strand is deterministic: the geometry of the can is fully determined once the radius is chosen, so calculus can find a single optimal shape. The second strand is random: individual cans vary slightly in fill volume and in whether they leak, so probability theory is needed to decide whether an observed sample is consistent with the factory standard.
因此，这个情景包含两条截然不同的数学主线，最终必须汇成一份连贯的商业建议。第一条主线是确定性的：一旦选定半径，罐子的几何形状就完全确定，因此可以用微积分求出唯一的最优形状。第二条主线是随机性的：单个罐子的灌装量和是否泄漏都存在微小波动，因此需要概率论来判断某个观测样本是否与工厂标准一致。
2. Modelling Assumptions | 建模假设
Before any calculus is attempted, the examiner expects a short list of assumptions. Every mathematical model is a simplified picture of reality, and marks are awarded for stating clearly what has been ignored.
在动笔做微积分之前，考官期待你列出简短的假设清单。任何数学模型都是对现实的简化，明确写出被忽略的因素本身就是得分点。
- The can is a perfect right circular cylinder with negligible wall thickness. | 罐体为理想直圆柱，罐壁厚度可忽略。
- The volume of metal used is proportional to the total surface area. | 金属用量与总表面积成正比。
- The cost of the top, the base and the curved side is the same per unit area at this stage. | 此阶段假设顶面、底面与侧面的单位面积成本相同。
- The can must contain exactly 330 cm³, since 1 ml = 1 cm³. | 罐体必须恰好容纳 330 立方厘米，因为 1 毫升 = 1 立方厘米。
- Fill volumes are independently and normally distributed with a known standard deviation. | 灌装量相互独立且服从正态分布，标准差已知。
- Each can either leaks or does not leak, with a constant probability for every can. | 每个罐子要么泄漏要么不泄漏，且每个罐子的泄漏概率恒定。
Each assumption points to a specific mathematical tool: the first three lead to differential calculus, the fourth links geometry to algebra, and the last two justify the normal and binomial models used later.
每条假设都对应一个具体的数学工具：前三条导向微分学，第四条把几何与代数联系起来，最后两条则为后文使用的正态模型与二项模型提供依据。
3. Setting Up the Optimisation Model | 建立优化模型
Let the radius of the base be r cm and the height be h cm. The company wants to minimise the total surface area S, because the surface area is proportional to the amount of aluminium used.
设底面半径为 r 厘米，高为 h 厘米。公司希望使总表面积 S 最小，因为表面积与铝材用量成正比。
V = πr²h = 330
S = 2πr² + 2πrh
The first equation is the volume constraint and the second is the objective function. The two variables r and h are linked by the constraint, so the model must be reduced to a single variable before differentiation. Rearranging the constraint gives h in terms of r.
第一个方程是体积约束，第二个是目标函数。两个变量 r 与 h 被约束条件绑定，因此必须先化为一元函数才能求导。整理约束条件，把 h 用 r 表示。
h = 330 ÷ (πr²)
Substituting this expression into the surface area formula eliminates h and produces a function of r alone. This substitution step is the single most important algebraic move in the whole question, and it is worth writing out in full rather than skipping straight to the derivative.
把这个表达式代入表面积公式消去 h，就得到只含 r 的函数。这一步代换是整个题目中最关键的代数操作，值得完整写出，而不是直接跳到求导。
S(r) = 2πr² + 2πr × 330 ÷ (πr²) = 2πr² + 660 ÷ r
S(r) → ∞ as r → ∞ because of the 2πr² term, so a minimum must exist somewhere in between.
定义域为 r > 0，因为半径不能为零或负。注意当 r → 0⁺ 时，由于 660 ÷ r 这一项，S(r) → ∞；当 r → ∞ 时，由于 2πr² 这一项，S(r) → ∞。因此最小值必然存在于两者之间。
4. Solving for the Stationary Point | 求驻点
Differentiate S with respect to r, treating each term separately. The power rule gives 4πr for the first term, and rewriting 660 ÷ r as 660r⁻¹ gives −660r⁻² for the second.
对 r 求导，逐项处理。第一项用幂法则得到 4πr；把 660 ÷ r 改写为 660r⁻¹，第二项得到 −660r⁻²。
dS/dr = 4πr − 660 ÷ r²
Setting the derivative equal to zero and multiplying through by r² produces a cubic equation in r, which can be solved exactly.
令导数为零，两边同乘 r²，得到一个关于 r 的三次方程，可以精确求解。
4πr = 660 ÷ r² → 4πr³ = 660 → r³ = 165 ÷ π
r = ∛(165 ÷ π) ≈ 3.744 cm
Substituting this radius back into the constraint gives the corresponding height.
把这个半径代回约束条件，得到对应的高度。
h = 330 ÷ (π × 3.744²) ≈ 330 ÷ 44.05 ≈ 7.49 cm
der has its height exactly equal to its diameter. This elegant relationship is almost always worth one or two marks on its own, and it can be proved generally by eliminating the numbers.
注意这个醒目的结果：h ≈ 7.49 厘米，而 2r ≈ 7.49 厘米，也就是说最优圆柱的高恰好等于直径。这个漂亮的关系通常本身就值一到两分，而且可以通过消去具体数值给出一般性证明。
V = πr²h = 2πr³ when h = 2r, so r³ = V ÷ (2π) and S = 6πr²
5. Verifying the Optimum | 验证最优解
A zero derivative only identifies a stationary point; it could be a maximum or a point of inflection. The second derivative test settles the matter quickly.
导数为零只能说明是驻点，它可能是极大值点或拐点。用二阶导数判别法可以迅速确定结论。
d²S/dr² = 4π + 1320 ÷ r³
tive. A positive second derivative confirms a local minimum, and because the function is convex over the whole domain r > 0, it is also the global minimum.
当 r ≈ 3.744 时，该值为 4π + 1320 ÷ 52.52 ≈ 12.57 + 25.13 ≈ 37.7，是正数。二阶导数为正说明是极小值；又因为函数在整个定义域 r > 0 上是凸的，所以它也是全局最小值。
The minimum surface area can now be evaluated in two ways, which acts as a useful check on the arithmetic.
现在可以用两种方法计算最小表面积，这相当于对运算做一个有用的检验。
S = 2π(3.744)² + 660 ÷ 3.744 ≈ 88.06 + 176.27 ≈ 264.3 cm²
S = 6πr² = 6π(3.744)² ≈ 264.3 cm²
Two independent methods giving 264.3 cm² is strong evidence that the algebra and the rounding are correct. Always finish an optimisation question with a short sentence interpreting the number in context, for example that each can requires about 264 cm² of aluminium sheet under the stated assumptions.
两种独立方法都得 264.3 平方厘米，这有力说明代数推导与取整都正确。优化题的最后一定要用一句话把数字放回情境中解释，例如在既定假设下每只罐子约需 264 平方厘米铝板。
6. Sensitivity, Cost Variation and Real Cans | 灵敏度、成本变化与真实罐体
small manufacturing errors in the radius barely change the surface area. Increasing r by 1% to about 3.782 cm gives S ≈ 264.4 cm², an increase of only about 0.02% ng property: production tolerances cost almost nothing in material terms near the optimum.
由于在最优点处 dS/dr = 0，半径上的微小制造误差几乎不会改变表面积。把 r 增大 1% 到约 3.782 厘米，S ≈ 264.4 平方厘米，仅增加约 0.02%。这种平坦性是一项真正有用的工程性质：在最优点附近，生产公差在材料成本上几乎不产生代价。
ring-pull. If the ends cost c times as much per unit area as the side, the effective cost becomes C = 2cπr² + 2πrh, and repeating the same method gives a general rule.
更贴近现实的模型会认为顶面和底面的单位面积成本高于侧面，因为罐端由更厚的板材冲压而成，还要加上拉环。若顶底单位面积成本是侧面的 c 倍，则有效成本为 C = 2cπr² + 2πrh，重复同样的方法可得一条一般规律。
C = 2cπr² + 660 ÷ r → dC/dr = 4cπr − 660 ÷ r² = 0 → h = 2c r
r and narrower can. This explains why real 330 ml cans are noticeably slimmer than the mathematical ideal: the ends really are more expensive, and there are further pressures such as stacking strength, shelf appearance, hand grip, pallet efficiency and the diameter limits of high-speed filling lines.
当 c = 1 时，回到 h = 2r。当 c = 1.3 时，最优解变成 h ≈ 2.6r，罐体更高更细。这就解释了为什么真实的 330 毫升罐子明显比数学理想形状更细长：罐端确实更贵，而且还存在堆叠强度、货架观感、握持手感、托盘运输效率以及高速灌装线直径限制等额外压力。
7. Statistical Quality Control: Testing the Mean Fill | 统计质量控制：检验平均灌装量
rd deviation σ = 1.5 ml. A random sample of n = 25 cans is taken each hour and the sample mean x̄ is calculated.
灌装机设定为输出 330 毫升，但没有机器是完美的。设每罐体积 X 服从均值为 μ、标准差 σ = 1.5 毫升的正态分布。每小时抽取 n = 25 罐的随机样本，计算样本均值 x̄。
From the sampling distribution of the mean, the standard error is σ ÷ √n = 1.5 ÷ √25 = 0.3 ml. The quality control team tests H₀: μ = 330 against H₁: μ ≠ 330 at the 5% significance level, a two-tailed test because both under-filling and over-filling matter.
由样本均值的抽样分布，标准误为 σ ÷ √n = 1.5 ÷ √25 = 0.3 毫升。质检团队在 5% 显著性水平下检验 H₀: μ = 330 对 H₁: μ ≠ 330，这是双尾检验，因为灌装不足与灌装过量都需要关注。
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