📚 Case Study Practical Exercises | 案例分析实战演练
This case study demonstrates how SQA Advanced Higher Mathematics can be applied to tackle interconnected real-world problems in a whisky distillery. We will work through an optimisation task to minimise packaging material, a regression analysis of production costs, and a hypothesis test for quality improvement. Each step blends calculus, statistics, and logical reasoning, reflecting the type of multi‑topic scrutiny expected in Year 13 assessments.
本案例展示了如何运用 SQA 高阶数学解决一家威士忌酒厂中相互关联的实际问题。我们将依次完成一个使包装材料最小化的优化任务、一个生产成本回归分析，以及一个质量改进的假设检验。每一步都融合微积分、统计学和逻辑推理，体现了13年级考核中要求的多主题综合应用。
1. Introduction to the Distillery Case Study | 蒸馏厂案例背景
A Highland distillery bottles its single malt in cylindrical tins holding exactly 700 cm³. Management wants to cut material costs by minimising the total surface area of each tin. At the same time, the production team needs to understand how batch size affects unit cost and whether a new filling machine genuinely reduces the defect rate. Mathematics offers a structured way to answer all three questions.
一家高地酒厂使用每罐恰好容纳 700 cm³ 的圆柱形金属罐来装瓶单一麦芽威士忌。管理层希望通过最小化每个罐子的总表面积来降低材料成本。同时，生产团队需要了解批量大小如何影响单位成本，以及新灌装机是否确实降低了次品率。数学为回答这三个问题提供了一种条理清晰的方法。
2. Formulating the Cylindrical Optimisation | 构建圆柱体优化问题
We let the radius be r cm and the height be h cm. The fixed volume is V = 700 cm³, so the constraint is πr²h = 700. Since the tin has a top and a bottom, the total surface area S is given by S = 2πr² + 2πrh. Expressing h in terms of r from the volume constraint yields h = 700 / (πr²).
令半径为 r cm，高度为 h cm。固定容积 V = 700 cm³，因此约束条件为 πr²h = 700。由于罐子有顶盖和底盖，总表面积 S 表示为 S = 2πr² + 2πrh。用容积约束将 h 用 r 表示，得到 h = 700 / (πr²)。
Substituting this into S eliminates h and gives a function of r alone:
S(r) = 2πr² + 2(700)/r = 2πr² + 1400 r⁻¹
代入 S 消去 h，得到仅依赖于 r 的函数：
S(r) = 2πr² + 2(700)/r = 2πr² + 1400 r⁻¹
3. Applying Differentiation to Find the Critical Radius | 用微分求临界半径
Differentiate S(r) with respect to r:
dS/dr = 4πr − 1400 r⁻²
对 S(r) 关于 r 求导：
dS/dr = 4πr − 1400 r⁻²
Set the derivative to zero for stationary points: 4πr − 1400 r⁻² = 0. Multiply through by r² to get 4πr³ = 1400, hence r³ = 1400 / (4π) = 350 / π. Therefore, the critical radius is
令导数为零求驻点：4πr − 1400 r⁻² = 0。两边乘以 r² 得 4πr³ = 1400，因此 r³ = 1400 / (4π) = 350 / π。
临界半径为
r = ∛(350 / π) ≈ ∛(350 / 3.1416) ≈ ∛111.4 ≈ 4.81 cm
Using the volume constraint, the corresponding height is h = 700 / (π × 4.81²) ≈ 9.63 cm. Notice that h = 2r when the surface area is minimised — a well‑known property of closed cylinders with a given volume.
利用容积约束，对应的罐高 h = 700 / (π × 4.81²) ≈ 9.63 cm。注意当表面积最小时有 h = 2r，这是定容密闭圆柱体的一个著名性质。
4. Confirming the Minimum with the Second Derivative | 用二阶导数确认极小值
The second derivative is found by differentiating dS/dr:
d²S/dr² = 4π + 2800 r⁻³
对 dS/dr 求导得到二阶导数：
d²S/dr² = 4π + 2800 r⁻³
For all r > 0, both terms are positive, so d²S/dr² > 0. At r ≈ 4.81 cm, this value is approximately 4π + 2800 / 111.4 ≈ 12.57 + 25.14 = 37.71 > 0. A positive second derivative confirms a local minimum. The minimum surface area is
对于所有 r > 0，两项均为正，因此 d²S/dr² > 0。当 r ≈ 4.81 cm 时，该值约为 4π + 2800 / 111.4 ≈ 12.57 + 25.14 = 37.71 > 0。正二阶导数确认了局部极小。最小表面积为
Smin = 2π(4.81)² + 1400/4.81 ≈ 145.2 + 291.1 = 436.3 cm²
The distillery can now specify tin dimensions of radius 4.81 cm and height 9.63 cm to use the least material.
酒厂现在可以将罐子的规格定为半径 4.81 cm、高度 9.63 cm，以使用最少的材料。
5. Gathering Data on Batch Production Costs | 收集批次生产成本数据
Moving to the economic side, the production manager recorded batch size x (in hundreds of bottles) and the corresponding unit cost y (pounds per bottle) over eight recent runs.
转向经济层面，生产经理记录了最近八个批次中的批量 x（以百瓶计）和相应的单位成本 y（英镑/瓶）。
| Batch Size (x) | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |
|---|---|---|---|---|---|---|---|---|
| Cost per bottle (y) £ | 3.80 | 2.95 | 2.45 | 2.10 | 1.90 | 1.75 | 1.65 | 1.60 |
A scatter plot would show a broadly descending pattern, but we need a formal model to predict cost and assess the relationship’s strength.
散点图会显示大致的下降趋势，但我们需要一个正式模型来预测成本并评估这一关系的强度。
6. Performing Linear Regression on Cost Data | 对成本数据执行线性回归
We aim to fit the model y = a + bx using least squares. Initial summaries: n = 8, mean of x = 22.5, mean of y = 2.15. The key sums are
Σx = 180, Σy = 17.20, Σx² = 5100, Σxy = 349.
我们旨在用最小二乘法拟合模型 y = a + bx。初步汇总：n = 8，x 的均值为 22.5，y 的均值为 2.15。关键求和为
Σx = 180，Σy = 17.20，Σx² = 5100，Σxy = 349。
Compute sums of squares: Sxx = Σx² − n x̄² = 5100 − 8×506.25 = 1050. Sxy = Σxy − n x̄ȳ = 349 − 8×22.5×2.15 = 349 − 387 = –38. Hence the slope
计算平方和：Sxx = 5100 − 8×506.25 = 1050。Sxy = 349 − 8×22.5×2.15 = 349 − 387 = –38。因此斜率为
b = Sxy / Sxx = –38 / 1050 ≈ –0.0362
The intercept: a = ȳ − b x̄ = 2.15 − (–0.0362)×22.5 ≈ 2.964. The regression equation is
截距：a = ȳ − b x̄ = 2.15 − (–0.0362)×22.5 ≈ 2.964。回归方程为
ŷ = 2.964 − 0.0362x
This suggests that an extra 100 bottles in batch size reduces the predicted cost by about £0.036 per bottle.
这意味着批量每增加 100 瓶，预测的每瓶成本约下降 £0.036。
7. Measuring Correlation Strength and Interpretation | 衡量相关性强度并解读
The Pearson correlation coefficient is r = Sxy / √(Sxx × Syy). We calculate Σy² = 45.51, thus Syy = 45.51 − 8×2.15² = 8.530. Then
皮尔逊相关系数为 r = Sxy / √(Sxx × Syy)。计算得 Σy² = 45.51，故 Syy = 45.51 − 8×2.15² = 8.530。于是
r = –38 / √(1050 × 8.530) ≈ –38 / 94.64 ≈ –0.402
The correlation is negative but only moderately strong. Since r² ≈ 0.161, only about 16% of the variation in unit cost is explained by batch size alone — other factors (like machine set‑up time or raw material price fluctuations) probably play larger roles. The model is a useful first approximation but could be enhanced by collecting more data or using multiple regression.
相关性虽为负，但仅具有中等强度。由于 r² ≈ 0.161，只有约 16% 的单位成本变异被批量大小所解释——其他因素（如机器调试时间或原料价格波动）可能起着更大作用。该模型是有效的初步近似，但可通过收集更多数据或使用多元回归加以改进。
8. Setting Up the Hypothesis Test for the New Machine | 建立新机器的假设检验
The old filling machine historically produced 5% defective bottles (p₀ = 0.05). After installing a newer model, a random sample of n = 500 bottles contained 16 defects, giving a sample proportion p̂ = 16/500 = 0.032. The distillery hopes to demonstrate that the defect rate has genuinely fallen.
旧灌装机历史上产生 5% 的次品瓶（p₀ = 0.05）。安装新型号后，随机抽取
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