📚 Year 11 SQA Advanced Mathematics: Case Study In-Depth Practice | Year 11 SQA 进阶数学：案例分析实战演练
Mastery of SQA Advanced Mathematics demands more than routine exercises; real-world problem-solving skills are essential. This case study demonstrates a step-by-step optimisation using differentiation, modelling a typical industrial design challenge. By working through the stages – formulating the model, applying calculus, verifying the optimum, and interpreting results – you will strengthen the ability to tackle unstructured problems independently.
掌握 SQA 进阶数学不仅需要常规练习，实际问题解决能力至关重要。本案例通过一个典型工业设计挑战，逐步展示微分优化建模方法。通过经历建立模型、应用微积分、验证最优解和解释结果等阶段，你将提升独立处理非结构化问题的能力。
1. Case Scenario: The Tin Can Problem | 案例场景：罐头问题
A manufacturer plans to produce cylindrical tin cans with a fixed capacity of 500 cm³. The cans are closed at both ends. The production cost is directly proportional to the amount of metal used, so the company wants to minimise the total surface area of the cylinder while keeping the volume exactly 500 cm³. Your task is to determine the optimal radius and height that achieve this minimum surface area.
某制造商计划生产容积固定为 500 cm³ 的圆柱形罐头，罐头两端均有盖子。生产成本与所用金属量成正比，因此公司希望在保持容积严格为 500 cm³ 的前提下，最小化圆柱体的总表面积。你的任务是找到实现最小表面积的最佳半径和高。
2. Identifying Variables and Objectives | 识别变量与目标
Let r (cm) be the base radius and h (cm) be the height of the cylinder. The goal is to minimise the surface area S (cm²) subject to the volume constraint V = 500 cm³. Both the objective function and the constraint must be expressed in terms of the same variables.
设底面半径为 r（cm），高为 h（cm）。目标是在体积约束 V = 500 cm³ 下最小化表面积 S（cm²）。目标函数和约束条件都必须用相同的变量表示。
3. Constructing the Surface Area Function | 构建表面积函数
For a closed cylinder, the surface area consists of two circular ends and the curved side: S = 2πr² + 2πr h. The volume is V = πr² h. Setting V = 500 gives πr² h = 500, so h = 500 / (πr²). Substitute this relation into S to obtain a function of r alone.
闭口圆柱的表面积由两个圆形底面和侧面组成：S = 2πr² + 2πr h。体积公式为 V = πr² h。令 V = 500 得到 πr² h = 500，故 h = 500 / (πr²)。将此关系代入 S，得到仅关于 r 的函数。
S(r) = 2πr² + 1000/r
This simplified expression shows how surface area depends on the radius, assuming the volume is fixed.
这个简化的表达式显示了在容积固定的假设下，表面积如何取决于半径。
4. Differentiating the Surface Area Function | 对表面积函数求导
aries. Using the power rule and the fact that d/dr (1/r) = -1/r², the first derivative is:
我们对 S(r) 关于 r 求导，以了解面积如何随半径变化。利用幂法则以及 d/dr (1/r) = -1/r²，一阶导数为：
dS/dr = 4πr – 1000/r²
This derivative will be zero at the critical point that possibly minimises S.
在可能使 S 最小化的临界点处，该导数将为零。
5. Locating Stationary Points | 寻找驻点
Set the first derivative equal to zero to identify stationary points:
令一阶导数等于零，以确定驻点：
4πr – 1000/r² = 0
Multiply through by r² to clear the denominator, yielding 4π r³ = 1000. Solve for r³:
方程两边同时乘以 r² 以消去分母，得到 4π r³ = 1000。解出 r³：
r³ = 250/π ≈ 79.577
Taking the cube root gives the critical radius. We will calculate its numerical value later.
开立方根即得临界半径。稍后我们将计算其数值。
6. Second Derivative Test for Nature | 二阶导数检验性质
To confirm that this stationary point yields a minimum, we examine the second derivative:
为确认该驻点产生最小值，我们考察二阶导数：
d²S/dr² = 4π + 2000/r³
Since r > 0, both terms are positive, so d²S/dr² > 0 for all permissible r. The function is convex, and the stationary point is a global minimum within the domain.
由于 r > 0，两项均为正，因此对所有允许的 r 都有 d²S/dr² > 0。函数是凸的，该驻点即为定义域内的全局最小值。
7. Computing the Optimal Dimensions | 计算最优尺寸
Now compute the numerical values. The critical radius is r = ∛ 16, r³ ≈ 79.577, so r ≈ 4.30 cm (to 2 decimal places). Substitute back to find the height:
现在计算数值。临界半径 r = ∛(250/π)。取 π ≈ 3.1416，r³ ≈ 79.577，因此 r ≈ 4.30 cm（精确到两位小数）。再代回求出高度：
h = 500 / (π r²) ≈ 500 / (π × 18.49) ≈ 500 / 58.09 ≈ 8.61 cm
0 cm and height 8.61 cm, which interestingly gives h ≈ 2r. The height equals the diameter.
最优罐头的半径为 4.30 cm，高为 8.61 cm，有趣的是 h ≈ 2r。高度恰好等于直径。
8. Verifying the Minimum Surface Area | 验证最小表面积
sign and compare it with a neighbouring shape to be certain. For r = 4.30 cm, h = 8.61 cm:
我们可以计算最优设计的表面积，并与邻近形状比较以确保无误。当 r = 4.30 cm，h = 8.61 cm 时：
S = 2π(4.30)² + 2π(4.30)(8.61) ≈ 116.18 + 232.70 = 348.88 cm²
00.53 + 250.07 = 350.60 cm², which is indeed larger. The calculus prediction is confirmed.
试取 r = 4.00 cm，得到 h = 500/(π×16) ≈ 9.95 cm，S ≈ 2π(16) + 2π(4)(9.95) ≈ 100.53 + 250.07 = 350.60 cm²，确实更大。微积分预测得到了验证。
9. Generalising the Result | 结果的一般化
is into the volume constraint gives h = 2r. Thus, irrespective of the specific volume, the most material-efficient closed cylinder always has height equal to its diameter. This classical result illustrates the power of calculus in design.
对于固定体积 V 的闭口圆柱，表面积函数为 S(r) = 2πr² + 2V/r。令 dS/dr = 0 可得 r³ = V/(2π)。将其代回体积约束得到 h = 2r。因此，无论体积的具体值是多少，最省料的闭口圆柱总是具有高等于直径的性质。这一经典结果展示了微积分在设计中的威力。
10. Sensitivity Analysis: Tolerance to Small Changes | 灵敏度分析：对小变化的容忍度
ns are impractical. How sensitive is the surface area to small deviations from the optimum? .40 cm while adjusting h to maintain volume V = 500 cm³. The area changes by less than 0.1% indicating a fairly flat minimum. This robustness is an important practical consideration.
在制造中，精确尺寸并不现实。表面积对偏离最优解的小变化有多敏感？计算 r = 4.20 cm 和 r = 4.40 cm 时的 S，同时调整 h 以维持 V = 500 cm³。面积变化不到 0.1%，表明最小值相当平坦。这种鲁棒性是一项重要的实际考量。
11. Real-World Constraints and Extensions | 现实约束与拓展
A pure mathematical model ignores seam overlaps, material thickness, and manufacturing limits. The optimal aspect ratio may change if the cost of the side metal differs from that of the ends, or if a fixed height-to-diameter ratio is required for stacking. Such modifications lead to multi-variable or constrained optimisation, typical of SQA extended-response questions.
纯数学模型忽略了接缝重叠、材料厚度和制造限制。如果侧面金属和底面的成本不同，或者为了堆叠而需要固定的高径比，最优形状比例可能会改变。这类修改引向多变量或约束优化，这是 SQA 扩展作答题的典型内容。
12. Key Takeaways from the Case Study | 案例分析的关键收获
- Model formulation translates a verbal scenario into mathematical functions and constraints – a fundamental skill in Advanced Mathematics.
- Differentiation ng dS/dr = 0 locate optimal points; always verify with a second derivative or sign change.
- Interpretation of the result (h = 2r) connects mathematical findings to real-world meaning.
- Sensitivity analysis assesses the practical reliability of the optimum.
- These steps form a general problem-solving framework applicable to area, volume, cost, and other optimisation contexts encountered in SQA examinations.
- 模型建立将文字描述转化为数学函数与约束——这是进阶数学的基本技能。
- 微分并求解 dS/dr = 0 可以定位最优点；务必用二阶导数或符号变化加以验证。
- 结果解释（h = 2r）将数学发现与现实意义联系起来。
- 灵敏度分析评估最优解的实际可靠性。
- 这些步骤构成了一个通用的问题解决框架，适用于 SQA 考试中出现的面积、体积、成本及其他优化情境。
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