IGCSE AQA Further Maths Case Study Drill | IGCSE AQA 进阶数学：案例分析实战演练

📚 IGCSE AQA Further Maths Case Study Drill | IGCSE AQA 进阶数学：案例分析实战演练

This article walks through a full, exam-style case study for AQA IGCSE Further Maths, showing how core topics such as differentiation, optimisation, and modelling come together in a single realistic scenario. By working step by step through the design of a soft‑drink can, you will learn how to translate a real‑world problem into a mathematical model, apply calculus to find an optimal solution, and interpret your answer critically.

本文通过一个完整的 AQA IGCSE 进阶数学考试风格案例研究，展示微分、优化和数学建模等核心主题如何在一个真实的场景中协同运用。我们将一步步设计一个汽水罐，学习如何将现实问题转化为数学模型，应用微积分寻找最优解，并批判性地解释答案。

1. Problem Statement | 问题陈述

A company wants to manufacture a cylindrical aluminium can to hold exactly 330 ml of soft drink. The cost of the metal is the same for the sides, the top, and the base. The production manager asks: “What dimensions (radius 𝑟 and height ℎ) will use the least amount of material?” This is a classic applied optimisation problem.

一家公司要生产一种圆柱形铝罐，恰好容纳 330 毫升软饮料。罐身、罐顶和罐底的金属成本相同。生产经理问：“什么尺寸（半径 𝑟 和高度 ℎ）最节省材料？” 这是一个经典的应用优化问题。

2. Building the Mathematical Model | 建立数学模型

Let the radius of the can be 𝑟 cm and the height be ℎ cm. The volume 𝑉 (in cm³) is fixed: 𝑉 = π𝑟²ℎ = 330. The surface area 𝐴 (in cm²) of the closed cylinder, which represents the amount of material, is given by: 𝐴 = 2π𝑟ℎ + 2π𝑟² (the curved side plus the two circular ends). We need to minimise 𝐴 subject to the fixed volume.

设罐的半径为 𝑟 cm，高为 ℎ cm。容积 𝑉（单位 cm³）是固定的：𝑉 = π𝑟²ℎ = 330。封闭圆柱体的表面积 𝐴（单位 cm²）代表材料用量，其表达式为：𝐴 = 2π𝑟ℎ + 2π𝑟²（侧面加上两个圆底面）。我们需要在体积固定的条件下最小化 𝐴。

This is a constrained optimisation problem. We will eliminate one variable using the constraint.

这是一个有约束的优化问题。我们将利用约束条件消去一个变量。

3. Reducing to a Single Variable | 化简为单变量函数

From the volume equation, express ℎ in terms of 𝑟: ℎ = 330 / (π𝑟²). Substitute this into the surface area formula:

从体积方程解出 ℎ，用 𝑟 表示：ℎ = 330 / (π𝑟²)。代入表面积公式：

𝐴(𝑟) = 2π𝑟 × (330 / (π𝑟²)) + 2π𝑟² = 660/𝑟 + 2π𝑟²

Now 𝐴 is a function of 𝑟 alone, defined for 𝑟 > 0. The problem has become: find the value of 𝑟 that minimises 𝐴(𝑟).

现在 𝐴 仅是 𝑟 的函数，定义域为 𝑟 > 0。问题变为：求使 𝐴(𝑟) 最小的 𝑟 值。

4. Differentiation | 微分求导

Differentiate 𝐴(𝑟) with respect to 𝑟 using standard rules:

使用基本求导法则对 𝐴(𝑟) 关于 𝑟 求导：

A'(𝑟) = −660/𝑟² + 4π𝑟

This derivative tells us the rate of change of surface area as the radius changes. To find the optimal radius, we set the derivative equal to zero.

这个导数反映了半径变化时表面积的变化率。要找到最佳半径，我们令导数等于零。

The term −660/𝑟² comes from differentiating 660𝑟⁻¹, giving −660𝑟⁻², and 2π𝑟² differentiates to 4π𝑟.

⁆660/𝑟² 项来自对 660𝑟⁻¹ 求导，得到 −660𝑟⁻²，而 2π𝑟² 求导得 4π𝑟。

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5. Finding the Stationary Point | 寻找驻点

Solve A'(𝑟) = 0:

解方程 A'(𝑟) = 0：

−660/𝑟² + 4π𝑟 = 0 ⇒ 4π𝑟 = 660/𝑟² ⇒ 𝑟³ = 660 / (4π) = 165/π

Hence the critical radius is:

因此临界半径为：

𝑟 = ∛(165/π) ≈ ∛(52.52) ≈ 3.74 cm

Now find the corresponding height using ℎ = 330 / (π𝑟²). With 𝑟 ≈ 3.74, 𝑟² ≈ 14.0, so ℎ ≈ 330 / (π × 14.0) ≈ 330 / 43.98 ≈ 7.50 cm.

现在用 ℎ = 330 / (π𝑟²) 计算对应的高度。𝑟 ≈ 3.74，𝑟² ≈ 14.0，所以 ℎ ≈ 330 / (π × 14.0) ≈ 330 / 43.98 ≈ 7.50 cm。

Notice that ℎ is approximately twice the radius: ℎ ≈ 2𝑟. This is a well‑known result for the most material‑efficient cylinder of given volume.

注意高度大约是半径的两倍：ℎ ≈ 2𝑟。对于给定体积下材料最省的圆柱体，这是一个众所周知的结果。

6. Second Derivative Test | 二阶导数判定

To confirm that this critical point gives a minimum, find the second derivative:

为了确认该临界点是最小值，求二阶导数：

A”(𝑟) = 1320/𝑟³ + 4π

For any 𝑟 > 0, A”(𝑟) is clearly positive. Since the second derivative is positive, the function is concave up, and the stationary point is a local minimum. In this problem, it is also the global minimum.

对于任意 𝑟 > 0，A”(𝑟) 显然为正。因为二阶导数为正，函数是下凸的，驻点是局部极小值。在这个问题中，它也是全局最小值。

Therefore the dimensions that minimise the surface area for a 330 ml can are approximately 𝑟 = 3.74 cm and ℎ = 7.50 cm.

因此，使 330 毫升罐表面积最小的尺寸约为半径 3.74 cm、高度 7.50 cm。

7. Extension: Cost with Different Materials | 拓展：不同材料的成本

In practice, the base and top may be made of thicker metal, doubling the cost per cm². Suppose the curved side costs £1 per cm², while the top and base each cost £2 per cm². The cost function becomes:

实际上，罐底和罐顶可能使用较厚的金属，每平方厘米成本是罐身的两倍。假设侧面成本为每平方厘米 1 英镑，而顶和底各为每平方厘米 2 英镑。成本函数变为：

C = (2π𝑟ℎ) × 1 + (2 × 2π𝑟²) = 2π𝑟ℎ + 4π𝑟²

Again eliminate ℎ using ℎ = 330/(π𝑟²):

再次用 ℎ = 330/(π𝑟²) 消去 ℎ：

C(𝑟) = 2π𝑟 × 330/(π𝑟²) + 4π𝑟² = 660/𝑟 + 4π𝑟²

Differentiate: C'(𝑟) = −660/𝑟² + 8π𝑟. Set to zero:

求导：C'(𝑟) = −660/𝑟² + 8π𝑟。设为零：

8π𝑟 = 660/𝑟² ⇒ 𝑟³ = 660/(8π) = 82.5/π

Thus 𝑟 = ∛(82.5/π) ≈ ∛(26.26) ≈ 2.97 cm, and then ℎ ≈ 330/(π × 8.82) ≈ 11.9 cm. The can becomes taller and narrower because the ends are more expensive.

因此 𝑟 = ∛(82.5/π) ≈ ∛(26.26) ≈ 2.97 cm，然后 ℎ ≈ 330/(π × 8.82) ≈ 11.9 cm。由于端面更贵，罐子变得更高更窄。

This shows how altering the cost parameters changes the optimal design—a key idea in further maths modelling.

这表明改变成本参数如何改变最优设计——这是进阶数学建模的一个关键理念。

8. Sensitivity Analysis | 敏感性分析

We can examine how small changes in the volume affect the optimal radius. If the volume is increased by 1%, from 330 to 333.3 cm³, the formula 𝑟 = ∛(𝑉/(2π)) for the equal‑cost case gives a radius increase of about 0.33%. This small sensitivity indicates the design is robust.

我们可以研究体积的微小变化如何影响最佳半径。在等成本情形中，若体积增加 1%，从 330 变为 333.3 cm³，公式 𝑟 = ∛(𝑉/(2π)) 给出的半径增大约 0.33%。这种低敏感性表明设计是稳健的。

For the unequal‑cost case, the exponent 1/3 still dampens the effect, but the constant factor changes—useful to discuss in a “Discussion” section of an investigation.

对于成本不等的情形，1/3 次幂仍然减弱了影响，但常数因子发生了变化——在探究报告的“讨论”部分中讨论这一点很有用。

9. Using Matrices to Transform the Can Design | 用矩阵变换罐体设计

Matrix transformations can help visualise how the can’s shape changes under scaling. If we apply a scaling matrix to the coordinates of the can profile, we can model stretching in height or width. For example, the matrix [[𝑘, 0], [0, 1]] scales the radius by factor 𝑘 while keeping height fixed. This connects to the geometry topic in IGCSE Further Maths.

矩阵变换可以帮助可视化罐体形状在缩放下的变化。如果我们对罐轮廓坐标应用缩放矩阵，可以模拟高度或宽度的拉伸。例如，矩阵 [[𝑘, 0], [0, 1]] 将半径缩放 𝑘 倍而高度不变。这与 IGCSE 进阶数学中的几何主题相联系。

Understanding transformations reinforces the idea that optimisation is deeply linked to geometry and algebra.

理解变换能强化一个观念：优化与几何和代数是紧密相连的。

10. Verifying with a Spreadsheet Model | 用电子表格模型验证

We can set up a spreadsheet with columns for radius, height, surface area, and cost. By varying 𝑟 in small increments, we can numerically confirm that the minimum occurs exactly where calculus predicted. This numerical approach is useful for checking work and for problems where differentiation is difficult.

我们可以建立一个电子表格，包含半径、高度、表面积和成本等列。通过以微小增量改变 𝑟，可以从数值上确认最小值正好出现在微积分预测的位置。这种数值方法对于检查工作以及对于难以求导的问题很有用。

In your IGCSE investigation, showing both algebraic and numerical verification can earn high marks.

在你的 IGCSE 探究中，同时展示代数和数值验证可以获得高分。

11. Common Pitfalls | 常见错误

• Forgetting the constraint: Some students differentiate the two‑variable surface area without substituting for ℎ. Always reduce to one variable first.
• 对于约束条件处理不当：一些学生直接对双变量表面积求导而没有代入 ℎ。务必先化为单变量函数。
• Sign errors in differentiation: Pay close attention when differentiating terms like 660/𝑟. It is 660𝑟⁻¹; derivative is −660𝑟⁻².
• 求导中的符号错误：在求 660/𝑟 这类项的导数时要格外小心。它是 660𝑟⁻¹，导数为 −660𝑟⁻²。
• Units: Mixing ml and cm³ can lead to confusion. 1 ml = 1 cm³, so always work in consistent units.
• 单位问题：混淆毫升和立方厘米会导致疑惑。1 ml = 1 cm³，因此始终使用一致的单位。
• Second derivative test omission: Always confirm the nature of the stationary point.
• 忽略二阶导数检验：务必确认驻点的性质。

12. Conclusion and Exam Tips | 总结与考试技巧

This case study demonstrates a powerful problem‑solving framework: (1) model the real situation with equations, (2) reduce to one variable using a constraint, (3) differentiate, (4) find and classify stationary points, (5) interpret the results. Such a structured approach works for a wide range of IGCSE Further Maths problems, from packaging to beam strength.

本案例研究展示了一个强大的问题解决框架：(1) 用方程对真实情景建模，(2) 利用约束化为单变量，(3) 求导，(4) 寻找并判定驻点，(5) 解释结果。这种结构化的方法适用于从包装到梁强度等广泛的 IGCSE 进阶数学问题。

Remember to show clear work, state any assumptions (e.g., ignoring the thickness of the metal), and check that your answer is physically reasonable. Good luck!

记住要展示清晰的步骤，说明任何假设（例如忽略金属厚度），并检查答案在物理上是否合理。祝你好运！

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