Case Study Practical Exercise in Advanced Higher Maths | SQA 进阶数学案例分析实战演练

📚 Case Study Practical Exercise in Advanced Higher Maths | SQA 进阶数学案例分析实战演练

Advanced Higher Mathematics demands far more than isolated technique recall; it requires the ability to model, analyse, and solve multi‑step problems in realistic contexts. Case studies provide an ideal framework for developing this integrative skill, bridging the gap between abstract theory and practical application. In this article, we work through four carefully chosen case studies from optimisation, vectors, differential equations, and matrices, demonstrating systematic problem‑solving and rigorous justification at every stage.

进阶数学远不止是孤立技巧的记忆，它要求学生能在真实情境中建模、分析并求解多步骤问题。案例分析为培养这种综合能力提供了理想的框架，在抽象理论与实际应用之间架起桥梁。本文将通过精心挑选的四个案例，涵盖最优化、向量、微分方程和矩阵，逐步展示系统的解题过程和严格的论证。

1. Case 1: Minimising Material Cost | 案例一：最小化材料成本

A manufacturer produces cylindrical tin cans that must hold exactly 1000 cm³ of product. The metal used for the top and bottom costs twice as much per square centimetre as the metal for the curved side. The factory needs to determine the radius r and height h that minimise the total material cost while satisfying the volume requirement.

某制造商生产的圆柱形罐头必须恰好容纳 1000 cm³ 的产品。罐顶和罐底所用金属的单位面积成本是侧面金属的两倍。工厂需要在满足容积要求的前提下，确定使总材料成本最低的半径 r 和高 h。

Let the cost per cm² for the side material be c pounds. Then the cost for top and bottom is 2c per cm². The volume constraint gives πr²h = 1000. The total cost C is the sum of the costs for the two circular ends and the curved rectangle: C = 2c(2πr²) + c(2πrh) = 4cπr² + 2cπrh.

设侧面材料每 cm² 的成本为 c 英镑，则顶部和底部的成本为每 cm² 2c。容积约束为 πr²h = 1000。总成本 C 是两个圆形端面和弯曲矩形侧面的成本之和：C = 2c(2πr²) + c(2πrh) = 4cπr² + 2cπrh。

To express C as a function of one variable, we substitute h = 1000/(πr²) into the cost equation:

为将 C 表示为单变量函数，我们将 h = 1000/(πr²) 代入成本方程：

C(r) = 4cπr² + 2cπr · (1000/(πr²)) = 4cπr² + 2000c r⁻¹

This simplification is crucial because it reduces the problem to a single‑variable calculus optimisation.

这一简化至关重要，因为它将问题归结为单变量微积分的最优化问题。

2. Formulating the Cost Function | 建立成本函数

Before differentiating, always check the domain: r must be positive. The cost function C(r) = 4cπr² + 2000c r⁻¹ is clearly defined for r > 0. Although the constant factor c does not affect the location of the minimum, we keep it to represent real monetary cost. Our aim is to find the stationary point of C and confirm that it gives a global minimum on (0, ∞).

在求导之前，务必检查定义域：r 必须为正。成本函数 C(r) = 4cπr² + 2000c r⁻¹ 在 r > 0 上有定义。虽然常数因子 c 不影响最小值的位置，但我们保留它以表示真实的货币成本。目标是找到 C 的驻点，并证明它在 (0, ∞) 上给出全局最小值。

Writing the second term as 2000c r⁻¹ makes differentiation straightforward using the power rule. Note how the problem translates a physical design constraint into a purely mathematical function ready for calculus tools.

将第二项写为 2000c r⁻¹，就可以直接用幂函数求导法则。注意该问题是如何将物理设计约束转化为纯粹数学函数，以便使用微积分工具。

3. Finding the Critical Point | 寻找驻点

We differentiate C(r) with respect to r:

对 C(r) 关于 r 求导：

dC/dr = 8cπr – 2000c r⁻²

Set dC/dr = 0 to locate stationary points:

令 dC/dr = 0 以确定驻点：

8cπr = 2000c / r² ⇒ 8πr³ = 2000 ⇒ r³ = 250/π

Because c > 0, it cancels without affecting the solution. The only positive critical value is r = ³√(250/π). In approximate form, r ≈ 4.30 cm. The corresponding height is h = 1000/(πr²). Substituting r³ = 250/π gives h = 1000/(πr²) = 1000/(π) · r/r³? A neater route: from volume, h = 1000/(πr²), and we already know r³ = 250/π, so h = 1000/(πr²) = (1000/π) × (1/r²). Notice that h/r = 1000/(πr³) = 1000/250 = 4, yielding the elegant relationship h = 4r.

由于 c > 0，它可约去且不影响解。唯一的正临界值为 r = ³√(250/π)，近似约为 4.30 cm。相应的高度 h = 1000/(πr²)。代入 r³ = 250/π 可得一个更直接的关系：由容积 h = 1000/(πr²)，发现 h/r = 1000/(πr³) = 1000/250 = 4，从而得到优美关系式 h = 4r。

4. Confirming Minimum and Final Dimensions | 确认最小值与最终尺寸

To classify the stationary point, we compute the second derivative:

为判定驻点类型，我们计算二阶导数：

d²C/dr² = 8cπ + 4000c r⁻³

Since r > 0 and c > 0, every term is positive, so d²C/dr² > 0 for all r in the domain. The stationary point is therefore a local minimum, and because the function tends to infinity as r → 0⁺ and as r → ∞, it is the global minimum. Thus the most economical dimensions are r = ³√(250/π) cm and h = 4r ≈ 17.2 cm.

由于 r > 0 且 c > 0，每一项均为正，故在整个定义域上 d²C/dr² > 0。因此驻点为局部极小值；又因为当 r → 0⁺ 和 r → ∞ 时函数均趋于无穷，故该点为全局最小值。因此最经济的尺寸为 r = ³√(250/π) cm，h = 4r ≈ 17.2 cm。

This case study illustrates the complete optimisation pathway: modelling a real‑world constraint, reducing to a single‑variable function, differentiation, solving, and rigorous second‑derivative verification. The result h = 4r is a memorable design rule arising directly from the cost structure.

该案例展示了完整的最优化路径：对真实约束建模、转化为单变量函数、求导、求解，以及严格的二阶导数验证。结果 h = 4r 是一个由成本结构直接产生的、便于记忆的设计准则。

5. Case 2: Concurrency of Medians in a Triangle | 案例二：三角形中线共点

Prove that the three medians of any triangle intersect at a single point (the centroid), and show that this point divides each median in the ratio 2:1. We use position vectors for a clean, coordinate‑free proof.

试证明任意三角形的三条中线相交于一点（即形心），并证明该点将每条中线分为 2:1 的比例。我们使用位置向量进行一个不依赖坐标系的简洁证明。

Let the vertices of triangle ABC have position vectors a, b, c relative to an origin O. The midpoints of BC, CA, and AB are respectively d = (1/2)(b+c), e = (1/2)(c+a), and f = (1/2)(a+b). Our goal is to find a point G that lies on all three medians AD, BE, and CF.

设三角形 ABC 的顶点关于原点 O 的位置向量为 a, b, c。BC, CA, AB 的中点分别为 d = (1/2)(b+c), e = (1/2)(c+a), f = (1/2)(a+b)。目标是找到一个同时位于三条中线 AD, BE 和 CF 上的点 G。

6. Vector Approach and Proof | 向量方法与证明

Consider a point G on median AD that divides it in the ratio 2:1 from vertex A. Using the section formula, the position vector of G is g = (a + 2d)/3. Substitute d:

考察中线 AD 上一点 G，它从顶点 A 出发将 AD 分为 2:1。利用定比分点公式，G 的位置向量为 g = (a + 2d)/3。代入 d：

g = ( a + 2·½(b+c) ) / 3 = ( a + b + c ) / 3

Now check whether the same point lies on median BE with the same ratio. The point dividing BE in the ratio 2:1 from B would have position vector (b + 2e)/3 = (b + 2·½(c+a))/3 = (a+b+c)/3, which is exactly g. Similarly, for median CF, (c+2f)/3 = (a+b+c)/3 = g. Hence the same point G lies on all three medians, proving concurrency, and the division ratio 2:1 follows directly from its construction.

现在检验同一点是否以相同比例位于中线 BE 上。从 B 点将 BE 分为 2:1 的点所具有的位置向量为 (b + 2e)/3 = (b + 2·½(c+a))/3 = (a+b+c)/3，这正是 g。类似地，对于中线 CF，(c+2f)/3 = (a+b+c

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Case Study Practical Exercise in Advanced Higher Maths | SQA 进阶数学案例分析实战演练