📚 Case Study in Action: Optimisation and Calculus | 案例分析实战演练：最优化与微积分
In Cambridge AS Level Mathematics, case study questions bridge pure theory and real-world application. This article walks through a complete optimisation problem – finding the maximum volume of a box made from a flat sheet – to demonstrate how calculus techniques are used in a structured, step‑by‑step investigation. You will see how to define variables, build a function, differentiate, test for maxima and interpret the answer in context.
在剑桥 AS 阶段数学中，案例分析题将纯理论与实际应用紧密连接。本文将通过一个完整的最优化问题——求由平面纸板制成的盒子的最大容积——展示如何有条理地运用微积分技巧逐步探究。你将看到如何定义变量、建立函数、求导、检验极大值，并结合实际解读答案。
1. Understanding the Problem Statement | 理解问题陈述
A rectangular card measuring 30 cm by 20 cm has squares of side length x cm cut from each corner. The flaps are folded up to form an open box. The challenge is to find the value of x that yields the maximum possible volume of the box.
一张 30 cm × 20 cm 的矩形卡纸，从每个角落剪去边长为 x cm 的正方形，然后将四边折起形成一个无盖盒子。任务是要找出使得盒子容积达到最大的 x 值。
Practical interpretation: x cannot be zero or exceed half of the shorter side, so 0 < x < 10, because the width 20 cm gives 20 − 2x > 0.
实际解释：x 不能为零，也不能超过短边的一半，因此 0 < x < 10，因为宽度 20 cm 必须满足 20 − 2x > 0。
2. Defining Variables and Constraints | 定义变量与约束
Let x be the side length of each cut‑out square (in cm). After cutting and folding, the dimensions of the box base become (30 − 2x) by (20 − 2x), and the height is x.
设 x 为每个被剪去正方形的边长（单位 cm）。剪切并折起后，盒子的底面尺寸变为 (30 − 2x) × (20 − 2x)，高度为 x。
The volume V (in cm³) can be expressed as:
体积 V（单位 cm³）可表示为：
V(x) = x(30 − 2x)(20 − 2x)
Domain constraint: 0 < x < 10. The variables are continuous within this interval.
定义域约束：0 < x < 10。该区间内变量是连续的。
3. Building the Mathematical Model | 建立数学模型
Expand the expression to obtain a polynomial form, which is easier to differentiate:
将表达式展开成多项式形式以便求导：
V(x) = x(600 − 100x + 4x²) = 4x³ − 100x² + 600x
This is a cubic function representing the volume. The task now becomes a pure calculus problem: find the absolute maximum of V on (0, 10) by locating stationary points.
这是一个表示体积的三次函数。现在任务变成了一个纯粹的微积分问题：通过寻找驻点，求 V 在 (0, 10) 上的绝对最大值。
4. Applying Differentiation | 求导运算
Differentiate V(x) with respect to x:
对 V(x) 关于 x 求导：
dV/dx = 12x² − 200x + 600
Set the derivative equal to zero to find stationary points:
令导数等于零以寻找驻点：
12x² − 200x + 600 = 0
Simplify by dividing through by 4:
两边除以 4 进行化简：
3x² − 50x + 150 = 0
5. Locating Critical Points | 寻找临界点
Solve the quadratic using the formula x = [−b ± √(b² − 4ac)] / (2a) with a = 3, b = −50, c = 150:
用公式 x = [−b ± √(b² − 4ac)] / (2a) 解二次方程，其中 a = 3, b = −50, c = 150：
Δ = (−50)² − 4×3×150 = 2500 − 1800 = 700
√Δ = √700 = 10√7 ≈ 26.46
x = (50 ± 26.46) / 6
The two solutions are:
两个解为：
- x₁ ≈ (50 + 26.46) / 6 ≈ 12.74 (outside the domain, x > 10, reject)
- x₂ ≈ (50 − 26.46) / 6 ≈ 3.92 (lies inside 0 < x < 10, accept)
- x₁ ≈ (50 + 26.46) / 6 ≈ 12.74（超出定义域，x > 10，舍弃）
- x₂ ≈ (50 − 26.46) / 6 ≈ 3.92（在 0 < x < 10 内，采纳）
The only relevant critical point is x ≈ 3.92 cm.
唯一相关的临界点是 x ≈ 3.92 cm。
6. Using the Second Derivative Test | 二次求导检验
To confirm that this critical point gives a maximum, find the second derivative:
为确认该临界点给出的是极大值，求二阶导数：
d²V/dx² = 24x − 200
Evaluate at x = 3.92:
代入 x = 3.92：
d²V/dx² ≈ 24×3.92 − 200 = 94.08 − 200 = −105.92 < 0
Since the second derivative is negative, the function is concave down, so the stationary point is a local maximum. Within the closed interval (0, 10), the volume also approaches zero at both ends, so this local maximum is the absolute maximum.
因为二阶导数为负，函数图像是凸向下（凹区间），因此该驻点是一个局部极大值。在开区间 (0, 10) 内，两端体积趋近于零，所以这个局部极大值也就是全局最大值。
7. Calculating the Maximum Volume | 计算最大容积
Substitute x = 3.92 back into the volume function:
将 x = 3.92 代回体积函数：
V ≈ 3.92 × (30 − 7.84) × (20 − 7.84)
≈ 3.92 × 22.16 × 12.16 ≈ 1056 cm³
Using the exact surd form, x = (25 − 5√7)/3 gives a more precise volume, but the approximate decimal value is sufficient for most practical purposes.
使用精确根式 x = (25 − 5√7)/3 可求得更精确的体积，但近似小数值足以满足大多数实际需求。
8. Interpreting the Result in Context | 结合实际解释结果
The optimal cut-out size is about 3.92 cm, producing a box of volume roughly 1056 cm³. If the manufacturer uses x = 3.9 cm or 4.0 cm, the volume will be slightly smaller. This demonstrates the sensitivity of the model to x.
最优剪切尺寸约为 3.92 cm，制成的盒子容积约 1056 cm³。如果制造商使用 x = 3.9 cm 或 4.0 cm，容积会略微减小。这显示出模型对 x 的敏感度。
The result makes physical sense: a very small x gives a shallow box, a very large x leaves a tiny base, and the optimum lies somewhere in between. Calculus locates that exact balance.
结果在物理上是合理的：太小的 x 导致盒子很浅，太大的 x 使底面极小，最优值恰好在两者之间。微积分精确找到了那个平衡点。
9. Extension: Generalising the Problem | 拓展：一般化问题
If the original card measures L by W (with L ≥ W), the general volume function is V = x(L − 2x)(W − 2x). One can show that the optimal x satisfies 12x² − 2(L + W)x + LW = 0, and the acceptable root lies inside 0 < x < W/2.
如果原始卡纸尺寸为 L × W（L ≥ W），一般体积函数为 V = x(L − 2x)(W − 2x)。可以证明最优 x 满足 12x² − 2(L + W)x + LW = 0，且可取的根落在 0 < x < W/2 内。
Such generalisation is a powerful skill in Cambridge case studies: you begin with a concrete example and abstract to a formula that holds for any rectangle.
这种一般化是剑桥案例分析题中的一项强大技能：从一个具体例子出发，抽象出一个适用于任何矩形的公式。
10. Summary of the Case Study Method | 案例分析方法总结
A structured approach to optimisation case studies looks like this:
一个结构化的最优化案例分析步骤如下：
- Read and interpret the scenario carefully. | 仔细阅读并理解情境。
- Identify the quantity to be maximised or minimised. | 识别需要最大化或最小化的量。
- Express that quantity as a function of one variable, noting restrictions. | 将该量表示为单变量函数，并注意限制条件。
- Differentiate and find stationary points. | 求导并找出驻点。
- Use the second derivative test or interval analysis to confirm maximum/minimum. | 用二次求导检验或区间分析确认极大/极小值。
- Substitute back for the optimum value and interpret the result. | 代回求最优值并解读结果。
- If appropriate, generalise the model or discuss limitations. | 在适当时，一般化模型或讨论局限性。
Mastering this pattern helps you confidently tackle any AS‑Level applied calculus question, whether about boxes, revenue, or kinematics.
掌握这一模式能帮助你自信地应对任何 AS 阶段的应用微积分问题，无论涉及盒子、收入还是运动学。
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