Case Study Practical Exercises in Year 12 Edexcel Maths | 案例分析实战演练：Year 12 Edexcel 数学

📚 Case Study Practical Exercises in Year 12 Edexcel Maths | 案例分析实战演练：Year 12 Edexcel 数学

In Year 12 Edexcel Mathematics, the ability to apply pure mathematical techniques to real-world contexts is a core skill. This case study practical exercise will take you through a complete problem-solving journey, from modelling a real situation with algebraic functions to using differentiation to optimise an outcome and interpreting the results in context. The worked example chosen is the classic ‘open box’ problem, which integrates algebra, coordinate geometry and calculus, and is typical of the structured questions found in the Edexcel AS exam. You will see how to construct a model, test its validity, and communicate your reasoning clearly – all essential for high marks in applied problem-solving.

在Year 12 Edexcel数学课程中，将纯数学技术应用于真实情境是一项核心能力。本案例分析实战演练将带你经历完整的解题旅程，从用代数函数对现实情境进行建模，到运用微分优化结果，再到结合情境解释结论。选用的是经典的“无盖纸盒”问题，它综合了代数、坐标几何和微积分，是Edexcel AS考试中典型的建构题。你将看到如何建立模型、检验其有效性并清晰地表达推理过程——这些都是应用类问题解决中获得高分的关键。

1. The Business Scenario and Problem Statement | 商业情境与问题陈述

A packaging company has a rectangular sheet of cardboard measuring 30 cm by 20 cm. They plan to manufacture an open-topped box by cutting out a square of side length x cm from each corner and folding up the flaps. The production manager wants to know the value of x that maximises the volume of the box, in order to use the minimum amount of material for a given volume, or to maximise the carrying capacity from a fixed sheet. As a mathematics consultant, you must model the situation, find the optimal cut-out size, and provide a clear report with justification and any restrictions on x.

一家包装公司有一张长30 cm、宽20 cm的长方形纸板。他们计划通过从每个角剪去边长为x cm的正方形，再将四周折起来，制作一个无盖盒子。生产经理想知道使盒子体积最大的x值，以便在给定材料下最大化载物能力。作为一名数学顾问，你需要对这一情境建立数学模型，求出最优裁剪尺寸，并提供一份附有理由和x取值限制的清晰报告。

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

Once the squares are removed and the flaps folded, the resulting box will have length (30 − 2x) cm, width (20 − 2x) cm and height x cm. Because all dimensions must be positive and the cardboard cannot be cut beyond its edges, the feasible domain for x is 0 < x < 10 (since the smaller side is 20 cm, x must be less than half of that). The volume V, in cubic centimetres, can be expressed as a function of x: V(x) = x(30 − 2x)(20 − 2x).

剪去正方形并折起后，所得盒子的长为(30 − 2x) cm，宽为(20 − 2x) cm，高为x cm。因为所有尺寸必须为正且裁剪不能超出边界，x的可行域为0 < x < 10（较小的边长为20 cm，x必须小于其一半）。体积V（立方厘米）可表达为x的函数：V(x) = x(30 − 2x)(20 − 2x)。

Expanding this expression gives a cubic polynomial: V(x) = x(600 − 60x − 40x + 4x²) = 4x³ − 100x² + 600x. This is the mathematical model we will work with. Note that if x = 0 or x = 10, the volume is zero, which makes physical sense (no height or no base).

展开该表达式得到一个三次多项式：V(x) = x(600 − 60x − 40x + 4x²) = 4x³ − 100x² + 600x。这就是我们要处理的数学模型。注意当x = 0或x = 10时体积为零，这符合物理意义（没有高度或没有底面）。

3. Differentiating to Find Stationary Points | 微分求驻点

To locate the maximum volume, we differentiate V with respect to x. Using the power rule: V'(x) = 12x² − 200x + 600. Setting the derivative equal to zero gives the equation 12x² − 200x + 600 = 0. Divide through by 4 to simplify: 3x² − 50x + 150 = 0.

为找出最大体积，我们对V关于x求导。运用幂法则：V'(x) = 12x² − 200x + 600。令导数等于零得到方程12x² − 200x + 600 = 0。两边同除以4以化简：3x² − 50x + 150 = 0。

Applying the quadratic formula, x = [50 ± √(2500 − 4 × 3 × 150)] / (2 × 3) = [50 ± √(2500 − 1800)] / 6 = [50 ± √700] / 6. Simplifying the surd, √700 = √(100 × 7) = 10√7, so x = (50 ± 10√7)/6 = (25 ± 5√7)/3.

运用二次公式，x = [50 ± √(2500 − 4 × 3 × 150)] / (2 × 3) = [50 ± √(2500 − 1800)] / 6 = [50 ± √700] / 6。化简根式，√700 = √(100 × 7) = 10√7，故x = (50 ± 10√7)/6 = (25 ± 5√7)/3。

These two stationary points are approximately x ≈ (25 + 13.23)/3 ≈ 12.74 and x ≈ (25 − 13.23)/3 ≈ 3.92. The larger value 12.74 lies outside the feasible domain (x < 10), so the only admissible critical point is x = (25 − 5√7)/3 ≈ 3.92 cm.

这两个驻点分别为x ≈ (25 + 13.23)/3 ≈ 12.74和x ≈ (25 − 13.23)/3 ≈ 3.92。较大值12.74落在可行域外（x < 10），因此唯一可取的临界点是x = (25 − 5√7)/3 ≈ 3.92 cm。

4. Using the Second Derivative to Confirm a Maximum | 利用二阶导数确认极大值

The second derivative V”(x) = 24x − 200 allows us to classify the stationary point. Evaluate V” at x ≈ 3.92: V”(3.92) = 24(3.92) − 200 ≈ 94.08 − 200 = −105.92. Since the second derivative is negative, the function is concave down at this point, confirming a local maximum.

二阶导数V”(x) = 24x − 200可用于判定驻点类型。代入x ≈ 3.92：V”(3.92) = 24(3.92) − 200 ≈ 94.08 − 200 = −105.92。由于二阶导数为负，函数在该点处为凸向下，确认这是局部极大值。

Because V(x) is continuous on the closed interval [0, 10] and vanishes at the endpoints, this local maximum is also the global maximum on the domain. Thus, the optimal cut-out size is (25 − 5√7)/3 cm, yielding a maximum volume of approximately 1056.3 cm³ (to 1 d.p.).

由于V(x)在闭区间[0, 10]上连续且端点处为零，此局部极大值也是定义域内的全局最大值。因此，最优裁剪尺寸为(25 − 5√7)/3 cm，对应最大体积约为1056.3 cm³（精确到一位小数）。

5. Interpreting the Result in Context | 情境中诠释结果

The mathematical solution means the company should cut out squares of side approximately 3.92 cm to achieve the greatest possible volume from each sheet. The production manager must now decide whether to adopt this exact measurement or to use a rounded figure like 3.9 cm or 4.0 cm, considering manufacturing tolerances. The volume is not extremely sensitive to small changes around this optimum, as the derivative near the critical point is zero, indicating relatively flat behaviour.

数学解意味着公司每次应从每个角剪去边长约3.92 cm的正方形，以从每张纸板得到最大体积。生产经理现在需要决定是采用该精确值，还是考虑制造公差而使用3.9 cm或4.0 cm这样的约数值。由于临界点附近的导数接近于零，说明变化平缓，体积对此最优值附近的小变动并不十分敏感。

It is good practice to present the answer in exact surd form where possible; the exact height x = (25 − 5√7)/3 cm demonstrates full algebraic accuracy, while the decimal approximation provides practical information for the workshop floor.

最佳做法是尽可能以精确根式形式给出答案；确切高度x = (25 − 5√7)/3 cm展现了完全的代数准确性，而小数近似则为生产车间提供了实用信息。

6. Validating the Model and Domain Constraints | 验证模型与定义域约束

Every mathematical model must be checked against reality. Here the domain 0 < x < 10 arises from the physical dimensions: x must be positive and less than half the shorter side. In addition, for the flaps to fold properly, x should probably not exceed half of the longer side either, but the tighter constraint naturally eliminates the extraneous root.

每一个数学模型都须经现实检验。此例中0 < x < 10的定义域来自物理尺寸：x必须为正且小于较短边的一半。此外，为使折页能正确折叠，x一般也不能超过较长边的一半，但更严格的约束自然排除了多余的根。

The model assumes the cardboard has negligible thickness and that folds are perfect right angles. In practice, slight material thickness might reduce the inner volume, but for AS level analysis the idealised model is sufficient and standard.

此模型假定纸板厚度忽略不计，且折叠为完美的直角。实际中，材料厚度可能略微减小内部容积，但在AS层次分析中，理想化模型已足够且符合标准做法。

7. Sensitivity Analysis and Marginal Change | 敏感性分析与边际变化

The derivative V'(x) represents the instantaneous rate of change of volume with respect to the cut size. At the optimum, V'(3.92) = 0, meaning a tiny increase in x has virtually no first-order effect on the volume. To estimate the effect of a small error Δx, we can use the second-order Taylor expansion: ΔV ≈ ½ V”(x*)(Δx)². Since V” is negative, any deviation reduces the volume.

导数V'(x)代表体积随裁剪尺寸变化的瞬时变化率。在最优处V'(3.92)=0，意味着x的微小增量对体积的一阶影响几乎为零。若要估算微小误差Δx的影响，可使用二阶泰勒展开：ΔV ≈ ½ V”(x*)(Δx)²。由于V”为负，任何偏离都会减少体积。

For instance, if the workforce cuts x = 4.0 cm instead of 3.92 cm (Δx ≈ 0.08), the volume change is ΔV ≈ ½ × (−105.9) × (0.08)² ≈ −0.34 cm³, a negligible loss. This confirms that the optimum is robust for practical purposes.

例如，若工人实际裁剪x = 4.0 cm而非3.92 cm（Δx ≈ 0.08），体积变化量为ΔV ≈ ½ × (−105.9) × (0.08)² ≈ −0.34 cm³，损失微乎其微。这证实此最优值在实际应用中具有鲁棒性。

8. Graphical Verification and Technology Use | 图形验证与技术工具使用

In the Edexcel specification, use of graphing software or a calculator with graphical display is encouraged. Plotting V(x) = 4x³ − 100x² + 600x for 0 ≤ x ≤ 10 clearly shows a peak at approximately x = 3.9. The graph also highlights that the volume rises sharply from zero and falls more gently, which matches the derivative analysis. Annotating key points on the graph demonstrates your ability to link algebraic and graphical representations.

在Edexcel考纲中，鼓励使用绘图软件或具备图形显示功能的计算器。绘制V(x) = 4x³ − 100x² + 600x在0 ≤ x ≤ 10的图像，可清晰看到约x = 3.9处有一峰值。图像还凸显出体积从零快速上升而后平缓下降，这与导数分析相符。在图形上标注关键点，能展示你关联代数与图形表征的能力。

Using a calculator’s solver or tracing facility, you can numerically verify the root of V'(x)=0 and confirm the maximum volume. This cross-check is a strong exam technique when time permits.

利用计算器的解方程或描迹功能，可数值验证V'(x)=0的根并确认最大体积。当时间允许时，这种交叉检验是一种有力的考试技巧。

9. Extension: What if the Cardboard is Square? | 拓展：若纸板为正方形？

If the cardboard sheet measured 20 cm by 20 cm instead, the volume function becomes V(x) = x(20 − 2x)² = 4x³ − 80x² + 400x. Setting the derivative 12x² − 160x + 400 = 0 and solving gives x = (160 ± √(25600 − 19200))/24 = (160 ± √6400)/24 = (160 ± 80)/24, so x = 10 or x = 10/3 ≈ 3.33. The feasible root is x = 10/3 cm, exactly one-sixth of the original side length. This is a neat result often used to introduce optimisation.

若纸板尺寸为20 cm × 20 cm的正方形，则体积函数变为V(x) = x(20 − 2x)² = 4x³ − 80x² + 400x。令导数12x² − 160x + 400 = 0并求解，得x = (160 ± √(25600 − 19200))/24 = (160 ± √6400)/24 = (160 ± 80)/24，故x = 10或x = 10/3 ≈ 3.33。可行根为x = 10/3 cm，恰为原边长度的六分之一。这是介绍最优化时常用的精巧结果。

Comparing the general rectangular case with the square case reinforces how the domain changes with the aspect ratio of the sheet, and why checking the validity of both stationary points is crucial.

将一般的矩形情形与正方形情形加以比较，可以加深对定义域如何随纸板长宽比变化的理解，并说明检验两个驻点的有效性为何至关重要。

10. Common Mistakes and How to Avoid Them | 常见错误及其避免方法

Forgetting to expand and simplify the volume expression before differentiating, leading to messy algebra. Always expand to a polynomial first.
未能先展开并简化体积表达式就求导，导致代数运算混乱。务必先展开为多项式。
Not identifying the correct domain: some students set V’=0, find two roots, and blindly choose the smaller one without considering x < 10. Always check the physical constraints.
未能识别正确的定义域：有些学生令V’=0，求出两个根后盲目选取较小者，而不考虑x < 10。始终要检查物理约束。
Misinterpreting the second derivative: simply stating ‘V” < 0 so maximum' without evaluating at the critical point loses marks. Show the substitution.
误解二阶导数：仅陈述“V”<0故为极大值”而未在临界点处取值会失分。需展示代入过程。
Giving only a decimal answer without the exact form when the question asks for it. Surd answers demonstrate precision.
题目要求精确值时只给出小数近似。根式答案体现精确性。

11. Summary of the Case Study Method | 案例分析方法总结

This practical exercise has modelled a real decision problem using Year 12 calculus. The structured approach is: (1) Define variables and construct a function; (2) Determine the feasible domain; (3) Differentiate and find stationary points; (4) Classify with second derivative test and check endpoints; (5) Interpret the solution in the original context; (6) Validate and discuss sensitivity. Master this framework, and you will be well-prepared for any application question on the Edexcel AS paper.

本实战演练运用Year 12微积分对真实决策问题进行了建模。结构化的方法为：(1) 定义变量并构造函数；(2) 确定可行域；(3) 求导并寻找驻点；(4) 用二阶导数检验分类并检查端点；(5) 在原情境中诠释解；(6) 验证并讨论敏感性。掌握此框架，你将能充分应对Edexcel AS试卷上的任何应用问题。

The open box problem is a classic not only because it tests differentiation and algebra, but because it teaches you to think critically about modelling assumptions and to communicate mathematically – skills that underpin the entire applied maths component of the course.

无盖纸盒问题之所以经典，不仅因为它检验了微分与代数，更因为它教会你批判性地思考模型假设并进行数学交流——正是这些能力支撑着课程全部应用数学内容。

12. Further Practice and Exam Tips | 进阶练习与考试技巧

To consolidate, try adapting the problem: use a 40 cm by 25 cm sheet, or impose a minimum height requirement of 5 cm. For each variation, work through all six steps and compare your solutions. In the Edexcel exam, always show the domain, the derivative equation in simplified form, and a clear statement linking the second derivative sign to a maximum or minimum. A sketched graph with the stationary point labelled can earn additional marks for communication.

为巩固所学，请尝试改编原题：使用40 cm × 25 cm的纸板，或要求最小高度为5 cm。对每种变式，套用全部六个步骤并比较解答。在Edexcel考试中，始终要展示定义域、化简后的导数方程，并清晰陈述二阶导数符号与极大值或极小值的联系。画出示意图并标注驻点，可在表达交流方面获得加分。

Regular practice with case studies builds confidence for the longer, unstructured problem-solving questions that appear near the end of the paper. Remember: a clear logical flow and correct mathematical notation are just as important as the final answer.

经常进行案例分析练习，可提升解答试卷末尾那些较长的、无明确步骤提示的问题时的信心。请记住：清晰的逻辑流程和正确的数学记号与最终答案同等重要。

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Case Study Practical Exercises in Year 12 Edexcel Maths | 案例分析实战演练：Year 12 Edexcel 数学