📚 Year 10 CAIE Additional Mathematics: Paper Writing Framework and Exemplar | Year 10 CAIE 进阶数学：论文写作框架与范文
Writing a mathematics paper is not a typical requirement for the CAIE IGCSE Additional Mathematics examination, yet many Year 10 programmes include an investigative project to help students develop analytical reasoning, communication, and structured problem-solving skills. This article provides a clear framework for composing such a paper, followed by a worked exemplar on optimising the volume of an open box, a classic application of differentiation from the CAIE Additional Mathematics syllabus. By following the step-by-step guide, learners can transform a mathematical exploration into a coherent and academically sound report.
撰写数学论文虽非CAIE IGCSE进阶数学考试的常规要求，但许多Year 10课程会设置探究项目，以帮助学生培养分析推理、沟通和结构化解决问题的能力。本文提供了一个撰写此类论文的清晰框架，并附上一篇关于优化无盖纸盒体积的范文——这是CAIE进阶数学大纲中微分应用的经典题目。学生按照这份分步指南，即可将数学探索转化为一篇条理清晰、学术规范的报告。
1. Understanding the Purpose of a Mathematics Paper | 理解数学论文的写作目的
A mathematics paper in the context of Year 10 CAIE Additional Mathematics goes beyond solving exercises. It requires students to select a topic, pose a question, apply relevant mathematical concepts, and present findings in a logical narrative. This process mirrors the way mathematicians communicate their work and prepares learners for the rigour of A Level coursework or extended essays. The paper should demonstrate a genuine inquiry, such as modelling a real-life situation with functions, investigating a geometric relationship, or solving an optimisation problem using calculus.
在Year 10 CAIE进阶数学中，数学论文不等于解答习题。学生需要自选主题、提出问题、运用相关数学概念，并以逻辑清晰的叙述展示研究成果。这一过程模拟了数学家交流成果的方式，也为学习者今后面对A Level课程作业或拓展论文的严谨要求做好准备。论文应展现真实的探究，比如用函数模拟现实情境、研究几何关系，或利用微积分求解最优化问题。
2. Overall Structure of the Paper | 论文的整体结构
A well-organised paper typically contains the following sections: Title, Abstract, Introduction, Mathematical Model or Methodology, Results, Discussion, Conclusion, and References. Each part serves a distinct role. The title and abstract convey the essence of the investigation; the introduction sets the context and states the aim; the methodology explains the mathematical tools used; the results present data, graphs, or equations; the discussion interprets the findings; and the conclusion summarises the work while suggesting extensions. Adhering to this structure ensures clarity and helps the reader follow the mathematical argument.
一篇条理清晰的论文通常包含以下几个部分：标题、摘要、引言、数学模型或方法、结果、讨论、结论和参考文献。每个部分各有其功能。标题和摘要传达研究的核心内容；引言交代背景并说明目的；方法部分解释所使用的数学工具；结果展示数据、图像或方程；讨论对发现进行解读；结论总结全文并提出延伸方向。遵循这一结构能确保论文清晰易读，帮助读者跟上数学论证的脉络。
3. Crafting the Title and Abstract | 撰写标题与摘要
The title should be concise yet informative, capturing the main idea of the investigation. For example, ‘Maximising the Volume of an Open Box Constructed from a Square Sheet’ immediately tells the reader what is being explored. The abstract—usually 80–120 words—briefly outlines the problem, the mathematical method used, key results, and the main conclusion. It enables a reader to decide whether the full paper is relevant. Write the abstract in the present tense or past tense, avoiding any content not present in the paper.
标题应当简洁而信息充分，能体现探究的主旨。例如，“最大化由正方形纸板制成的无盖盒体积”这个标题立刻让读者明白探究内容。摘要通常为80–120词，简要概述问题、使用的数学方法、关键结果和主要结论。它帮助读者判断全文是否相关。摘要用现在时或过去时撰写，避免出现论文中未包含的内容。
4. Writing an Effective Introduction | 写出有效的引言
The introduction sets the scene. Begin by explaining why the topic is interesting or useful, perhaps linking it to real-world applications such as packaging design or engineering constraints. Clearly state the research question or objective, e.g., ‘What dimensions of the cut-out squares will yield the maximum volume for the open box?’ Also mention the mathematical areas you will draw upon—differentiation, quadratic functions, or algebraic manipulation. A strong introduction motivates the investigation and prepares the reader for the method to follow.
引言铺设背景。先解释该主题为何有趣或实用，可联系实际应用，如包装设计或工程约束。然后明确陈述研究问题或目标，例如：“剪去的正方形边长取何值时，无盖盒的体积达到最大？”同时提及将用到的数学领域——微分、二次函数或代数处理。有力的引言能激发读者兴趣，并为后面的方法做好铺垫。
5. Building the Mathematical Model | 建立数学模型
In this section, translate the physical problem into algebraic language. Define all variables and provide a clear diagram if possible. For the open box problem, let the original square sheet have side length L cm, and let x cm be the side length of each square cut from the corners. The resulting box has height x cm and a base of side length (L–2x) cm. The volume V can then be expressed as:
V(x) = x(L – 2x)²
在本节中，将实际问题转化为代数语言。定义所有变量，并尽量提供清晰的示意图。以无盖盒问题为例，设原正方形纸板边长为L cm，每个角剪去的正方形边长为x cm。所得盒子的高为x cm，底边长为(L–2x) cm。于是体积V可表达为：
V(x) = x(L – 2x)²
State any reasonable domain for the variable, such as 0 < x < L/2, ensuring the box physically exists. This model forms the foundation for the subsequent calculus work and should be explained step by step.
同时指明变量的合理取值范围，如 0 < x < L/2，以保证盒子实际存在。该模型是后续微积分运算的基础，应逐步解释清楚。
6. Applying Mathematical Techniques | 运用数学方法
With the volume function established, proceed to find its maximum using differentiation, a key topic in CAIE IGCSE Additional Mathematics. Expand the expression if necessary and then differentiate with respect to x. For a sheet with L = 30 cm, the function becomes:
V(x) = x(30 – 2x)² = 4x³ – 120x² + 900x
Differentiating gives:
dV/dx = 12x² – 240x + 900
Set dV/dx = 0 to locate stationary points. Solving the quadratic equation 12x² – 240x + 900 = 0 yields the possible values of x. Use the second derivative test or the sign of dV/dx on either side to confirm which stationary point gives a maximum. Show all algebraic steps clearly, as this demonstrates the application of syllabus content under the topic ‘Applications of Differentiation’.
建立体积函数后，利用微分求其最大值，这是CAIE IGCSE进阶数学的核心主题。必要时先展开表达式，然后对x求导。对于边长为L = 30 cm的纸板，函数变为：
V(x) = x(30 – 2x)² = 4x³ – 120x² + 900x
求导得：
dV/dx = 12x² – 240x + 900
令 dV/dx = 0 以确定驻点。解二次方程 12x² – 240x + 900 = 0 得到x的可能值。利用二阶导数检验或驻点两侧一阶导数的符号，确认哪个驻点对应最大值。清晰展示所有代数步骤，这体现了“微分的应用”这一教学内容的实际运用。
7. Presenting the Results | 展示结果
Results should be presented in an organised manner. Construct a table showing values of x, the corresponding base dimensions, and the volume V for key points. For the 30 cm sheet example:
| x (cm) | Base side (cm) | Volume V (cm³) |
|---|---|---|
| 0 | 30 | 0 |
| 5 | 20 | 2000 |
| 7.5 | 15 | 1687.5 |
| 10 | 10 | 1000 |
| 15 | 0 | 0 |
Include a graph of V against x, clearly labelling the maximum point. The visual representation reinforces the analytical finding. State the optimal dimensions explicitly: for L = 30 cm, the maximum volume of 2000 cm³ occurs when x = 5 cm, giving a base of 20 cm × 20 cm × 5 cm height.
结果应条理清晰地呈现。构建表格，列出关键点的x值、对应的底边尺寸和体积V。以30 cm纸板为例：
| x (cm) | 底边长 (cm) | 体积 V (cm³) |
|---|---|---|
| 0 | 30 | 0 |
| 5 | 20 | 2000 |
| 7.5 | 15 | 1687.5 |
| 10 | 10 | 1000 |
| 15 | 0 | 0 |
绘制V随x变化的图像，并清晰地标出最大值点。直观的图像能强化分析结论。明确写出最优尺寸：当L=30 cm时，x=5 cm时体积达到最大2000 cm³，此时盒子尺寸为20 cm × 20 cm × 5 cm。
8. Discussion and Interpretation | 讨论与诠释
Interpret the results in the context of the original problem. Explain why x = 5 cm gives a larger volume than x = 7.5 cm or x = 2.5 cm, referring to the balance between height and base area. Discuss any limitations: the model assumes perfect construction with no material thickness, and the sheet is perfectly square. Mention what would happen if the sheet were rectangular or if a lid were added—this shows deeper thinking. If any simplifying assumptions were made, acknowledge them and explain their impact.
将结果放回原问题背景下进行解读。解释为何x=5 cm能比x=7.5 cm或x=2.5 cm产生更大的体积，关联高度与底面积之间的平衡。讨论模型的局限性：例如假设制作完美、材料厚度为零，以及纸板为完全正方形。可提及若纸板为矩形或加上盖子结果会如何变化——这展现了更深入的思考。如有任何简化假设，应明确指出并解释其影响。
9. Conclusion and Extensions | 结论与拓展
The conclusion should summarise the investigation succinctly: the maximum volume for a 30 cm square sheet is 2000 cm³ when squares of 5 cm are removed. Restate the mathematical significance—the problem illustrates how differentiation transforms a physical constraint into an optimised solution. Suggest possible extensions: using a rectangular sheet, minimising surface area for a fixed volume, or applying the same method to cylindrical containers. This signals that the investigation is complete yet opens avenues for further inquiry.
结论应简洁地总结探究：对于30 cm的正方形纸板，剪去5 cm正方形时最大体积为2000 cm³。重申其数学意义——该问题展示了微分如何将物理约束转化为最优解。建议可能的拓展方向：使用长方形纸板、在体积固定的前提下最小化表面积、或将同样方法应用于圆柱形容器。这表明探究虽已完整，但为后续研究打开了空间。
10. Exemplar Paper: Introduction to the Open Box Problem | 范文展示：无盖盒问题的引言
Below is a worked excerpt that integrates the framework into a continuous paper. The investigation is ‘Maximising the Volume of an Open Box Constructed from a Square Sheet’, designed for a Year 10 CAIE Additional Mathematics project.
以下是一篇将上述框架融为一体的范文节选。探究课题为“最大化由正方形纸板制成的无盖盒体积”，适用于Year 10 CAIE进阶数学项目。
Problem statement: In many packaging industries, reducing waste while maximising container capacity is crucial. This investigation considers a square cardboard sheet of side length 30 cm. By cutting equal squares from each corner and folding up the sides, an open box is formed. The aim is to determine the cut size x that maximises the volume of the box, using differentiation and algebraic methods from the CAIE Additional Mathematics syllabus.
问题陈述：在许多包装行业中，减少浪费并最大化容器容量至关重要。本次探究考虑一张边长为30 cm的正方形纸板。剪去四个角上相同的小正方形，折叠各边即可制成一个无盖盒。目标是通过CAIE进阶数学大纲中的微分与代数方法，求出使盒子体积最大化的剪去尺寸x。
11. Exemplar Paper: Derivation and Optimisation | 范文展示：推导与优化过程
Let x cm be the side length of each cut-out square, where 0 < x < 15. After cutting, the base of the box is a square of side (30 – 2x) cm, and the height is x cm. Therefore, the volume V in cubic centimetres is:
V(x) = x(30 – 2x)² = 4x³ – 120x² + 900x
设剪去的正方形边长为x cm，其中 0 < x < 15。剪去后，盒底为边长(30–2x) cm的正方形，高为x cm。因此体积V（单位：立方厘米）为：
V(x) = x(30 – 2x)² = 4x³ – 120x² + 900x
Differentiating with respect to x yields dV/dx = 12x² – 240x + 900. Setting dV/dx = 0 gives 12x² – 240x + 900 = 0 → x² – 20x + 75 = 0. Solving yields (x – 5)(x – 15) = 0, so x = 5 or x = 15. Since x = 15 lies outside the domain (the base would vanish), we consider x = 5. The second derivative d²V/dx² = 24x – 240. At x = 5, d²V/dx² = 120 – 240 = –120 < 0, confirming a maximum. Substituting x = 5 into V(x) gives V = 5 × (20)² = 2000 cm³.
对x求导得 dV/dx = 12x² – 240x + 900。令 dV/dx = 0 得 12x² – 240x + 900 = 0 → x² – 20x + 75 = 0。解得 (x – 5)(x – 15) = 0，故 x = 5 或 x = 15。由于 x = 15 超出定义域（底边将为零），只需考虑 x = 5。二阶导数 d²V/dx² = 24x – 240，当 x = 5 时，d²V/dx² = 120 – 240 = –120 < 0，确认为最大值。将 x = 5 代入 V(x) 得 V = 5 × (20)² = 2000 cm³。
12. Exemplar Paper: Concluding Remarks and Reflections | 范文展示：结论与反思
Hence, the maximum volume achievable is 2000 cm³ when 5 cm squares are removed from each corner. The investigation successfully applied differentiation to a practical optimisation task. The result aligns with the expectation that an optimal balance exists between height and base area. If the original sheet were scaled by a factor k, the optimal x would scale proportionally, which could be a natural extension to explore. The paper demonstrates a clear progression from problem to model to solution, consistent with the framework described above.
因此，从每个角剪去5 cm的正方形时，可达到最大体积2000 cm³。本次探究成功地将微分应用于实际优化问题。结果符合预期，即高度与底面积之间存在最佳平衡点。如果原始纸板按比例k缩放，最优x也会按比例变化，这可以作为一个自然的延伸探究。整篇论文展示了从问题到模型再到解决方案的清晰推进过程，与上述框架完全一致。
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