📚 Quadratic Modelling – Example 2.6.2 | 二次函数建模例题 2.6.2 精讲
Example 2.6.2 in the AQA A-Level Mathematics Year 1 course explores how a real-world situation can be converted into a quadratic model, and how completing the square reveals its maximum value. This is one of the most frequently tested modelling skills in Pure Mathematics Paper 1.
AQA 数学一年级课程中的例题 2.6.2 展示了如何将实际问题转化为二次函数模型，并运用配方法求出其最大值。这是纯数学试卷一中最常考查的建模技能之一。
1. The Problem | 题目重现
Suppose a farmer has 80 metres of fencing and wishes to form a rectangular enclosure along a straight river bank. The side along the river needs no fencing. Let the width of the enclosure be x metres and the length along the river be L metres.
假设某农夫有 80 米篱笆，想沿平直的河岸围出一块矩形场地。临河的一边不需要篱笆。设场地的宽度（垂直于河岸）为 x 米，沿河方向的长度为 L 米。
The question asks: which width x produces the largest possible area for the enclosure?
题目要求：宽度 x 取何值时，围场面积最大？
2. Expressing the Constraint | 建立约束关系
Because only three sides are fenced, the total length of fencing satisfies 2x + L = 80.
由于只需围三边，篱笆总长满足 2x + L = 80。
2x + L = 80 → L = 80 − 2x
The domain is restricted: x > 0 and L > 0 together give 0 < x < 40.
定义域受到限制：x > 0 且 L > 0，可得 0 < x < 40。
3. Building the Area Function | 构建面积函数
The area A of a rectangle is width multiplied by length:
矩形面积 A 等于宽乘以长：
A = x × L = x(80 − 2x) = 80x − 2x²
This quadratic has a negative coefficient of x², so its graph is a downward-opening parabola and therefore possesses a maximum value.
该二次函数的 x² 项系数为负，因此图像是开口向下的抛物线，所以必定存在最大值。
4. Completing the Square | 配方求解
To locate the maximum, rewrite A in completed-square form:
为了求出最大值，将 A 写成顶点式：
A = x(80 − 2x) = −2x² + 80x = −2(x − 20)² + 800
The step-by-step working is shown below:
下面逐步展示计算过程：
Factor out −2: A = −2(x² − 40x)
提取公因数 −2：A = −2(x² − 40x)
Complete the square inside: x² − 40x = (x − 20)² − 400
对内层配方：x² − 40x = (x − 20)² − 400
Multiply back: A = −2[(x − 20)² − 400] = −2(x − 20)² + 800
代回并展开：A = −2[(x − 20)² − 400] = −2(x − 20)² + 800
5. Interpreting the Vertex | 解读顶点
occurs at x = 20. Since (x − 20)² ≥ 0 for every real x, the term −2(x − 20)² is at most 0.
在 A = −2(x − 20)² + 800 中，抛物线的顶点位于 x = 20 处。由于 (x − 20)² ≥ 0 对所有实数 x 恒成立，所以 −2(x − 20)² 的最大值为 0。
Therefore, the maximum area is A = 800 m² when x = 20 m.
因此，当 x = 20 米时，最大面积为 A = 800 平方米。
The corresponding length is L = 80 − 2 × 20 = 40 m.
对应的长度为 L = 80 − 2 × 20 = 40 米。
6. Verification | 验证结果
Test a few values in the model to confirm that x = 20 is sensible.
在模型中代入几个数值，检验 x = 20 是否合理。
| Width x (m) | Length L = 80 − 2x (m) | Area A = xL (m²) |
| 10 | 60 | 600 |
| 20 | 40 | 800 |
| 30 | 20 | 600 |
The symmetrical pattern confirms that the maximum occurs at the centre of the domain.
数据呈现出对称性，进一步确认最大值出现在定义域的正中央。
7. Alternative Method: Differentiation | 另一种方法：微分
The same maximum can be found using calculus. Differentiate A with respect to x:
同样可以用微积分求最大值。对 x 求导：
dA/dx = 80 − 4x
Setting dA/dx = 0 gives 80 − 4x = 0 → x = 20.
令 dA/dx = 0，得 80 − 4x = 0 → x = 20。
The second derivative d²A/dx² = −4 < 0 confirms that x = 20 corresponds to a maximum.
二阶导数 d²A/dx² = −4 < 0 确认 x = 20 对应最大值。
8. Common Mistakes and Exam Tips | 常见错误与考试要点
at the river side needs no fencing, and incorrectly writing the perimeter as 2x + 2L = 80.
忘记河岸一边无需围栏，错误地写成 2x + 2L = 80。
Omitting the domain 0 < x < 40; AQA mark schemes often award a mark for stating a sensible domain.
遗漏定义域 0 < x < 40；AQA 评分标准通常会为写出合理定义域给分。
ing at x = 20 without finding the maximum area A = 800 or the corresponding length L = 40.
求出 x = 20 后不再继续，漏掉最大面积 A = 800 或对应长度 L = 40。
Making a sign error when completing the square; always expand −2(x − 20)² + 800 to check it returns −2x² + 80x.
配方时出现符号错误；务必展开 −2(x − 20)² + 800，验证是否还原为 −2x² + 80x。
Example 2.6.2 demonstrates the complete modelling cycle: define variables, construct the model, apply algebra, interpret the result, and verify. Mastering this cycle is essential for securing top marks in AQA Pure Mathematics.
例题 2.6.2 完整展示了建模流程：定义变量、建立模型、运用代数、解释结果、验证答案。掌握这一流程是在 AQA 纯数学考试中冲击高分的关键。
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