📚 Case Study in Quadratic Optimisation for CCEA Further Mathematics | CCEA进阶数学案例研究：二次函数优化
This case study is designed for Year 10 students following the CCEA Further Mathematics course. It guides you through a practical optimisation problem, showing how quadratic functions can model real-life situations and how to find maximum or minimum values using algebraic, graphical and calculus-based methods. By working through the steps, you will strengthen your problem-solving skills and prepare for exam-style questions.
本案例研究面向CCEA进阶数学课程十年级学生。它将引导你完成一个实际的最优化问题，展示二次函数如何模拟现实生活场景，以及如何使用代数、图形和微积分方法求最大值或最小值。通过逐步演练，你将强化解题技巧，并为考试题型做好准备。
1. Problem Statement | 问题陈述
A farmer plans to fence a rectangular field along the bank of a straight river. He has exactly 100 metres of fencing material. The river bank acts as a natural boundary, so fencing is needed only for the three remaining sides – the two widths and one length. The farmer’s goal is to choose the dimensions of the field so that the enclosed area is as large as possible. What dimensions should he use?
一位农民计划在笔直的河岸边围一个矩形田地。他有恰好100米围栏材料。河岸作为天然边界，因此只需要为其余三边——两条宽边和一条长边——设置围栏。农民的目标是选择田地的尺寸，使围合的面积尽可能大。他应该使用哪些尺寸？
2. Setting Up Variables and a Diagram | 设定变量并画出示意图
It is always helpful to draw a simple sketch. Let the width of the field (the side perpendicular to the river) be x metres. Because two widths and one length are fenced, the total fencing used is 2x + length = 100. Rearranging gives the length parallel to the river as (100 − 2x) metres. The enclosed area A is then the product of width and length, so A = x(100 − 2x).
画一个简单的示意图总是有帮助的。设田地的宽度（垂直于河岸的边）为 x 米。由于围栏覆盖了两条宽边和一条长边，围栏总用量为 2x + 长边 = 100。整理可得平行于河岸的长度为 (100 − 2x) 米。围合面积 A 便是宽与长的乘积，即 A = x(100 − 2x)。
3. Forming the Quadratic Function | 构造二次函数
Expand the area expression: A = 100x − 2x². This is a quadratic function of the form A(x) = −2x² + 100x, where the coefficient of x² is negative (−2). A negative leading coefficient tells us the graph is an inverted parabola, so the function has a maximum value rather than a minimum. The variable x represents the width, and it must be positive, with 0 < x < 50 so that the length remains positive.
展开面积表达式：A = 100x − 2x²。这是一个形如 A(x) = −2x² + 100x 的二次函数，其中 x² 的系数为负 (−2)。负的二次项系数告诉我们图像是开口向下的抛物线，因此函数有最大值而不是最小值。变量 x 代表宽度，且必须为正，要求 0 < x < 50 以使长度保持为正。
4. Finding the Maximum Area by Completing the Square | 通过配方法求最大面积
To locate the maximum precisely, we rewrite the quadratic in vertex form by completing the square. Factor out −2 from the terms containing x: A = −2(x² − 50x). Inside the brackets, we complete the square: x² − 50x = (x − 25)² − 625. Substitute back: A = −2[(x − 25)² − 625] = −2(x − 25)² + 1250. The expression −2(x − 25)² is always ≤ 0, so A is maximised when (x − 25)² = 0, giving x = 25 metres. The maximum area is Amax = 1250 m². The corresponding length is 100 − 2(25) = 50 metres.
为了精确找到最大值，我们通过配方法将二次函数改写为顶点形式。从含 x 的项中提取公因数 −2：A = −2(x² − 50x)。在括号内进行配方：x² − 50x = (x − 25)² − 625。代回原式：A = −2[(x − 25)² − 625] = −2(x − 25)² + 1250。表达式 −2(x − 25)² 总是 ≤ 0，因此当 (x − 25)² = 0，即 x = 25 米时，A 取得最大值。最大面积为 Amax = 1250 平方米。对应的长度为 100 − 2(25) = 50 米。
5. Alternative Method: Axis of Symmetry | 替代方法：对称轴
For any quadratic y = ax² + bx + c, the x-coordinate of the vertex is given by x = −b / (2a). Here a = −2 and b = 100, so x = −100 / (2 × −2) = 25. This approach quickly yields the optimal width without completing the square. Substituting x = 25 into the area formula confirms the maximum area of 1250 m². In CCEA examinations, this shortcut is very useful when you need speed.
对于任何二次函数 y = ax² + bx + c，其顶点的 x 坐标由 x = −b / (2a) 给出。这里 a = −2, b = 100，所以 x = −100 / (2 × −2) = 25。此方法无需配方即可快速得到最优宽度。将 x = 25 代入面积公式便可验证最大面积为 1250 平方米。在CCEA考试中，当你需要快速求解时，这个捷径非常有用。
6. Graphical Interpretation and a Table of Values | 图像解释与数值表
Plotting area against width produces a symmetrical inverted parabola. The vertex (25, 1250) is the peak of the graph, while the x-intercepts occur at x = 0 and x = 50, where the area is zero. To visualise the behaviour, we can construct a table of values:
绘制面积关于宽度的图像，得到一条对称的开口向下抛物线。顶点 (25, 1250) 是图像的最高点，而 x 轴截距出现在 x = 0 和 x = 50，此时面积为零。为直观显示函数的变化，我们可以建立一个数值表：
| Width x (m) | Length (100−2x) (m) | Area A (m²) |
|---|---|---|
| 0 | 100 | 0 |
| 10 | 80 | 800 |
| 20 | 60 | 1200 |
| 25 | 50 | 1250 |
| 30 | 40 | 1200 |
| 40 | 20 | 800 |
| 50 | 0 | 0 |
The symmetry around x = 25 is clear, and the area increases to a peak before falling again. This graphical check can help catch algebraic mistakes.
围绕 x = 25 的对称性十分明显，面积在达到峰值后再次下降。这种图像验证有助于发现代数错误。
7. Verification Using Differential Calculus | 使用微分进行验证
CCEA Further Mathematics introduces basic differentiation. We can find dA/dx directly from A = 100x − 2x²: the derivative is dA/dx = 100 − 4x. Setting the derivative equal to zero gives 100 − 4x = 0 → x = 25. The second derivative d²A/dx² = −4 is negative, confirming a maximum. This method reinforces the algebraic result and bridges to more advanced optimisation techniques.
CCEA进阶数学引入了基本微分知识。我们可以直接从 A = 100x − 2x² 求导：dA/dx = 100 − 4x。令导数为零，得 100 − 4x = 0 → x = 25。二阶导数 d²A/dx² = −4 为负，确认是最大值。这一方法巩固了代数结果，并连接到更高级的优化技巧。
8. Sensitivity Analysis: Varying the Fence Length | 灵敏度分析：改变围栏长度
Suppose the farmer has a flexible budget and can buy L metres of fencing. Replacing 100 with a general constant L, the area becomes A = x(L − 2x) = −2x² + Lx. The optimal width is x = L/4, and the maximum area is Amax = L²/8. For L = 100, this gives 1250 m²; for L = 200, the maximum area jumps to 200²/8 = 5000 m². The result scales with the square of the fence length – doubling the fence quadruples the maximum area.
假设农民有灵活的预算，可以购买 L 米围栏。将100替换为一般常数 L，面积变为 A = x(L − 2x) = −2x² + Lx。最优宽度为 x = L/4，最大面积为 Amax = L²/8。当 L = 100 时，得到 1250 平方米；当 L = 200 时，最大面积跃升至 200²/8 = 5000 平方米。这一结果与围栏长度的平方成正比——围栏长度翻倍，最大面积变为四倍。
9. Extending the Problem: Adding a Cost Constraint | 扩展问题：增加成本约束
In reality, fencing materials are not uniform in price. Imagine the river-side fencing costs £8 per metre due to the need for stronger posts, while the two width sides cost £5 per metre. The farmer’s total budget is £400. We now have a cost equation: 5(2x) + 8y = 400, where y is the length along the river. Solve for y: y = (400 − 10x)/8 = 50 − 1.25x. The area becomes A = x(50 − 1.25x) = 50x − 1.25x². Using x = −b/(2a) with a = −1.25, b = 50 gives x = 20 m. The length is y = 25 m, and the maximum area under the budget is 500 m². This variation shows how constraints alter the optimal solution and introduces linear programming concepts.
现实中，围栏材料价格并不统一。假设沿河围栏由于需要更坚固的柱子，每米价格为 £8，而两侧宽边每米 £5。农民的总预算是 £400。现在我们有一个成本方程：5(2x) + 8y = 400，其中 y 是沿河长度。解出 y：y = (400 − 10x)/8 = 50 − 1.25x。面积变为 A = x(50 − 1.25x) = 50x − 1.25x²。利用 x = −b/(2a)，其中 a = −1.25, b = 50，得 x = 20 米。长度为 y = 25 米，在该预算下的最大面积为 500 平方米。这个变式展示了约束条件如何改变最优解，并引入了线性规划的概念。
10. Real-World Connections and Exam Tips | 实际联系与考试技巧
Optimisation problems appear in many fields – agriculture, engineering, economics and product design. In CCEA Further Mathematics papers, you can expect a structured question that leads you through modelling, algebra and interpretation. Always define your variables clearly, write down the expression to be maximised or minimised, and reduce it to one variable. Mastering both completing the square and the symmetric axis method gives you flexibility. Finally, check that your answer makes sense in context (e.g. positive dimensions) and draw a quick graph if time allows.
优化问题出现在许多领域——农业、工程、经济学和产品设计中。在CCEA进阶数学试卷中，你可以预期将遇到一道引导你进行建模、代数和解释的结构化试题。始终
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