A Case Study: Maximising Area with a Fixed Fence | 案例分析：固定篱笆下围出最大面积

📚 A Case Study: Maximising Area with a Fixed Fence | 案例分析：固定篱笆下围出最大面积

In this case study, we will apply quadratic functions and optimisation techniques from Year 9 CIE Further Mathematics to a real‑world problem. Follow the step‑by‑step analysis to see how abstract algebra turns into a practical tool for making the best possible decision.

在本案例中，我们将运用 Year 9 CIE 进阶数学中的二次函数与最优化方法，解决一个实际问题。跟随分步分析，看看抽象的代数如何变成做出最佳决策的实用工具。

1. Problem Statement | 问题陈述

A farmer has 100 metres of straight fencing. He wants to enclose a rectangular field against a long, straight wall, using the wall as one side of the rectangle. The other three sides will be made from the fencing. What dimensions of the rectangle will give the maximum possible area, and what is that area?

一位农民有 100 米直的篱笆。他想靠着一堵很长的直墙围出一块矩形场地，以墙作为矩形的一边。其余三边用篱笆围成。问矩形的边长取何值时，围出的面积最大？最大面积是多少？

2. Understanding and Assumptions | 理解与假设

We assume that the wall is infinitely long, so we do not have to worry about its length limiting the rectangle. The fencing is used only for the two widths and the one length opposite the wall. The field is perfectly rectangular, and we ignore any practical issues such as gates, posts or uneven ground.

我们假设墙无限长，因此无需担心墙的长度限制矩形的另一边。篱笆仅用于构成两个宽边和一个与墙相对的长边。场地是完美的矩形，我们忽略门、桩或地面不平等实际问题。

3. Building a Mathematical Model | 建立数学模型

Let x be the width of the rectangle perpendicular to the wall (the two equal sides made of fencing). Then the length of the fence used for these two sides is 2x. The remaining fencing forms the length y that is parallel to the wall. Since the total fencing is 100 m, we have:

设 x 为矩形垂直于墙的宽度（由篱笆构成的两条等长边）。这两边所用的篱笆长度为 2x。剩下的篱笆构成平行于墙的长边 y。因为篱笆总长为 100 m，可得：

2x + y = 100 => y = 100 - 2x

The variables must be non‑negative, and y must also be non‑negative, so x can only take values between 0 and 50 metres: 0 ≤ x ≤ 50.

变量必须为非负数，且 y 也必须为非负数，因此 x 只能在 0 到 50 米之间取值：0 ≤ x ≤ 50。

4. Expressing the Area Function | 表达面积函数

The area A of the rectangle is the product of the width and the length. Substituting y gives a quadratic function in x:

矩形面积 A 为宽与长的乘积。代入 y 后得到关于 x 的二次函数：

A(x) = x · y = x (100 - 2x) = -2x² + 100x

This is a concave‑down quadratic (the coefficient of x² is negative), so it has a maximum value at its vertex.

这是一个开口向下的二次函数（x² 的系数为负），因此其顶点对应最大值。

5. Finding the Maximum by Completing the Square | 配方法求最大值

Rewrite the quadratic in vertex form by completing the square. Factor out the coefficient of x² from the x terms:

通过配方法将二次函数写成顶点式。从含 x 的项中提出 x² 的系数：

A(x) = -2 (x² - 50x)

Inside the bracket, add and subtract (50/2)² = 625:

在括号内加上并减去 (50/2)² = 625：

A(x) = -2 [x² - 50x + 625 - 625] = -2 [(x - 25)² - 625]

Distribute the -2:

展开 -2：

A(x) = -2 (x - 25)² + 1250

The maximum area occurs when the squared term is zero, i.e. when x = 25. At this x, the maximum area is 1250 m².

当平方项为零时，即 x = 25，面积达到最大值。此时最大面积为 1250 m²。

6. Verifying with Differentiation (Extension) | 导数验证（拓展）

In further mathematics we often use calculus as an alternative. Differentiate A(x) with respect to x:

在进阶数学中，我们常用微积分作为另一种方法。对 A(x) 关于 x 求导：

dA/dx = -4x + 100

Set the derivative to zero to find stationary points:

令导数为零以求稳定点：

-4x + 100 = 0 => x = 25

The second derivative is d²A/dx² = -4 < 0, confirming a maximum. Both methods agree.

二阶导数 d²A/dx² = -4 < 0，确认是最大值。两种方法结果一致。

7. Optimum Dimensions | 最优尺寸

Substitute x = 25 back into the expression for y:

将 x = 25 代回 y 的表达式：

y = 100 - 2(25) = 50 m

So the rectangle that maximises the area is 25 m wide (perpendicular to the wall) and 50 m long (parallel to the wall). The enclosed area is 1250 m².

因此，使面积最大的矩形宽 25 米（垂直于墙），长 50 米（平行于墙）。围出的面积为 1250 平方米。

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8. Verifying the Result | 结果检验

check boundary values and a few interior points to ensure the maximum really is at x = 25.

我们应检查边界值及几个内部点，确保最大值确实在 x = 25 处。

x (m)	y (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

pattern with the peak at x = 25. Both algebraic and numerical checks confirm the optimum.

表格清晰地展示了以 x = 25 为峰值的对称模式。代数与数值检验均证实了最优解。

9. Practical Considerations and Limitations | 实际考虑与局限性

gion, and the optimum might become x = 35, giving y = 30, both on the boundary. Also, corners may need reinforcement, and the fence might sag, slightly reducing the effective straight length. Such real‑ monstrate why mathematical models must always be validated against the physical situation.

实际上，墙的长度可能是有限的。如果可用的墙只有 30 米长，那么 y 必须 ≤ 30。这将改变可行域，最优解可能变成 x = 35，y = 30，两者都在边界上。此外，角落可能需要加固，篱笆可能下垂，略微减少有效的直线长度。这些现实约束说明为什么数学模型始终需要根据实际情况进行验证。

10. Extensions and Variations | 拓展与变体

lution gives x = L/4, y = L/2, and maximum area L²/8. What if the field is not a rectangle? more expensive and less convenient. Comparing shapes is an excellent further exploration.

同样的推理可应用于任意固定篱笆长度 L。一般解为 x = L/4，y = L/2，最大面积 L²/8。若场地不是矩形呢？一个以墙为直径的半圆将围出面积 ½ π (100/π)² ≈ 1592 m²，面积更大，但弧形篱笆成本更高且不便。比较不同形状是一个非常有益的拓展探索。

11. Common Mistakes to Avoid | 常见错误分析

x + y = 100, forgetting that there are two widths. When completing the square, mistakes with signs are frequent – out justification does not work here; the symmetry of the rectangle is broken by the wall.

学生经常忘记说明定义域 0 ≤ x ≤ 50，这可能导致无意义的负 y 值。另一个错误是将篱笆方程写成 x + y = 100，忘记有两个宽边。配方法中经常出现符号错误——始终通过展开回代进行检验。最后，在没有依据的情况下直接令 x = y 在此处不适用；墙的存在破坏了矩形的对称性。

12. Summary and Reflection | 总结与反思

This case study illustrates the full cycle of mathematical modelling: from a verbal description to an algebraic model, to solving it with pure mathematics (quadratic vertex or calculus), and then interpreting the solution in context and discussing its limitations. Year 9 Further Mathematics gives you the tools to handle such optimisation problems confidently. Practise by changing the numbers, introducing new constraints, or exploring different shapes – each variation strengthens your modelling skills.

本案例展示了数学建模的完整循环：从文字描述到代数模型，用纯数学方法（二次函数顶点或微积分）求解，然后将解置于实际背景中解释并讨论其局限性。Year 9 进阶数学为你提供了自信处理这类最优化问题的工具。通过改变数字、引入新约束或探索不同形状来进行练习——每种变化都会增强你的建模能力。

Published by TutorHao | Further Mathematics Revision Series | aleveler.com

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