Case Study Practice: Optimisation with Constraints | 案例分析实战演练：约束条件下的优化

📚 Case Study Practice: Optimisation with Constraints | 案例分析实战演练：约束条件下的优化

Real-world applications of advanced mathematics often require you to combine multiple techniques – algebra, calculus, and logical reasoning – to solve a single complex problem. This case study takes you step-by-step through an optimisation problem typical of the WJEC Year 11 Further Mathematics syllabus, helping you bridge the gap between isolated skills and integrated problem-solving. By carefully modelling a situation, differentiating, and interpreting the results, you will develop the confidence to tackle exam-style case studies with precision.

进阶数学的实际应用常常要求你将代数、微积分和逻辑推理结合起来，解决一个复杂的综合问题。本篇案例研究将带你逐步完成一个 WJEC 十一年级进阶数学大纲中典型的优化问题，帮助你弥合孤立技能与综合解题之间的差距。通过仔细建立模型、求导并解释结果，你将掌握应对考试风格案例分析的信心和准确性。

1. Problem Statement | 问题陈述

A farmer has 100 metres of fencing and wishes to enclose a rectangular area against a long straight wall. No fencing is needed along the wall itself. Find the dimensions of the rectangle that give the maximum enclosed area, and calculate that maximum area.

一位农场主有100米长的围栏，他想沿着一道长长的直墙围出一个矩形区域，靠墙的一侧无需围栏。求能围出的最大矩形面积，并给出此时矩形的尺寸。

This classic problem exemplifies constrained optimisation, where a resource (fencing length) limits the variable geometry. It appears frequently in WJEC examinations and requires clear variable definition and careful differentiation.

这个经典问题展示了资源（围栏长度）限制几何形状时的约束优化，在WJEC考试中频繁出现，要求定义清晰的变量并进行谨慎的求导。

2. Understanding the Constraints | 理解约束条件

There are only three sides to fence: two identical widths perpendicular to the wall and one length parallel to the wall. The total fencing is therefore: 2 × width + length = 100. This relationship will later allow us to reduce the number of variables.

需要围住的只有三条边：两段垂直于墙的等宽边和一段平行于墙的长边。因此围栏总长为：2 × 宽 + 长 = 100。这一关系式可以让我们减少变量的数量。

The wall itself provides the fourth boundary without cost. Recognising such real-world features is key to setting up a correct mathematical model. Always sketch a diagram and label it with the chosen variables before writing any equations.

墙壁本身提供了第四道边界而无需成本。识别这类现实特征对建立正确的数学模型至关重要。在写任何方程之前，一定要画出草图并用所选变量标注。

3. Defining Variables and Notation | 定义变量与符号

Let the width of the rectangle (the sides perpendicular to the wall) be x metres. Let the length parallel to the wall be y metres. Both x and y must be positive. The area to be maximised is A = x y (in m²).

设矩形的宽（垂直于墙的边）为 x 米，平行于墙的长为 y 米。x 和 y 都必须是正数。要最大化的面积 A = x y（单位为平方米）。

Choosing sensible variable names makes the working easier to follow. In WJEC exams, clearly state what each variable represents at the start of your solution – this gains method marks even if arithmetic slips occur later.

选择合理的变量名称能让解题过程更容易理解。在WJEC考试中，解答开头要明确指出每个变量代表什么——即使后续计算有小错误，这样也能获得方法分。

4. Formulating the Objective Function | 建立目标函数

From the fencing constraint: 2x + y = 100, so y = 100 − 2x. Substitute into the area expression: A(x) = x (100 − 2x) = 100x − 2x². Now the area depends solely on x, turning the problem into a single-variable optimisation.

根据围栏约束：2x + y = 100，于是 y = 100 − 2x。代入面积表达式：A(x) = x (100 − 2x) = 100x − 2x²。现在面积只依赖于 x，问题转化为了单变量优化。

This reduction step is crucial: always use the constraint to express one variable in terms of the other. The resulting function is called the objective function – the quantity you wish to maximise or minimise.

这种化简步骤至关重要：始终利用约束条件将一个变量用另一个变量表示。得到的函数叫作目标函数，也就是你希望最大化或最小化的量。

5. Finding the Derivative | 求导数

Differentiate A(x) = 100x − 2x² with respect to x: dA/dx = 100 − 4x. This gives the rate of change of area as width varies. In WJEC, standard differentiation rules for polynomials are assumed: derivative of axⁿ is naxⁿ⁻¹.

对 A(x) = 100x − 2x² 关于 x 求导：dA/dx = 100 − 4x。这给出了面积随宽度变化的速率。在WJEC中，需要掌握多项式的标准求导法则：axⁿ 的导数是 naxⁿ⁻¹。

vative of −2x² is −2 × 2 × x¹ = −4x. This step sets up the equation for stationary points.

确保正确应用幂函数求导法则：100x¹ 的导数是 100 × 1 × x⁰ = 100，−2x² 的导数是 −2 × 2 × x¹ = −4x。这一步为求驻点方程做好了准备。

6. Locating Critical Points | 确定临界点

Set dA/dx = 0 to find stationary points: 100 − 4x = 0 ⇒ 4x = 100 ⇒ x = 25. When x = 25, the corresponding length is y = 100 − 2(25) = 50. The area is A = 25 × 50 = 1250 m².

令 dA/dx = 0 来求驻点：100 − 4x = 0 ⇒ 4x = 100 ⇒ x = 25。当 x = 25 时，对应的长度为 y = 100 − 2(25) = 50，面积 A = 25 × 50 = 1250 m²。

asible domain (0 < x < 50, because y must be positive). Always check the domain boundaries – here, x = 0 or x = 50 give zero area, so the optimum must lie inside.

这个唯一的驻点是可行域内（0 < x < 50，因为 y 必须为正）的最大值或最小值的唯一候选。务必检查定义域的边界——这里 x = 0 或 x = 50 时面积均为零，因此最优解一定在内部。

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7. Confirming the Maximum Nature | 确认最大性

Use the second derivative test: d²A/dx² = −4. Since this is negative for all x, the stationary point is a maximum. The area function is concave down, guaranteeing that x = 25 yields the global maximum over the feasible interval.

用二阶导数检验：d²A/dx² = −4。因为该值对所有 x 均为负，该驻点是一个极大值点。面积函数是下凹的，这保证了在可行区间内 x = 25 取得全局最大值。

WJEC examiners expect you to verify the nature of the stationary point, not just find it. A quick sign analysis of the first derivative around x = 25 is also acceptable: to the left, dA/dx > 0 (increasing); to the right, dA/dx < 0 (decreasing).

WJEC考官期望你验证驻点的性质，而不仅仅是找到它。对一阶导数在 x = 25 附近进行符号分析也可以接受：左侧 dA/dx > 0（递增）；右侧 dA/dx < 0（递减）。

8. Interpretation and Practical Meaning | 解释与实际意义

The maximum area of 1250 m² is achieved when the width perpendicular to the wall is 25 m and the length parallel to the wall is 50 m. Notice that the optimal rectangle is twice as long as it is wide. This 2:1 ratio is a direct consequence of the one-sided wall constraint.

当垂直于墙的宽度为25米、平行于墙的长度为50米时，能得到最大面积1250平方米。注意，最优矩形的长恰好是宽的两倍。这个 2:1 的比例是单侧靠墙约束的直接结果。

In a full enclosure with four sides, the equal-side square would maximise area for a given perimeter. The presence of the wall changes the symmetry, illustrating how constraints modify the mathematical structure of solutions.

如果是一个四边闭合的围栏，对于给定周长，等边正方形能最大化面积。墙的存在改变了对称性，这说明了约束条件是如何改变解的数学结构的。

9. Alternative Approach: Completing the Square | 替代方法：配方法

The objective function A(x) = −2x² + 100x can be maximised without calculus. Rewrite as A = −2(x² − 50x) = −2[(x − 25)² − 625] = 1250 − 2(x − 25)². The squared term is non-negative, so the maximum of 1250 occurs when x = 25.

目标函数 A(x) = −2x² + 100x 无需微积分也可求最大值。重写为 A = −2(x² − 50x) = −2[(x − 25)² − 625] = 1250 − 2(x − 25)²。平方项非负，因此当 x = 25 时取得最大值1250。

This algebraic method provides a valuable cross-check and is sometimes required in non-calculus WJEC units. Recognising the quadratic form and applying completing the square can be faster when differentiation is not allowed.

这种代数方法提供了一种有用的交叉检验方式，有时 WJEC 非微积分单元也会要求使用。当不允许使用微分时，识别二次型并应用配方法可能更快捷。

10. Extension: Partial Wall or Different Costs | 扩展：部分墙壁或不同成本

Suppose the wall is only 40 m long, and the farmer cannot extend the fenced length beyond it. Now y ≤ 40 imposes a new restriction. The unconstrained optimum had y = 50, so this new constraint binds. The maximum area would occur at y = 40, giving 2x + 40 = 100 ⇒ x = 30, area = 1200 m².

假设墙只有40米长，农场主的围栏长度不能超出墙的范围。现在 y ≤ 40 产生了新的限制。无约束最优解中 y = 50，因此这个新约束起作用了。最大面积将出现在 y = 40 时，得到 2x + 40 = 100 ⇒ x = 30，面积为1200平方米。

Constraints can be inequalities, and checking boundaries becomes essential. WJEC case studies may add such twists to test deeper understanding of feasibility. Always revisit the domain after solving the unconstrained problem.

约束可以是不等式，检查边界变得至关重要。WJEC 案例分析可能会添加这类变化来考查对可行性的更深理解。在求解无约束问题之后，一定要重新检查定义域。

11. Common Pitfalls in Such Problems | 此类问题的常见错误

One frequent mistake is misidentifying the constraint equation, for example writing x + 2y = 100 instead of 2x + y = 100. Careless labelling of the diagram is often the root cause. Another pitfall is forgetting to check the second derivative or domain endpoints, leading to an incorrect conclusion about the maximum.

一个常见错误是错误地列出约束方程，比如将 2x + y = 100 写成了 x + 2y = 100。图示标注马虎往往是根本原因。另一个陷阱是忘记检查二阶导数或定义域端点，从而得出关于最大值的错误结论。

Also, students sometimes leave the answer as x = 25 without calculating the corresponding y and the maximum area. The question usually asks for dimensions and area, so ensure all required values are stated clearly with units.

此外，学生有时只给出 x = 25 而不计算对应的 y 和最大面积。题目通常要求给出尺寸和面积，因此要确保清晰陈述所有需要的值并带上单位。

12. Practice Case Study for You | 供你练习的案例研究

A company produces open-topped boxes from square sheets of cardboard 30 cm by 30 cm by cutting equal squares of side x cm from each corner and folding up the sides. Formulate the volume V as a function of x, find the value of x that maximises volume, and verify your result with a second derivative test.

一家公司用边长为30厘米的正方形纸板制作开口盒，方法是从每个角剪去边长为 x 厘米的小正方形，然后将四边折起。将体积 V 表示为 x 的函数，求出使体积最大的 x 值，并用二阶导数检验验证你的结论。

Key steps to attempt: Express box dimensions in terms of x, write volume V = x(30 − 2x)² tiate, solve dV/dx = 0, check domain 0 < x < 15, and confirm maximum nature. The answer is x = 5 cm giving a volume of 2000 cm³.

需要尝试的关键步骤：用 x 表示盒子的尺寸，写出体积 V = x(30 − 2x)²，展开，求导，解 dV/dx = 0，检查定义域 0 < x < 15，并确认极大值性质。答案是 x = 5 cm，最大体积为2000立方厘米。

Attempt this independently, then review against the methodology above. Consistent practice with such structured case studies will sharpen your mathematical modelling and calculus skills, exactly as required for top marks in WJEC Further Mathematics.

请独立尝试，然后对照上述方法回顾。持续练习这种结构化的案例研究将提升你的数学建模和微积分技能，恰好符合 WJEC 进阶数学取得高分的要求。

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