📚 Case Study Practice: Maximising an Open-Box Volume with Calculus | 案例分析实战演练：用微积分最大化无盖盒子的容积
In WJEC Year 12 Mathematics, you are often presented with real-world modelling problems that require you to translate a scenario into mathematical language, apply calculus, and interpret the results. This case study walks you through a classic optimisation problem: finding the dimensions of a square-based open box that maximise its volume when constructed from a given sheet of card. Follow the step-by-step reasoning to reinforce your problem-solving skills.
在 WJEC 12 年级数学中，你经常会遇到现实世界的建模问题，要求你将场景转化为数学语言，应用微积分并解释结果。本案例研究将带你完成一个经典的最优化问题：在给定一张纸板的情况下，求出一个正方形底无盖盒子的尺寸，使其容积最大化。请跟随逐步推理，巩固你的问题解决能力。
1. Introducing the Real-World Scenario | 介绍实际场景
A manufacturer wishes to produce open-top boxes from square sheets of card measuring 10 cm by 10 cm. The plan is to cut out equal squares from each corner and then fold up the flaps. The challenge is to choose the size of the corner squares so that the finished box holds the greatest possible volume.
一家制造商想用边长为 10 cm 的正方形纸板制作无盖盒子。计划是从每个角剪去相同的小正方形，然后把边折起来。挑战在于选择角上小正方形的尺寸，使最终盒子的容积尽可能大。
2. Defining the Variables | 定义变量
Let x be the side length (in cm) of the small squares cut from each corner. After the cuts, the base of the resulting box will be a square with side length (10 − 2x), and the height of the box will be x. We assume 0 < x < 5 so that the base remains positive.
设 x 为每个角剪去的小正方形的边长（单位 cm）。裁剪后，所得盒子的底面是一个边长为 (10 − 2x) 的正方形，盒子的高为 x。我们假定 0 < x < 5 以使底面边长仍为正。
3. Formulating the Volume Function | 建立容积函数
The volume V of the open box is given by the product of the base area and the height. The base area is (10 − 2x)², and the height is x. Hence the volume function is:
无盖盒子的容积 V 等于底面积乘以高。底面积为 (10 − 2x)²，高为 x。因此容积函数为：
V(x) = x(10 − 2x)²
This expression defines a cubic model that we will analyse using differentiation.
这个表达式定义了一个三次模型，我们将利用微分对其进行分析。
4. Expanding the Volume Expression | 展开容积表达式
To make differentiation easier, expand the function step by step. First, expand the squared term: (10 − 2x)² = 100 − 40x + 4x². Multiply by x:
为便于求导，逐步展开函数。首先展开平方项：(10 − 2x)² = 100 − 40x + 4x²。然后乘以 x：
V(x) = 4x³ − 40x² + 100x
This polynomial is valid for 0 < x < 5 and is smooth, guaranteeing that any maximum in this interval can be found with the derivative.
该多项式在 0 < x < 5 上有效且光滑，确保了该区间内的任何最大值都可通过导数求得。
5. Differentiating to Find Stationary Points | 求导以寻找驻点
Differentiate V with respect to x using standard rules. The first derivative tells us the rate of change of volume with respect to the corner size:
使用标准法则对 V 关于 x 求导。一阶导数告诉我们容积相对于角料尺寸的变化率：
dV/dx = 12x² − 80x + 100
Stationary points occur when dV/dx = 0. Set the derivative equal to zero and simplify by dividing each term by 4:
当 dV/dx = 0 时出现驻点。令导数等于零，并将每一项除以 4 以化简：
3x² − 20x + 25 = 0
6. Solving the Quadratic Equation | 解二次方程
The quadratic 3x² − 20x + 25 = 0 can be factorised as (3x − 5)(x − 5) = 0. Thus the critical values are x = 5/3 (approximately 1.667) and x = 5. Since the domain restricts x < 5, we discard x = 5 (this would give a volume of zero). The only feasible stationary point is x = 5/3 cm.
二次方程 3x² − 20x + 25 = 0 可分解为 (3x − 5)(x − 5) = 0。因此临界值为 x = 5/3（约 1.667）和 x = 5。由于定义域限制 x < 5，我们舍去 x = 5（此时容积为零）。唯一可行的驻点为 x = 5/3 cm。
7. Justifying the Maximum – Second Derivative | 用二阶导数验证最大值
To confirm that this stationary point gives a maximum, compute the second derivative: d²V/dx² = 24x − 80. Evaluate at x = 5/3:
为确认该驻点对应最大值，计算二阶导数：d²V/dx² = 24x − 80。代入 x = 5/3：
d²V/dx² = 24(5/3) − 80 = 40 − 80 = −40
Since the second derivative is negative, the function is concave down, guaranteeing a local maximum. Given the domain endpoints both yield zero volume, this local maximum is the global maximum.
由于二阶导数为负，函数上凸，保证了局部最大值。考虑到定义域端点处容积均为零，该局部最大值即为全局最大值。
8. Calculating the Maximum Volume | 计算最大容积
Substitute x = 5/3 into the original volume function. The base side length becomes 10 − 2(5/3) = 20/3 cm, and the height is 5/3 cm.
将 x = 5/3 代入原容积函数。底面边长变为 10 − 2(5/3) = 20/3 cm，高为 5/3 cm。
Vmax = (5/3)×(20/3)² = (5/3)×(400/9) = 2000/27 ≈ 74.07 cm³
The box with the greatest capacity will have a height of approximately 1.67 cm and a square base of about 6.67 cm, achieving nearly 74 cubic centimetres.
容积最大的盒子高约为 1.67 cm，正方形底面边长约为 6.67 cm，可获得约 74 立方厘米的容积。
9. Supporting the Conclusion with a Table of Values | 用数值表格佐证结论
Creating a table of volume values for different x can illustrate the maximum well. Below is a small numerical check around the optimal x.
制作不同 x 对应的容积表格可以很好地展示最大值。以下是在最优解附近的数值验证。
| x (cm) | Base side (cm) | Volume V (cm³) |
|---|---|---|
| 1.4 | 7.2 | 72.576 |
| 1.6 | 6.8 | 73.984 |
| 1.667 | 6.666 | 74.074 (approx) |
| 1.8 | 6.4 | 73.728 |
The table confirms that the volume peaks close to x = 1.667, consistent with the algebraic result.
表格证实容积在 x = 1.667 附近达到峰值，与代数结果一致。
10. Generalising the Approach | 方法的推广
This optimisation technique is not limited to box‑making. The same steps – define variables, build a function, differentiate, find critical points, and verify with a second derivative or table – apply to cost minimisation, profit maximisation, or area problems. Always check the domain and interpret the mathematical solution in the original physical context.
这种最优化方法不仅限于制作盒子。同样的步骤——定义变量、建立函数、求导、寻找临界点并用二阶导数或表格验证——适用于成本最小化、利润最大化或面积问题。务必检查定义域，并在原始物理情境中解释数学解。
11. Common Pitfalls to Avoid | 常见陷阱
Many students forget to confirm that a stationary point is actually a maximum, or they use the second derivative incorrectly. Also ensure that the expression for the objective function reflects the constraints accurately. In this case study, using (10 − x)² instead of (10 − 2x)² is a typical error. Practice re‑reading the problem statement and sketching a diagram before writing any equations.
许多学生忘记确认驻点确实是最大值，或者错误地使用二阶导数。还要确保目标函数的表达式准确反映约束条件。在本案例中，使用 (10 − x)² 而不是 (10 − 2x)² 是典型错误。在写任何方程之前，务必练习重读问题陈述并画出示意图。
12. Summary and Exam Tips | 总结与应考提示
In WJEC AS Mathematics, optimisation case studies reward clear logical structure. Show all your working, state the domain, give the differentiated steps, and finish with a statement that answers the original question in context. Practising a handful of such problems will build confidence for the exam.
在 WJEC AS 数学中，最优化案例研究要求展示清晰的逻辑结构。展示所有的计算过程，注明定义域，给出求导步骤，最后用一句根据情境回答原始问题的话作结。练习若干个此类问题将为考试建立信心。
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