Example 6.4.3: Finding the Maximum Volume of an Open Box | 示例6.4.3：求开口盒子的最大体积

📚 Example 6.4.3: Finding the Maximum Volume of an Open Box | 示例6.4.3：求开口盒子的最大体积

In this example, we will explore a classic optimisation problem from pure mathematics. We are given a rectangular sheet of cardboard and we need to cut squares from each corner to form an open-top box. Our goal is to determine the size of the cut-out squares that maximises the volume of the box.

在这个示例中，我们将探讨纯数学中的一个经典最优化问题。假设我们有一张矩形纸板，需要从每个角剪去正方形，然后折叠成一个无盖盒子。我们的目标是确定剪去正方形的边长，使盒子的体积最大。

1. Understanding the Problem | 理解问题

We start with a rectangular piece of cardboard that is 15 cm long and 8 cm wide. From each of the four corners, we cut out a square of side length \(x\) cm (we will avoid LaTeX, so we write x cm). After cutting, we fold up the four flaps to form an open-top box. The variable x is unknown and must be chosen so that the resulting box has the largest possible volume.

我们首先有一张长为15厘米、宽为8厘米的矩形纸板。从四个角分别剪去边长为x厘米的小正方形（这里我们直接用x表示）。剪完后，将四边向上折叠，形成一个无盖盒子。变量x未知，需要选取合适的值，使得最终形成的盒子体积尽可能大。

Because the sheet is 8 cm wide, the value of x is constrained by the width: we need 2x < 8, so x < 4. Also x must be positive because we cannot cut a negative length. Thus the domain is \(0 < x < 4\).

由于纸板的宽为8厘米，x的值受到宽度限制：需要2x < 8，因此x < 4。同时，x必须为正数，因为我们不能剪出负的长度。因此定义域为 0 < x < 4。

2. Setting Up the Volume Function | 建立体积函数

After cutting out squares of side x from each corner, the base of the box has length \(15 – 2x\) and width \(8 – 2x\). The height of the box is exactly x. Therefore the volume V(x) is the product of these three dimensions:

从每个角剪去边长为x的小正方形后，盒子的底面长度为 15 − 2x，宽度为 8 − 2x。盒子的高恰好为x。因此，体积V(x)是这三个尺寸的乘积：

V(x) = x(15 − 2x)(8 − 2x)

We can expand this expression to obtain a polynomial. First multiply the two binomials:

我们可以展开这个表达式以得到一个多项式。首先将两个二项式相乘：

(15 − 2x)(8 − 2x) = 120 − 30x − 16x + 4x² = 120 − 46x + 4x²

Then multiply by x:

然后乘以x：

V(x) = x(120 − 46x + 4x²) = 120x − 46x² + 4x³

This cubic function describes the volume for any valid x in the interval (0, 4).

这个三次函数描述了在区间(0, 4)内任意有效x对应的体积。

3. Differentiation | 求导

To find the maximum volume, we need to locate the stationary points of V(x). A stationary point occurs where the first derivative is zero. Let us differentiate V(x) with respect to x.

为了找到最大体积，我们需要确定V(x)的驻点。驻点出现在一阶导数为零的地方。让我们对V(x)关于x求导。

V′(x) = d/dx (120x − 46x² + 4x³) = 120 − 92x + 12x²

We can also write the derivative in standard quadratic form:

我们也可以用标准二次形式写出导数：

V′(x) = 12x² − 92x + 120

Notice that V′(x) is defined for all x in the domain, so the only candidates for extrema are the roots of V′(x) = 0.

注意V′(x)在定义域内处处有定义，因此极值的唯一候选点就是方程V′(x) = 0的根。

4. Solving the Stationary Points | 求解驻点

We set the derivative equal to zero and solve the quadratic equation:

我们令导数等于零，并求解这个二次方程：

12x² − 92x + 120 = 0

To simplify, divide every term by 4:

为了方便，将所有项除以4：

3x² − 23x + 30 = 0

Now we factor or use the quadratic formula. The discriminant is:

现在我们因式分解或使用二次公式。判别式为：

Δ = (−23)² − 4 × 3 × 30 = 529 − 360 = 169

Since Δ = 169 = 13², the roots are:

由于Δ = 169 = 13²，根为：

x = (23 ± √169) / (2 × 3) = (23 ± 13) / 6

This gives two solutions:

这给出两个解：

x = (23 + 13)/6 = 36/6 = 6

x = (23 − 13)/6 = 10/6 = 5/3

So the stationary points occur at x = 6 and x = 5/3.

所以驻点出现在x = 6和x = 5/3处。

5. Validating the Domain | 验证定义域

We must check whether these solutions lie within our allowed interval 0 < x < 4. Clearly x = 6 is larger than 4, so it is not physically possible because the width of the cardboard would become negative: 8 − 2(6) = −4. We reject x = 6.

我们必须检查这些解是否落在允许的区间0 < x < 4内。显然x = 6大于4，在物理上不可能，因为纸板的宽度会变成负值：8 − 2×6 = −4。我们舍弃x = 6。

The other solution x = 5/3 ≈ 1.667 lies in (0, 4). Hence this is the only valid stationary point in the domain.

另一个解x = 5/3 ≈ 1.667落在区间(0, 4)内。因此这是定义域内唯一有效的驻点。

To confirm that x = 5/3 gives a maximum rather than a minimum, we use the second derivative test.

为了确认x = 5/3给出的是最大值而不是最小值，我们使用二阶导数检验。

6. Determining the Maximum | 判断极大值

We compute the second derivative of V(x):

我们计算V(x)的二阶导数：

V″(x) = d/dx (12x² − 92x + 120) = 24x − 92

Now evaluate V″ at x = 5/3:

现在在x = 5/3处计算V″：

V″(5/3) = 24 × (5/3) − 92 = 40 − 92 = −52

Because V″(5/3) < 0, the point x = 5/3 is a local maximum by the second derivative test. Since it is the only stationary point in the interval, it is also the global maximum on that interval.

因为V″(5/3) < 0，根据二阶导数检验，x = 5/3是局部极大值点。由于它是区间内唯一的驻点，因此也是该区间上的全局最大值。

7. Calculating the Maximum Volume | 计算最大体积

To find the corresponding maximum volume, we substitute x = 5/3 back into the volume function V(x) = 120x − 46x² + 4x³.

为了找到对应的最大体积，我们将x = 5/3代回体积函数V(x) = 120x − 46x² + 4x³。

V(5/3) = 120 × (5/3) − 46 × (5/3)² + 4 × (5/3)³

We compute each term:

我们逐项计算：

120 × (5/3) = 200

120 × (5/3) = 200
46 × (5/3)² = 46 × 25/9 = 1150/9

46 × (5/3)² = 46 × 25/9 = 1150/9
4 × (5/3)³ = 4 × 125/27 = 500/27

4 × (5/3)³ = 4 × 125/27 = 500/27

To add these fractions, use a common denominator of 27:

为了将这些分数相加，使用公分母27：

V(5/3) = 200 − 1150/9 + 500/27 = 5400/27 − 3450/27 + 500/27 = 2450/27

Therefore the maximum volume is:

因此最大体积为：

V_max = 2450/27 ≈ 90.74 cm³

Thus the optimal square side length is x = 5/3 cm ≈ 1.67 cm, giving a box of volume approximately 90.74 cm³.

因此最优的小正方形边长为x = 5/3 厘米 ≈ 1.67厘米，此时盒子体积约为90.74立方厘米。

8. Checking the Endpoints | 检查端点

Because the function V(x) is continuous on the closed interval [0, 4], we can check the endpoints to ensure the stationary point truly gives the global maximum. At the endpoints, we expect the volume to be zero.

因为函数V(x)在闭区间[0, 4]上连续，我们可以检查端点以确保驻点确实给出全局最大值。在端点处，我们预期体积为零。

When x = 0, no corners are cut, so the box has no height and its volume is 0.

当x = 0时，没有剪去任何角，因此盒子没有高度，体积为0。

When x = 4, the width becomes 8 − 8 = 0, so again the volume is 0.

当x = 4时，宽度变为8 − 8 = 0，所以体积同样为0。

The following table summarises the volume at the key points:

下表总结了关键点处的体积：

x	V(x)
0	0
5/3	2450/27 ≈ 90.74
4	0

This confirms that the maximum occurs in the interior, not at the endpoints.

这证实了最大值出现在区间内部，而不是在端点处。

9. Graphical Interpretation | 图像解释

The cubic graph of V(x) from x = 0 to x = 4 rises from zero, reaches a maximum, and then decreases back to zero. The shape is a simple arch. The derivative V′(x) changes from positive before x = 5/3 to negative after x = 5/3, which is the signature of a local maximum.

V(x)的三次函数图像从x = 0到x = 4，从零上升，达到最大值，然后下降回零。图像呈现一个简单的拱形。导数V′(x)在x = 5/3之前为正，之后为负，这正是局部极大值的特征。

If we plotted V′(x), we would see that it is a positive quadratic that crosses the x-axis at x = 5/3 and x = 6. In the domain (0, 4), V′(x) is positive for x < 5/3 and negative for x > 5/3, confirming the maximum.

如果我们绘制V′(x)的图像，会看到一个开口向上的二次函数，它与x轴相交于x = 5/3和x = 6。在定义域(0, 4)内，当x < 5/3时V′(x)为正，当x > 5/3时V′(x)为负，这进一步确认了最大值。

10. Common Mistakes | 常见错误

Students often make several mistakes when solving such optimisation problems. Let us list them to help you avoid them.

学生在解决这类最优化问题时常常会犯几个错误。让我们列出这些易错点，帮助你避免。

Forgetting the domain constraints. Always remember that x must satisfy both 2x < 15 and 2x < 8, so x < 4. Ignoring this can lead to accepting x = 6 as a valid answer.

忘记定义域限制。务必记住x必须满足2x < 15且2x < 8，因此x < 4。忽略这一点可能导致接受x = 6作为有效答案。
Using the first derivative test incorrectly. A stationary point is not necessarily a maximum; you must verify using the second derivative or sign chart.

错误使用一阶导数检验。驻点不一定是最大值；你必须使用二阶导数或符号表进行验证。
Making algebraic errors when expanding the volume expression. Carefully expand (15 − 2x)(8 − 2x) before multiplying by x.

展开体积表达式时出现代数错误。在乘以x之前，先仔细展开(15 − 2x)(8 − 2x)。
Forgetting to include units in the final answer. The maximum volume should be given in cubic centimetres (cm³).

最终答案忘记写单位。最大体积应以立方厘米（cm³）为单位。

11. Key Takeaways | 要点总结

This example illustrates the complete procedure for solving a constrained optimisation problem:

这个示例说明了解决约束最优化问题的完整流程：

Identify the quantity to be maximised and write it as a function of one variable.

确定要最大化的量，并将其写成单变量函数。
State the domain of the variable based on the physical constraints.

根据物理约束给出变量的定义域。
Differentiate the function to find the critical points.

对函数求导以找到临界点。
Eliminate any critical points outside the domain and use the second derivative test to classify the remaining ones.

排除定义域之外的临界点，并使用二阶导数检验对剩余点进行分类。
Evaluate the function at the critical point and at the endpoints to determine the global maximum.

在临界点和端点处评估函数值，以确定全局最大值。

By following these steps, you can solve a wide range of real-world problems that involve finding a maximum or minimum value.

遵循这些步骤，你可以解决许多涉及最大值或最小值的现实世界问题。

12. Practice Problem | 练习问题

To reinforce your understanding, try this similar problem:

为了巩固你的理解，尝试一个类似的问题：

A box is to be made from a square sheet of cardboard with side length 12 cm by cutting squares of side x cm from each corner and folding up the sides. Find the value of x that gives the maximum volume, and calculate the maximum volume.

一个盒子由边长为12厘米的正方形纸板制成，从每个角剪去边长为x厘米的小正方形并向上折叠。求使体积最大的x值，并计算最大体积。

Answer: The volume is V(x) = x(12 − 2x)² = 144x − 48x² + 4x³. The valid stationary point is x = 2 cm, and the maximum volume is 128 cm³.

答案：体积为V(x) = x(12 − 2x)² = 144x − 48x² + 4x³。有效驻点为x = 2厘米，最大体积为128立方厘米。

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Example 6.4.3: Finding the Maximum Volume of an Open Box | 示例6.4.3：求开口盒子的最大体积