Solving Optimization Problems in IB Mathematics | IB数学：优化问题的求解策略

📚 Solving Optimization Problems in IB Mathematics | IB数学：优化问题的求解策略

Optimization is one of the most practical and frequently tested topics in IB Mathematics, appearing in both Analysis and Approaches (AA) and Applications and Interpretation (AI). The essence of an optimization problem is to find the maximum or minimum value of a function under given constraints, using differentiation and critical-point analysis.

优化问题是IB数学中最实用且最常考的主题之一，出现在分析与方法（AA）和应用与解释（AI）两门课程中。优化问题的本质是在给定约束条件下，利用微分和临界点分析，寻找函数的最大值或最小值。

1. Understanding the Core Idea | 理解核心思想

At its heart, optimization asks: “What is the best possible outcome?” In mathematics, this translates into identifying where a function attains its greatest or least value. For a differentiable function f(x), local extrema occur where the derivative f'(x) equals zero or does not exist. These points are called critical points.

优化问题的核心问题是：“最佳结果是什么？”在数学中，这转化为确定函数在何处取得最大值或最小值。对于可微函数 f(x)，局部极值出现在导数 f'(x) 等于零或不存在的地方。这些点被称为临界点。

However, not every critical point is an extremum. A horizontal point of inflection, such as at x = 0 for f(x) = x³, has f'(0) = 0 but is neither a maximum nor a minimum. Therefore, we must use the first or second derivative test to classify each critical point correctly.

然而，并非每个临界点都是极值点。例如，f(x) = x³ 在 x = 0 处的水平拐点，虽然 f'(0) = 0，但它既不是最大值也不是最小值。因此，我们必须使用一阶或二阶导数判定法来正确分类每个临界点。

2. Step-by-Step Strategy | 分步解题策略

A reliable strategy for solving any optimization problem involves the following steps:

解决任何优化问题都有一套可靠的策略，包括以下步骤：

Read the problem carefully and identify the quantity to be optimized (maximum or minimum).

仔细阅读题目，确定需要优化的量（最大值或最小值）。
Introduce variables and write down the constraint equation if one exists.

引入变量，如果存在约束条件，则写出约束方程。
Express the quantity to be optimized as a single-variable function using the constraint.

利用约束条件，将待优化的量表示为一个单变量函数。
Find the first derivative, set it to zero, and solve for the critical points.

求一阶导数，令其为零，并解出临界点。
Use the first or second derivative test to confirm the nature of each critical point.

使用一阶或二阶导数判定法，确认每个临界点的性质。
Check endpoints of the domain if the interval is closed, since extrema may occur there.

如果定义域是闭区间，则需检查端点，因为极值也可能在端点处取得。
State the final answer with appropriate units and a clear conclusion.

给出最终答案，注明单位并清晰总结结论。

3. The First Derivative Test | 一阶导数判定法

The first derivative test examines the sign change of f'(x) around a critical point x = c. If f'(x) changes from positive to negative, then f(c) is a local maximum. If it changes from negative to positive, then f(c) is a local minimum. If there is no sign change, the point is neither.

一阶导数判定法检查 f'(x) 在临界点 x = c 附近的符号变化。如果 f'(x) 从正变负，则 f(c) 是局部最大值；如果从负变正，则 f(c) 是局部最小值；如果符号没有变化，则它不是极值点。

f'(x) changes from + to − ⇒ local maximum
f'(x) changes from − to + ⇒ local minimum

This test is especially useful when the second derivative is difficult to compute or when f”(c) = 0. For example, consider f(x) = x⁴ − 4x³. Then f'(x) = 4x³ − 12x² = 4x²(x − 3). The critical points are x = 0 and x = 3. At x = 0, f'(x) does not change sign, so it is not an extremum. At x = 3, f'(x) changes from negative to positive, so f(3) = −27 is a local minimum.

该判定法在二阶导数难以计算或 f”(c) = 0 时尤其有用。例如，考虑 f(x) = x⁴ − 4x³，则 f'(x) = 4x³ − 12x² = 4x²(x − 3)。临界点为 x = 0 和 x = 3。在 x = 0 处，f'(x) 符号不变，因此不是极值点；在 x = 3 处，f'(x) 由负变正，因此 f(3) = −27 是局部最小值。

4. The Second Derivative Test | 二阶导数判定法

The second derivative test uses the concavity of the function at a critical point. If f'(c) = 0 and f”(c) > 0, then f(c) is a local minimum because the graph is concave up. If f'(c) = 0 and f”(c) < 0, then f(c) is a local maximum because the graph is concave down.

二阶导数判定法利用函数在临界点处的凹凸性。如果 f'(c) = 0 且 f”(c) > 0，则图形上凹，f(c) 是局部最小值；如果 f'(c) = 0 且 f”(c) < 0，则图形下凹，f(c) 是局部最大值。

f”(c) > 0 ⇒ local minimum
f”(c) < 0 ⇒ local maximum
f”(c) = 0 ⇒ test inconclusive

If f”(c) = 0, the second derivative test fails, and we must fall back on the first derivative test. For example, f(x) = x⁴ has f'(0) = 0 and f”(0) = 0, but it clearly has a local minimum at x = 0. Only the first derivative test, which shows f'(x) changing from negative to positive, reveals this correctly.

如果 f”(c) = 0，二阶导数判定法失效，我们必须回到一阶导数判定法。例如，f(x) = x⁴ 满足 f'(0) = 0 且 f”(0) = 0，但显然在 x = 0 处有局部最小值。只有一阶导数判定法，显示 f'(x) 从负变正，才能正确揭示这一点。

5. Closed-Interval Optimization | 闭区间上的优化

When the domain of the function is a closed interval [a, b], the Extreme Value Theorem guarantees that a continuous function attains both a maximum and a minimum on that interval. These extrema must occur either at critical points inside the interval or at the endpoints a and b.

当函数的定义域是闭区间 [a, b] 时，极值定理保证了连续函数在该区间上一定存在最大值和最小值。这些极值可能出现在区间内部的临界点，也可能出现在端点 a 和 b 处。

Therefore, the procedure is: compute all critical points in (a, b), evaluate f at these points and at a and b, and select the largest and smallest values. This method does not require derivative tests, as direct comparison of values gives the absolute extrema.

因此，步骤如下：计算 (a, b) 内所有临界点，计算这些点以及 a 和 b 处的函数值，然后比较选出最大值和最小值。这种方法不需要导数判定法，因为直接比较函数值即可得到绝对极值。

For instance, let f(x) = x³ − 3x + 2 on [0, 2]. Since f'(x) = 3x² − 3 = 3(x − 1)(x + 1), the only critical point in the interval is x = 1. Evaluating: f(0) = 2, f(1) = 0, f(2) = 4. Hence the maximum is 4 at x = 2, and the minimum is 0 at x = 1.

例如，设 f(x) = x³ − 3x + 2 在 [0, 2] 上。由于 f'(x) = 3x² − 3 = 3(x − 1)(x + 1)，区间内唯一的临界点是 x = 1。计算得 f(0) = 2, f(1) = 0, f(2) = 4。因此最大值在 x = 2 处为 4，最小值在 x = 1 处为 0。

6. Maximizing Area | 面积最大化问题

A classic optimization problem asks: “A farmer has 100 metres of fencing and wants to enclose a rectangular field with the largest possible area. What dimensions should be used?” Let the length be x and the width be y. The constraint is 2x + 2y = 100, so x + y = 50, hence y = 50 − x.

一个经典优化问题是：“农夫有100米围栏，想要围出一个面积最大的矩形场地，应使用什么尺寸？”设长为 x，宽为 y。约束条件为 2x + 2y = 100，即 x + y = 50，因此 y = 50 − x。

The area is A = xy = x(50 − x) = 50x − x². Differentiating gives A'(x) = 50 − 2x. Setting A'(x) = 0 yields x = 25. Then y = 25. Since A”(x) = −2 < 0, this is a maximum. The maximum area is 25 × 25 = 625 m², achieved by a square.

面积为 A = xy = x(50 − x) = 50x − x²。求导得 A'(x) = 50 − 2x。令 A'(x) = 0，解得 x = 25，于是 y = 25。由于 A”(x) = −2 < 0，这是最大值。最大面积为 25 × 25 = 625 平方米，此时为正方形。

This result illustrates a general principle: for a fixed perimeter, the rectangle with maximum area is a square. A similar principle applies in many geometric optimization problems.

这个结果说明了一个普遍原理：在周长固定时，面积最大的矩形是正方形。类似的原理适用于许多几何优化问题。

7. Minimizing Cost | 成本最小化问题

Optimization also appears in economics and manufacturing. Suppose a company produces cylindrical cans. For a fixed volume V, what radius r and height h minimise the total surface area (and thus material cost)? The volume constraint is V = πr²h, so h = V / (πr²).

优化也出现在经济学和制造业中。假设一家公司生产圆柱形罐子。在体积 V 固定时，什么半径 r 和高度 h 能使总表面积（即材料成本）最小？体积约束为 V = πr²h，因此 h = V / (πr²)。

The total surface area is S = 2πr² + 2πrh. Substituting h gives S(r) = 2πr² + 2V / r. Differentiating with respect to r:

总表面积为 S = 2πr² + 2πrh。代入 h 得到 S(r) = 2πr² + 2V / r。对 r 求导：

S'(r) = 4πr − 2V / r²

Setting S'(r) = 0 gives 4πr = 2V / r², so r³ = V / (2π). Using h = V / (πr²), we find h = 2r. Thus the optimal can has height equal to its diameter. This “economical can” result is a famous application of calculus.

令 S'(r) = 0，得 4πr = 2V / r²，因此 r³ = V / (2π)。利用 h = V / (πr²)，可得 h = 2r。因此最优罐子的高度等于其直径。这个“经济罐”结果是微积分的一个著名应用。

8. Maximizing Volume | 体积最大化问题

Another common problem involves creating an open-top box by cutting equal squares of side x from each corner of a rectangular sheet of dimensions 30 cm by 20 cm, then folding up the sides. The resulting box has volume V(x) = x(30 − 2x)(20 − 2x).

另一个常见问题是从一张长30厘米、宽20厘米的矩形纸板的四个角各剪去边长为 x 的小正方形，然后折叠成一个无盖盒子。所得盒子的体积为 V(x) = x(30 − 2x)(20 − 2x)。

Expanding: V(x) = 4x³ − 100x² + 600x. The domain is 0 < x < 10, since 20 − 2x must be positive. Differentiating:

展开得 V(x) = 4x³ − 100x² + 600x。定义域为 0 < x < 10，因为需要 20 − 2x > 0。求导：

V'(x) = 12x² − 200x + 600

Set V'(x) = 0 and divide by 4: 3x² − 50x + 150 = 0. Using the quadratic formula:

令 V'(x) = 0 并除以4：3x² − 50x + 150 = 0。使用二次公式：

x = [50 ± √(2500 − 1800)] / 6 = [50 ± √700] / 6 ≈ 3.92 or 12.74

Only x ≈ 3.92 lies in the domain. Since V”(x) = 24x − 200, we have V”(3.92) ≈ −105.9 < 0, confirming a maximum. The maximum volume is approximately 3.92 × 22.16 × 12.16 ≈ 1056 cm³.

只有 x ≈ 3.92 在定义域内。由于 V”(x) = 24x − 200，V”(3.92) ≈ −105.9 < 0，确认为最大值。最大体积约为 3.92 × 22.16 × 12.16 ≈ 1056 立方厘米。

9. Handling Constraints with Implicit Differentiation | 用隐函数求导处理约束

Some optimization problems involve more complex relationships where the quantity to be optimized is not easily expressed as a single-variable function. In such cases, implicit differentiation can be powerful. For example, minimize x² + y² subject to xy = 4. Here y = 4/x, so f(x) = x² + 16/x².

有些优化问题涉及更复杂的关系，待优化的量不容易表示为单变量函数。在这种情况下，隐函数求导非常有用。例如，在 xy = 4 的条件下最小化 x² + y²。这里 y = 4/x，因此 f(x) = x² + 16/x²。

We could also keep the relationship implicit. Let F = x² + y² and differentiate with respect to x: dF/dx = 2x + 2y(dy/dx). From xy = 4, implicit differentiation gives y + x(dy/dx) = 0, so dy/dx = −y/x. Substituting yields dF/dx = 2x − 2y²/x. Setting this to zero gives x² = y².

我们也可以保持关系为隐式。设 F = x² + y²，对 x 求导：dF/dx = 2x + 2y(dy/dx)。由 xy = 4 隐式求导得 y + x(dy/dx) = 0，所以 dy/dx = −y/x。代入得 dF/dx = 2x − 2y²/x。令其为零，得 x² = y²。

With xy = 4, we get x = ±2, y = ±2. The minimum value is 4 + 4 = 8. This demonstrates how implicit differentiation can simplify constrained optimization, especially when solving for one variable explicitly is awkward.

结合 xy = 4，得 x = ±2, y = ±2。最小值为 4 + 4 = 8。这展示了隐函数求导如何简化约束优化，特别是在显式求解一个变量较麻烦时。

10. Applications in Kinematics | 运动学中的应用

In IB Physics and Mathematics, optimization often appears in kinematics. Consider a particle moving along a straight line with displacement s(t) = t³ − 6t² + 9t (in metres). To find when the particle’s speed is maximum or minimum, we work with velocity v(t) = s'(t) = 3t² − 12t + 9 and acceleration a(t) = v'(t) = 6t − 12.

在IB物理和数学中，优化经常出现在运动学中。考虑一个沿直线运动的粒子，位移为 s(t) = t³ − 6t² + 9t（米）。为了找到粒子速度的最大或最小值，我们研究速度 v(t) = s'(t) = 3t² − 12t + 9 和加速度 a(t) = v'(t) = 6t − 12。

Setting v'(t) = 6t − 12 = 0 gives t = 2 as the only critical point of velocity. Since v”(t) = 6 > 0, v(2) is a local minimum. Indeed v(2) = 12 − 24 + 9 = −3 m/s. If speed is taken as the absolute value, |v(2)| = 3 m/s, which is actually the minimum speed in the interval [0, 3]. The maximum speed occurs at an endpoint.

令 v'(t) = 6t − 12 = 0，得 t = 2 是速度的唯一临界点。由于 v”(t) = 6 > 0，v(2) 是局部最小值。实际上 v(2) = 12 − 24 + 9 = −3 m/s。如果速度取绝对值，|v(2)| = 3 m/s，在 [0, 3] 区间内这确实是最小速率。最大速率出现在端点处。

11. Common Mistakes and Pitfalls | 常见错误与陷阱

Many students lose marks on optimization problems due to avoidable errors. One common mistake is forgetting to check the domain of the function. For example, in a geometric problem, lengths must be positive; a critical point outside the domain is meaningless and must be discarded.

许多学生在优化问题上失分是因为一些可以避免的错误。一个常见错误是忘记检查函数的定义域。例如，在几何问题中，长度必须为正；超出定义域的临界点毫无意义，必须舍弃。

Another frequent error is confusing local extrema with absolute extrema. Unless the domain is explicitly restricted to a closed interval, a critical point found by setting f'(x) = 0 only guarantees a local extremum. To claim absolute extremum, you must examine the behaviour of the function as x approaches the boundaries of its domain.

另一个常见错误是混淆局部极值与绝对极值。除非定义域明确限制为闭区间，否则令 f'(x) = 0 找到的临界点只能保证是局部极值。要断言绝对极值，必须考察函数在定义域边界附近的变化行为。

Students also often forget to justify that the critical point is indeed a maximum or minimum. Always use either the first or second derivative test, and state your conclusion clearly. Finally, be careful with units: if the question asks for dimensions, give lengths; if it asks for the optimized value, give the value itself.

学生还经常忘记说明临界点确实是最大值或最小值。务必使用一阶或二阶导数判定法，并清晰陈述结论。最后，注意单位：如果题目问尺寸，就给出长度；如果问优化后的值，就给出该值本身。

12. Practice Questions for IB | IB练习题

To master optimization, practice is essential. Here are three exam-style questions worth attempting:

要掌握优化，练习必不可少。以下是三道值得尝试的考试风格题目：

Find the minimum value of the sum of a positive number and its reciprocal, i.e., minimize f(x) = x + 1/x for x > 0.

求一个正数与其倒数之和的最小值，即对于 x > 0，最小化 f(x) = x + 1/x。
A cylindrical can with an open top must hold 500 cm³ of liquid. Find the radius and height that minimise the amount of material used.

一个无盖圆柱形罐子需要盛装500立方厘米液体。求使材料用量最小的半径和高度。
A wire of length 40 cm is cut into two pieces. One piece is bent into a square, and the other into a circle. How should the wire be cut to maximize the total enclosed area?

将一根长40厘米的导线剪成两段，一段弯成正方形，另一段弯成圆形。应如何切割才能最大化总封闭面积？

For the first problem, f'(x) = 1 − 1/x², giving x = 1, and f”(1) = 2 > 0, so the minimum value is 2. The second and third problems require constructing the correct function from the geometric constraints, then following the standard optimisation procedure.

对于第一题，f'(x) = 1 − 1/x²，得 x = 1，且 f”(1) = 2 > 0，因此最小值为2。第二题和第三题需要从几何约束中构建正确的函数，然后遵循标准优化步骤。

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Solving Optimization Problems in IB Mathematics | IB数学：优化问题的求解策略