Maxima and Minima Problems | 最大值与最小值问题

📚 Maxima and Minima Problems | 最大值与最小值问题

Maxima and minima problems are a core application of differentiation in Edexcel A-Level Mathematics. They require you to identify stationary points, classify them, and interpret their meaning in practical contexts. This article covers first and second derivative tests, global extrema, modelling methods, and exam technique.

最大值与最小值问题是爱德思 A-Level 数学中微分应用的核心内容。这类问题要求你找出驻点、判定驻点类型，并结合实际情境解释结果。本文涵盖一阶与二阶导数判定法、全局极值、建模方法以及考试技巧。

1. What Are Maxima and Minima? | 什么是最大值与最小值？

A maximum point is where a function reaches a peak; a minimum point is where it reaches a trough. At a local maximum, the function value is greater than all nearby values. At a local minimum, it is smaller than all nearby values.

最大值点（极大值点）是函数达到峰值的点；最小值点（极小值点）是函数达到谷值的点。在局部极大值处，函数值大于附近所有值；在局部极小值处，函数值小于附近所有值。

The gradient of the tangent is zero at a smooth turning point. This gives a simple starting point: solve dy/dx = 0 to locate possible turning points.

在光滑的转折点处，切线的斜率为零。这提供了一个简单的起点：解方程 dy/dx = 0 以找到可能的转折点。

2. Stationary Points and dy/dx = 0 | 驻点与 dy/dx = 0

A stationary point occurs where the first derivative is zero. For a function y = f(x), solve f'(x) = 0 to find the x-values of possible maxima or minima.

驻点出现在一阶导数为零的位置。对于函数 y = f(x)，解 f'(x) = 0 可以找到可能的最大值或最小值的 x 坐标。

dy/dx = 0 at a stationary point

For example, if y = x³ – 3x² + 4, then dy/dx = 3x² – 6x = 3x(x – 2). Setting dy/dx = 0 gives x = 0 and x = 2.

例如，若 y = x³ – 3x² + 4，则 dy/dx = 3x² – 6x = 3x(x – 2)。令 dy/dx = 0，得 x = 0 和 x = 2。

These values are candidates only. A stationary point can be a maximum, a minimum, or a point of inflection. You must classify it before stating the final answer.

这些值只是候选点。驻点可能是极大值点、极小值点或拐点。你必须在给出最终答案前判定其类型。

3. Classifying Stationary Points: First Derivative Test | 利用一阶导数判定驻点

The first derivative test examines the sign of dy/dx immediately to the left and right of the stationary point. If the sign changes from positive to negative, the point is a local maximum. If it changes from negative to positive, it is a local minimum.

一阶导数判定法考察驻点左侧和右侧 dy/dx 的符号。如果符号由正变负，该点为局部极大值；如果由负变正，则为局部极小值。

For dy/dx = 3x(x – 2), at x = 0 the sign changes from positive to negative, so (0, 4) is a local maximum. At x = 2 the sign changes from negative to positive, so (2, 0) is a local minimum.

对于 dy/dx = 3x(x – 2)，在 x = 0 处符号由正变负，因此 (0, 4) 为局部极大值；在 x = 2 处符号由负变正，因此 (2, 0) 为局部极小值。

This test is reliable and is especially useful when the second derivative is messy. However, it requires checking derivative values in the intervals around the stationary point.

该方法可靠，尤其在一阶导数表达式复杂时很有用。但它需要检查驻点附近区间内导数的符号。

4. Classifying Stationary Points: Second Derivative Test | 利用二阶导数判定驻点

Compute d²y/dx² at the stationary point. If d²y/dx² > 0, the point is a local minimum. If d²y/dx² < 0, it is a local maximum. If d²y/dx² = 0, the test is inconclusive and you should use the first derivative test.

计算驻点处的 d²y/dx²。如果 d²y/dx² > 0，该点为局部极小值；如果 d²y/dx² < 0，该点为局部极大值；如果 d²y/dx² = 0，则无法判定，应使用一阶导数判定法。

d²y/dx² > 0 ⇒ local minimum; d²y/dx² < 0 ⇒ local maximum

For example, y = x³ – 3x² + 4 gives d²y/dx² = 6x – 6. At x = 0, d²y/dx² = -6 < 0, so it is a maximum. At x = 2, d²y/dx² = 6 > 0, so it is a minimum.

例如，y = x³ – 3x² + 4 的二阶导数为 d²y/dx² = 6x – 6。在 x = 0 处，d²y/dx² = -6 < 0，所以为极大值；在 x = 2 处，d²y/dx

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