Solving Maxima and Minima Problems | 最大值最小值问题求解

📚 Solving Maxima and Minima Problems | 最大值最小值问题求解

Finding maximum and minimum values of functions is one of the most fundamental and frequently tested topics in A-Level Mathematics. Whether you are working with quadratics, trigonometric expressions, or differentiable functions, the ability to locate and classify stationary points is essential for success in both pure mathematics and applied contexts.

求解函数的最大值与最小值是A-Level数学中最基础且最常考的内容之一。无论你面对的是二次函数、三角函数表达式，还是可导函数，能够确定并分类驻点，对于纯数学与应用情境的考试都至关重要。

1. Completing the Square for Quadratics | 二次函数配方法

For a quadratic function of the form f(x) = ax² + bx + c, the turning point can be found by completing the square. Write it as f(x) = a(x – h)² + k, where (h, k) is the vertex. If a > 0, the vertex is a minimum; if a < 0, it is a maximum.

对于形如 f(x) = ax² + bx + c 的二次函数，可以通过配方法找到其极值点。将其写成 f(x) = a(x – h)² + k 的形式，其中 (h, k) 即为顶点坐标。若 a > 0，顶点处为最小值；若 a < 0，顶点处为最大值。

Example: Find the minimum value of f(x) = 2x² – 8x + 7.

示例：求 f(x) = 2x² – 8x + 7 的最小值。

2x² – 8x + 7 = 2(x² – 4x) + 7 = 2[(x – 2)² – 4] + 7 = 2(x – 2)² – 1

The vertex is at (2, -1), and since a = 2 > 0, the minimum value is -1.

顶点坐标为 (2, -1)，由于 a = 2 > 0，因此最小值为 -1。

2. The Discriminant Method | 判别式法

Another approach for quadratic maximum/minimum problems involves the discriminant. If we set y = ax² + bx + c, then for real x, the discriminant of the equation ax² + bx + (c – y) = 0 must satisfy Δ ≥ 0. This gives a range of possible y-values, from which we can read off the maximum or minimum.

求解二次函数最值的另一方法是判别式法。若设 y = ax² + bx + c，那么对于实数 x，方程 ax² + bx + (c – y) = 0 的判别式需满足 Δ ≥ 0。由此可得 y 的取值范围，从而直接读出最大值或最小值。

Example: Find the maximum value of y = 3 + 4x – x² using the discriminant.

示例：用判别式法求 y = 3 + 4x – x² 的最大值。

Rewrite as x² – 4x + (y – 3) = 0. For real x, Δ = (-4)² – 4(1)(y – 3) ≥ 0, so 16 – 4y + 12 ≥ 0, thus y ≤ 7. The maximum value is 7.

将其改写为 x² – 4x + (y – 3) = 0。由于 x 为实数，Δ = (-4)² – 4(1)(y – 3) ≥ 0，即 16 – 4y + 12 ≥ 0，所以 y ≤ 7。最大值为 7。

3. Trigonometric Maxima and Minima | 三角函数的最值

Expressions of the form a sin θ + b cos θ can be rewritten as R sin(θ + α) or R cos(θ – α), where R = √(a² + b²). Since sin and cos are always between -1 and 1, the maximum value of a sin θ + b cos θ is R, and the minimum value is -R.

形如 a sin θ + b cos θ 的表达式可以改写为 R sin(θ + α) 或 R cos(θ – α)，其中 R = √(a² + b²)。由于 sin 和 cos 的取值范围始终在 -1 与 1 之间，因此 a sin θ + b cos θ 的最大值为 R，最小值为 -R。

Example: Find the maximum value of f(θ) = 3 sin θ + 4 cos θ.

示例：求 f(θ) = 3 sin θ + 4 cos θ 的最大值。

Here R = √(3² + 4²) = 5, so the maximum value is 5, occurring when sin(θ + α) = 1.

这里 R = √(3² + 4²) = 5，因此最大值为 5，在 sin(θ + α) = 1 时取得。

Using R sin(θ + α) form: 3 sin θ + 4 cos θ = 5 sin(θ + 53.13°), where α = arctan(4/3). The max is 5, the min is -5.

利用 R sin(θ + α) 形式：3 sin θ + 4 cos θ = 5 sin(θ + 53.13°)，其中 α = arctan(4/3)。最大值为 5，最小值为 -5。

4. Stationary Points via Differentiation | 利用导数求驻点

For a differentiable function f(x), stationary points occur where f'(x) = 0. These points may be local maxima, local minima, or points of inflection. To determine the nature of a stationary point, we can use the first derivative test or the second derivative test.

对于可导函数 f(x)，驻点出现在 f'(x) = 0 处。这些点可能是局部最大值、局部最小值或拐点。要判断驻点的性质，可以使用一阶导数测试法或二阶导数测试法。

Example: Find the stationary points of f(x) = x³ – 3x² + 2 and classify them.

示例：求 f(x) = x³ – 3x² + 2 的驻点并分类。

First, f'(x) = 3x² – 6x = 3x(x – 2). Setting f'(x) = 0 gives x = 0 or x = 2.

首先，f'(x) = 3x² – 6x = 3x(x – 2)。令 f'(x) = 0，解得 x = 0 或 x = 2。

Substituting back: f(0) = 2, f(2) = 8 – 12 + 2 = -2. The stationary points are (0, 2) and (2, -2).

代回原函数：f(0) = 2，f(2) = 8 – 12 + 2 = -2。因此驻点为 (0, 2) 和 (2, -2)。

5. First Derivative Test | 一阶导数符号判别法

The first derivative test examines the sign of f'(x) immediately to the left and right of a stationary point. If f'(x) changes from positive to negative, the point is a local maximum. If f'(x) changes from negative to positive, the point is a local minimum. If there is no sign change, it is a point of inflection.

一阶导数测试法考察驻点左右两侧 f'(x) 的符号变化。若 f'(x) 由正变负，则该点为局部最大值；若 f'(x) 由负变正，则为局部最小值；若符号无变化，则为拐点。

For f(x) = x³ – 3x² + 2 above, we examine f'(x) on intervals:

对于上文的 f(x) = x³ – 3x² + 2，我们考察 f'(x) 在各区间上的符号：

When x < 0 (e.g., x = -1): f'(-1) = 3(-1)(-3) = 9 > 0, positive.

当 x < 0（如 x = -1）：f'(-1) = 3(-1)(-3) = 9 > 0，为正。
When 0 < x < 2 (e.g., x = 1): f'(1) = 3(1)(-1) = -3 < 0, negative.

当 0 < x < 2（如 x = 1）：f'(1) = 3(1)(-1) = -3 < 0，为负。
When x > 2 (e.g., x = 3): f'(3) = 3(3)(1) = 9 > 0, positive.

当 x > 2（如 x = 3）：f'(3) = 3(3)(1) = 9 > 0，为正。

Since f'(x) changes from positive to negative at x = 0, (0, 2) is a local maximum. Since f'(x) changes from negative to positive at x = 2, (2, -2) is a local minimum.

由于 f'(x) 在 x = 0 处由正变负，所以 (0, 2) 为局部最大值；f'(x) 在 x = 2 处由负变正，所以 (2, -2) 为局部最小值。

6. Second Derivative Test | 二阶导数判别法

The second derivative test is often quicker. Compute f”(x) at each stationary point x = a. If f”(a) > 0, the point is a local minimum. If f”(a) < 0, the point is a local maximum. If f”(a) = 0, the test is inconclusive and you should use the first derivative test.

二阶导数测试法通常更快捷。在每个驻点 x = a 处计算 f”(x)。若 f”(a) > 0，则该点为局部最小值；若 f”(a) < 0，则为局部最大值；若 f''(a) = 0，该测试失效，应改用一阶导数测试法。

For f(x) = x³ – 3x² + 2, we have f”(x) = 6x – 6.

对于 f(x) = x³ – 3x² + 2，我们有 f”(x) = 6x – 6。

At x = 0: f”(0) = -6 < 0, so (0, 2) is a local maximum.

在 x = 0 处：f”(0) = -6 < 0，因此 (0, 2) 为局部最大值。
At x = 2: f”(2) = 6 > 0, so (2, -2) is a local minimum.

在 x = 2 处：f”(2) = 6 > 0，因此 (2, -2) 为局部最小值。

This matches the conclusion from the first derivative test.

这与一阶导数测试法得出的结论完全一致。

7. Maxima and Minima on a Closed Interval | 闭区间上的最值

When a function is restricted to a closed interval [a, b], the absolute maximum and minimum values occur either at stationary points inside the interval or at the endpoints. The procedure is: find all stationary points in the interval, evaluate f at each stationary point and at both endpoints, then compare the values.

当函数被限制在闭区间 [a, b] 上时，其绝对最大值和最小值要么出现在区间内部的驻点处，要么出现在端点处。求解步骤为：先找出区间内所有驻点，然后计算各驻点及两个端点处的函数值，最后比较大小。

Example: Find the maximum and minimum values of f(x) = x² – 4x + 5 on the interval [0, 3].

示例：求 f(x) = x² – 4x + 5 在区间 [0, 3] 上的最大值与最小值。

f'(x) = 2x – 4 = 0 gives x = 2. Since 2 is inside [0, 3], we evaluate:

f'(x) = 2x – 4 = 0 解得 x = 2。由于 2 在区间 [0, 3] 内，我们计算各点函数值：

x	f(x)
0	5
2	4 – 8 + 5 = 1
3	9 – 12 + 5 = 2

The minimum value is 1 at x = 2, and the maximum value is 5 at x = 0.

最小值为 1（在 x = 2 处取得），最大值为 5（在 x = 0 处取得）。

8. Practical Optimisation Problems | 实际优化问题

Many examination questions frame maxima/minima problems in real-world contexts. The key steps are: (1) define the variable and write down the quantity to be maximised or minimised; (2) express it as a function of one variable, using any given constraints; (3) differentiate, set f'(x) = 0, and solve; (4) verify whether the stationary point is a maximum or minimum; (5) check endpoints or domain restrictions.

许多考试题目会将最值问题置于实际生活情境中。关键步骤为：(1) 定义变量并写出需要最大化或最小化的量；(2) 利用题目给定的约束条件，将其表示为单变量函数；(3) 求导，令 f'(x) = 0 并求解；(4) 判断该驻点是最大值还是最小值；(5) 检查端点或定义域限制。

Example: A farmer has 100 metres of fencing and wants to enclose a rectangular field against an existing wall, so only three sides need fencing. Find the maximum possible area.

示例：一位农民有 100 米长的篱笆，想靠着一面已有的墙围出一块矩形场地，因此只需要围三面。求最大可能的面积。

Let the length perpendicular to the wall be x, and the length parallel to the wall be y. The fencing constraint is 2x + y = 100, so y = 100 – 2x. The area is A = xy = x(100 – 2x) = 100x – 2x².

设垂直于墙的边长为 x，平行于墙的边长为 y。篱笆总长为 2x + y = 100，因此 y = 100 – 2x。面积为 A = xy = x(100 – 2x) = 100x – 2x²。

Differentiating: A'(x) = 100 – 4x. Setting A'(x) = 0 gives x = 25. Then y = 100 – 50 = 50, and A = 25 × 50 = 1250 m².

求导：A'(x) = 100 – 4x。令 A'(x) = 0，得 x = 25。此时 y = 100 – 50 = 50，面积 A = 25 × 50 = 1250 平方米。

Since A”(x) = -4 < 0, this is a maximum. The maximum area is 1250 m².

由于 A”(x) = -4 < 0，该点为最大值点。最大面积为 1250 平方米。

9. Optimisation with Algebraic Constraints | 含代数约束的优化问题

In some problems, the quantity to be optimised involves more than one variable, but a constraint equation allows us to reduce it to a single-variable function. Common examples include maximising the volume of a box made from a rectangular sheet, or minimising the surface area of a cylinder of fixed volume.

在某些问题中，需要优化的量涉及多个变量，但约束方程可以将问题化简为单变量函数。常见题型包括：用矩形纸板制作盒子并求最大体积，或在固定体积下求圆柱的最小表面积。

Example: An open-top box is made by cutting squares of side x from each corner of a 30 cm by 40 cm rectangular sheet and folding up the sides. Find the value of x that maximises the volume.

示例：从一块 30 cm × 40 cm 的矩形纸板的四个角各剪去边长为 x 的小正方形，然后折起四边做成无盖盒子。求使体积最大的 x 值。

The dimensions of the box are (40 – 2x) by (30 – 2x) by x, so V(x) = x(40 – 2x)(30 – 2x) = 4x³ – 140x² + 1200x.

盒子的尺寸为 (40 – 2x) × (30 – 2x) × x，因此 V(x) = x(40 – 2x)(30 – 2x) = 4x³ – 140x² + 1200x。

V'(x) = 12x² – 280x + 1200 = 4(3x² – 70x + 300). Setting V'(x) = 0: 3x² – 70x + 300 = 0.

V'(x) = 12x² – 280x + 1200 = 4(3x² – 70x + 300)。令 V'(x) = 0：3x² – 70x + 300 = 0。

Using the quadratic formula: x = [70 ± √(4900 – 3600)] / 6 = [70 ± √1300] / 6 = [70 ± 10√13] / 6.

用求根公式：x = [70 ± √(4900 – 3600)] / 6 = [70 ± √1300] / 6 = [70 ± 10√13] / 6。

The two roots are x ≈ 17.7 cm and x ≈ 5.7 cm. Since x must satisfy 0 < x < 15 (so that 30 – 2x > 0), only x ≈ 5.7 cm is valid. V”(5.7) < 0 confirms a maximum.

两个根约为 x ≈ 17.7 cm 和 x ≈ 5.7 cm。由于 x 必须满足 0 < x < 15（即 30 - 2x > 0），只有 x ≈ 5.7 cm 有效。V”(5.7) < 0 确认该点为最大值。

10. Economics and Marginal Analysis | 经济学中的边际分析

In economics, profit maximisation is a classic application. If profit P(x) is a function of production level x, then the maximum profit occurs where P'(x) = 0, i.e., where marginal cost equals marginal revenue. The second derivative P”(x) < 0 confirms a maximum.

在经济学中，利润最大化是经典应用。若利润 P(x) 是产量 x 的函数，则利润最大化的条件为 P'(x) = 0，即边际成本等于边际收益。二阶导数 P”(x) < 0 可确认为最大值。

Example: A company models its daily profit in pounds as P(x) = -0.5x² + 40x – 300, where x is the number of units produced in hundreds. Find the production level that maximises profit.

示例：某公司将其日利润（英镑）建模为 P(x) = -0.5x² + 40x – 300，其中 x 是以百件计的产量。求使利润最大的产量。

P'(x) = -x + 40 = 0 gives x = 40. P”(x) = -1 < 0, so this is a maximum. The maximum profit is P(40) = -0.5(1600) + 40(40) – 300 = -800 + 1600 – 300 = £500.

P'(x) = -x + 40 = 0 解得 x = 40。P”(x) = -1 < 0，因此为最大值。最大利润为 P(40) = -0.5(1600) + 40(40) - 300 = -800 + 1600 - 300 = 500 英镑。

11. Common Mistakes and Exam Tips | 常见错误与考试技巧

There are several common pitfalls that students should be careful to avoid when solving maxima and minima problems.

在求解最大值最小值问题时，学生需要特别注意避免以下几种常见陷阱。

Forgetting to check endpoints in interval problems. Always compare stationary-point values with endpoint values.

在区间问题上忘记检查端点。务必比较驻点值与端点值的大小。
Using f”(x) = 0 as a conclusion. When f”(x) = 0, the test is inconclusive; use the first derivative test instead.

将 f”(x) = 0 直接作为结论。当 f”(x) = 0 时，二阶导数测试无效，应改用一阶导数测试。
Ignoring the domain of the variable in practical problems. Physical constraints (lengths, areas, time) must be positive.

在应用题中忽略变量的定义域。物理量（长度、面积、时间）必须为正数。
Mixing up maximum and minimum. Always compute f”(x) or use a sign table to verify the nature of the stationary point.

混淆最大值与最小值。始终计算 f”(x) 或使用符号表来判断驻点的性质。
Not simplifying expressions before differentiating. Factorising or expanding first often makes differentiation easier and reduces errors.

求导前不化简表达式。先展开或因式分解通常能让求导更简便，同时减少错误。

When time is limited in an exam, remember the three core steps: find stationary points (f'(x) = 0), determine their nature (f”(x) or sign table), and check boundaries (endpoints or domain limits).

考试时间紧张时，牢记三步核心流程：找驻点（令 f'(x) = 0），判断性质（用 f”(x) 或符号表），检查边界（端点或定义域限制）。

12. Practice Questions | 强化练习题

The following exercises cover the key techniques from this article. Attempt them without looking at the solutions first, in exam-style conditions.

以下练习覆盖了本文的所有核心技巧。建议在模拟考试的状态下独立完成，再对照答案检查。

Find the minimum value of f(x) = 3x² + 12x – 5 by completing the square.

用配方法求 f(x) = 3x² + 12x – 5 的最小值。
Find the maximum value of f(θ) = 5 sin θ – 12 cos θ.

求 f(θ) = 5 sin θ – 12 cos θ 的最大值。
Given f(x) = x⁴ – 4x³ + 2, find all stationary points and classify them.

已知 f(x) = x⁴ – 4x³ + 2，求所有驻点并判断其性质。
A cylindrical tin has a fixed volume of 1000 cm³. Find the radius that minimises the total surface area.

一个圆柱形罐子的体积固定为 1000 cm³。求使总表面积最小的半径。
Find the maximum and minimum values of f(x) = x³ – 6x² + 9x + 1 on the interval [0, 5].

求 f(x) = x³ – 6x² + 9x + 1 在区间 [0, 5] 上的最大值与最小值。

Solutions highlight: Q1 minimum -17; Q2 maximum 13; Q3 (0, 2) point of inflection, (3, -25) minimum; Q4 r = (500/π)^(1/3) cm; Q5 maximum 21 at x = 5, minimum -1 at x = 2.

答案提示：第1题最小值 -17；第2题最大值 13；第3题 (0, 2) 为拐点，(3, -25) 为最小值；第4题 r = (500/π)^(1/3) cm；第5题在 x = 5 处有最大值 21，在 x = 2 处有最小值 -1。

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Solving Maxima and Minima Problems | 最大值最小值问题求解