📚 Maxima and Minima Problems | 极大值与极小值问题
In A-Level Mathematics, particularly in the Edexcel specification, maxima and minima problems form a core part of calculus applications. These problems involve finding the highest or lowest values of a function, often in real-world contexts such as maximising area, minimising cost, or optimising physical quantities. By using differentiation, students learn to identify critical points and determine their nature, a skill essential for modelling and problem-solving.
在A-Level数学中，特别是Edexcel考试大纲中，极大值与极小值问题是微积分应用的核心内容。这些问题涉及寻找函数的最高值或最低值，通常应用于实际场景，如面积最大化、成本最小化或物理量优化。通过使用微分，学生学会识别关键点并确定其性质，这是建模和解决问题的必要技能。
1. Introduction to Maxima and Minima | 极大值与极小值简介
A maximum point on a curve is where the function attains its highest value in a particular interval, while a minimum point is where it attains its lowest value. These points are collectively known as extrema. In calculus, we use the derivative to locate points where the gradient of the tangent is zero, called stationary points. Not all stationary points are maxima or minima; some may be points of inflection.
曲线上的极大值点是函数在特定区间内达到最高值的位置，而极小值点则是达到最低值的位置。这些点统称为极值。在微积分中，我们利用导数找到切线斜率为零的点，这些点称为驻点。并非所有驻点都是极大值或极小值；有些可能是拐点。
For a function y = f(x), stationary points occur when f'(x) = 0.
对于函数 y = f(x)，当 f'(x) = 0 时出现驻点。
2. Stationary Points and Derivatives | 驻点与导数
To find maxima and minima, we first compute the first derivative f'(x) and set it equal to zero. Solving f'(x) = 0 yields the x-coordinates of stationary points. The corresponding y-values are found by substituting back into the original function. These stationary points are candidates for local maxima, local minima, or points of inflection. We then apply either the first derivative test or the second derivative test to classify them.
为找到极大值和极小值，我们首先计算一阶导数 f'(x) 并将其设为零。解方程 f'(x) = 0 可得到驻点的 x 坐标。将 x 代入原函数即可得到对应的 y 值。这些驻点可能是局部极大值、局部极小值或拐点。然后我们应用一阶导数检验或二阶导数检验来对其进行分类。
3. The First Derivative Test | 一阶导数检验
The first derivative test examines the sign of f'(x) just before and just after 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, it is a local minimum. If the sign does not change, the stationary point is a point of inflection. This method is reliable but requires evaluating the derivative at points around the stationary value.
一阶导数检验检查驻点紧邻左右两侧 f'(x) 的符号。如果 f'(x) 由正变负，则该点为局部极大值。如果 f'(x) 由负变正，则为局部极小值。如果符号不变，则该驻点是拐点。这种方法可靠，但需要在驻点附近取点计算导数值。
- f'(x) changes from positive to negative → Local maximum | f'(x) 由正变负 → 局部极大值
- f'(x) changes from negative to positive → Local minimum | f'(x) 由负变正 → 局部极小值
- No sign change → Point of inflection | 符号不变 → 拐点
4. The Second Derivative Test | 二阶导数检验
The second derivative test uses the value of f”(x) at the stationary point. If f”(x) < 0, the point is a local maximum. If f''(x) > 0, it is a local minimum. If f”(x) = 0, the test is inconclusive and we must use the first derivative test. This test is often quicker but relies on the existence and simplicity of the second derivative.
二阶导数检验使用驻点处的 f”(x) 值。如果 f”(x) < 0，该点为局部极大值。如果 f''(x) > 0，则为局部极小值。如果 f”(x) = 0，则该检验无法判定，必须使用一阶导数检验。这一方法通常更快捷，但依赖于二阶导数的存在且易于计算。
If f”(x) > 0 → Local minimum; if f”(x) < 0 → Local maximum.
若 f”(x) > 0 → 局部极小值；若 f”(x) < 0 → 局部极大值。
5. Types of Stationary Points | 驻点的类型
Summarising, a stationary point can be a local maximum, a local minimum, or a point of inflection. A local maximum is the peak of a curve in its neighbourhood; a local minimum is a trough. A point of inflection is where the curve changes concavity but the gradient is zero. It is important to distinguish between stationary points of inflection and non-stationary points of inflection, though in maxima and minima problems we focus on the stationary ones.
总结起来，驻点可以是局部极大值、局部极小值或拐点。局部极大值是其邻域内的曲线顶点；局部极小值是谷底。拐点则是曲线凹凸性改变且斜率为零的点。区分驻点拐点和非驻点拐点很重要，但在极大极小值问题中我们主要关注驻点。
| Point Type | f'(x) | f”(x) | 中文 |
|---|---|---|---|
| Local Maximum | 0 | f”(x) < 0 | 局部极大值 |
| Local Minimum | 0 | f”(x) > 0 | 局部极小值 |
| Stationary Point of Inflection | 0 | f”(x) = 0 (but needs sign change) | 驻点拐点 |
6. Global vs Local Extrema | 全局极值与局部极值
Maxima and minima problems often require distinguishing between local (relative) and global (absolute) extrema. A local maximum is the highest point in a small interval, but a global maximum is the highest value over the entire domain. For closed intervals [a, b], the global maximum may occur at a stationary point or at an endpoint. Thus, we evaluate f(x) at stationary points and at the endpoints to find the absolute extrema.
极大极小值问题常常需要区分局部（相对）极值和全局（绝对）极值。局部极大值是在小区间内的最高点，而全局极大值是在整个定义域上的最高值。对于闭区间 [a, b]，全局极大值可能出现在驻点或端点处。因此，我们需要计算驻点和端点处的函数值 f(x)，以确定绝对极值。
7. Optimisation Problems – Framework | 优化问题的框架
Many exam questions present real-life scenarios: maximise the volume of a box, minimise the surface area, or optimise profit. The general strategy is: (1) identify the quantity to be optimised (objective function), (2) write it in terms of a single variable using given constraints, (3) find the derivative and set to zero to locate stationary points, (4) determine the nature of the stationary points using the second derivative or first derivative test, (5) check endpoints if the domain is restricted, and (6) state the solution in context, including units.
许多考题会呈现真实场景：最大化箱子的体积、最小化表面积或优化利润。一般策略为：(1) 确定要优化的量（目标函数）；(2) 利用给定约束将其表示为单变量函数；(3) 求导并令导数为零，找到驻点；(4) 使用二阶导数或一阶导数检验确定驻点性质；(5) 如果定义域受限，检查端点； (6) 结合上下文给出答案，包括单位。
8. Worked Example: Maximising Area | 例题：面积最大化
A farmer has 100 m of fencing and wants to enclose a rectangular field against a straight river, using the river as one side. Find the dimensions that maximise the area. Let the side parallel to the river be y and the two sides perpendicular be x. Then total fencing used: 2x + y = 100, so y = 100 – 2x. Area A = x x y = x(100 – 2x) = 100x – 2x². Differentiate: A'(x) = 100 – 4x. Set A'(x) = 0 to give x = 25. Then y = 50. Second derivative: A”(x) = -4 < 0, confirming a maximum. The maximum area is 25 x 50 = 1250 m².
一位农民有100米围栏，想靠一条笔直的河流围成一个矩形田地，以河岸为一边。求使面积最大的尺寸。设平行于河流的边长为 y，垂直的两边长为 x。所用围栏总计：2x + y = 100，因此 y = 100 – 2x。面积 A = x x y = x(100 – 2x) = 100x – 2x²。求导：A'(x) = 100 – 4x。令 A'(x) = 0 得 x = 25。则 y = 50。二阶导数：A”(x) = -4 < 0，确认为极大值。最大面积为 25 x 50 = 1250 m²。
9. Worked Example: Minimising Cost | 例题：成本最小化
A company wants to produce a cylindrical can with volume 1000 cm³. The cost per cm² of the top and bottom is 2 pence, while the curved side costs 1 pence per cm². Find the radius r and height h that minimise the total cost. Volume constraint: πr²h = 1000, so h = 1000/(πr²). Cost C = 2 x (cost of two ends) + 1 x (curved surface). Area of one end: πr², so two ends: 2πr². Curved surface area: 2πrh. So C = 2 x (2πr²) + 1 x (2πrh) = 4πr² + 2πrh. Substitute h: C = 4πr² + 2πr(1000/(πr²)) = 4πr² + 2000/r. Differentiate: dC/dr = 8πr – 2000/r². Set to 0: 8πr = 2000/r² implies 8πr³ = 2000, hence r³ = 250/π, so r = (250/π)^(1/3). Calculate approximate r ≈ 4.30 cm. Then h ≈ 1000/(π x (4.30)²) ≈ 17.2 cm. The second derivative d²C/dr² = 8π + 4000/r³ is positive for r > 0, confirming a minimum cost.
一家公司要生产容积为1000 cm³的圆柱形罐子。顶部和底部的成本为每平方厘米2便士，弯曲侧面成本为1便士每平方厘米。求使总成本最小的半径 r 和高 h。体积约束：πr²h = 1000，因此 h = 1000/(πr²)。成本 C = 2 x (两端成本) + 1 x (曲面)。一端面积：πr²，两端共2πr²。曲面面积：2πrh。因此 C = 2 x (2πr²) + 1 x (2πrh) = 4πr² + 2πrh。代入 h：C = 4πr² + 2πr(1000/(πr²)) = 4πr² + 2000/r。求导：dC/dr = 8πr – 2000/r²。设为零：8πr = 2000/r² 可得 8πr³ = 2000，因此 r³ = 250/π，所以 r = (250/π)^(1/3)。计算近似 r ≈ 4.30 cm。然后 h ≈ 1000/(π x (4.30)²) ≈ 17.2 cm。二阶导数 d²C/dr² = 8π + 4000/r³ 在 r > 0 时为正值，确认为最小成本。
10. Common Pitfalls and Tips | 常见陷阱与提示
Students often forget to verify whether a stationary point is a maximum or minimum, assuming that solving f'(x)=0 gives the answer. Always use the second derivative test or first derivative test to classify. Another frequent mistake is ignoring the domain: lengths and volumes must be positive, so discard negative solutions
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