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IB Mathematics: Extrema and Stationary Points of Multivariable Functions | IB数学:多元函数的极值与驻点

2026-08-30 17:39 · 本文部分内容由AI辅助创作

📚 IB Mathematics: Extrema and Stationary Points of Multivariable Functions | IB数学:多元函数的极值与驻点

Many quantities in economics, physics, and engineering depend on more than one variable. The temperature at a point on a metal plate depends on its coordinates (x,y); the elevation of a mountain depends on longitude and latitude. To find the highest or lowest value of such a function, we must extend the ideas of maxima and minima from single-variable calculus to functions of two or more variables.

经济学、物理和工程中的许多量都依赖于不止一个变量。金属板上某点的温度取决于坐标 (x,y);山的高度取决于经度和纬度。要找出这类函数的最大值或最小值,我们必须把单变量微积分中的极大值与极小值概念推广到两个或更多变量的函数。


1. Partial Derivatives and the Gradient | 偏导数与梯度

For a function f(x,y), the partial derivative fₓ is the derivative with respect to x while y is held constant; fᵧ is defined similarly. The gradient vector, denoted ∇f, collects both partial derivatives into a single vector.

对于函数 f(x,y),偏导数 fₓ 是在保持 y 不变的情况下对 x 求导;fᵧ 的定义类似。梯度向量,记为 ∇f,将两个偏导数收集到一个向量中。

∇f(x,y) = ( ∂f/∂x , ∂f/∂y ) = ( fₓ , fᵧ )

The gradient points in the direction of steepest increase, and its magnitude measures the rate of increase. This vector plays a central role in locating extrema, because at a maximum, minimum, or saddle point, the gradient must vanish.

梯度指向函数增长最快的方向,其大小衡量增长速率。这个向量在寻找极值中起核心作用,因为在最大值、最小值或鞍点处,梯度必须为零。


2. Stationary Points | 驻点

A stationary point of a differentiable function f(x,y) is a point where the gradient is the zero vector.

可微函数 f(x,y) 的驻点是梯度为零向量的点。

fₓ(a,b) = 0 and fᵧ(a,b) = 0

Geometrically, the tangent plane at a stationary point is horizontal. For a smooth function, any local maximum or minimum in the interior of the domain must occur at a stationary point. However, not every stationary point is an extremum; some are saddle points.

几何上,驻点处的切平面是水平的。对于光滑函数,定义域内部的任何局部最大值或最小值必定出现在驻点处。然而,并非每个驻点都是极值点;有些是鞍点。


3. Local Extrema and Saddle Points | 局部极值与鞍点

tationary point that is neither a local maximum nor a local minimum is called a saddle point.

若在所有足够接近 (a,b) 的点 (x,y) 处都有 f(a,b) ≥ f(x,y),则 (a,b) 是局部最大值点;若附近总有 f(a,b) ≤ f(x,y),则是局部最小值点。既不是局部最大值也不是局部最小值的驻点称为鞍点。

ing in the x-direction makes f increase, while moving in the y-direction makes f decrease.

鞍点最简单的例子是函数 f(x,y) = x² − y² 的原点。沿 x 方向移动时 f 增大,而沿 y 方向移动时 f 减小。

f(x,y) = x² − y² ⇒ saddle point at (0,0)


4. The Second Derivative Test | 二阶导数判别法

To classify a stationary point, we compute the second-order partial derivatives and form the Hessian matrix.

要对驻点进行分类,我们计算二阶偏导数并构造黑塞矩阵。

H = [ fₓₓ fₓᵧ ; fₓᵧ fᵧᵧ ]

The discriminant D, also called the Hessian determinant, is defined by

判别式 D,也称为黑塞行列式,定义为:

D = fₓₓ fᵧᵧ − (fₓᵧ)²

Condition Conclusion
D > 0 and fₓₓ > 0 Local minimum | 局部最小值
D > 0 and fₓₓ < 0 Local maximum | 局部最大值
D < 0 Saddle point | 鞍点
D = 0 Inconclusive | 判别法失效,需进一步分析

Intuitively, D > 0 means the surface bends the same way in all directions; the sign of fₓₓ then tells whether the bend is upward or downward. D < 0 means the surface bends upward in one direction and downward in another, creating a saddle.

直观上,D > 0 表示曲面在所有方向上都以相同方式弯曲;fₓₓ 的符号则告诉我们弯曲是向上还是向下。D < 0 表示曲面在一个方向向上弯曲、在另一个方向向下弯曲,从而形成鞍点。


5. Worked Example 1 | 例题一

Find and classify the stationary points of f(x,y) = x³ − 3x + y².

求函数 f(x,y) = x³ − 3x + y² 的驻点并判断其类型。

First find the partial derivatives.

先求偏导数。

fₓ = 3x² − 3, fᵧ = 2y

Set fₓ = 0 and fᵧ = 0:

令 fₓ = 0 且 fᵧ = 0:

3x² − 3 = 0 ⇒ x = ±1; 2y = 0 ⇒ y = 0

The stationary points are (1,0) and (−1,0). Now compute the second derivatives.

驻点为 (1,0) 和 (−1,0)。现在计算二阶导数。

fₓₓ = 6x, fᵧᵧ = 2, fₓᵧ = 0

0) is a local minimum. At (−1,0), D = (−6) × 2 − 0 = −12 < 0, so (−1,0) is a saddle point.

在 (1,0) 处,D = fₓₓ fᵧᵧ − (fₓᵧ)² = 6 × 2 − 0 = 12 > 0 且 fₓₓ = 6 > 0,因此 (1,0) 是局部最小值点。在 (−1,0) 处,D = (−6) × 2 − 0 = −12 < 0,因此 (−1,0) 是鞍点。


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6. Worked Example 2: The Inconclusive Case | 例题二:判别法失效的情形

Consider f(x,y) = x⁴ + y⁴. The gradient equations give

考虑函数 f(x,y) = x⁴ + y⁴。梯度方程为

fₓ = 4x³ = 0, fᵧ = 4y³ = 0 ⇒ (0,0)

2y², and fₓᵧ = 0, so D = 0 at the origin. The second derivative test gives no information.

二阶导数为 fₓₓ = 12x²,fᵧᵧ = 12y²,fₓᵧ = 0,因此原点处 D = 0。二阶导数判别法无法给出结论。

a saddle point because values are positive along the x-axis and negative along the y-axis.

然而,f(x,y) = x⁴ + y⁴ ≥ 0 且 f(0,0) = 0,所以原点实际上是全局最小值点。相比之下,f(x,y) = x⁴ − y⁴ 在原点处也有 D = 0,但它是鞍点,因为沿 x 轴函数值为正,沿 y 轴函数值为负。


7. Global Extrema on Closed Domains | 闭区域上的全局极值

When a function is defined on a closed and bounded region, the Extreme Value Theorem guarantees that it attains both a global maximum and a global minimum. These values can occur at interior stationary points or on the boundary of the region.

当函数定义在闭且有界区域上时,极值定理保证函数一定取得全局最大值和全局最小值。这些值可能出现在区域内部的驻点处,也可能出现在区域的边界上。

nd boundary values x² + y² = 1. Hence the global minimum is 0 and the global maximum is 1.

例如,在圆盘 x² + y² ≤ 1 上,函数 f(x,y) = x² + y² 在内部有驻点 (0,0),值为 0;在边界上 x² + y² = 1。因此全局最小值为 0,全局最大值为 1。


8. Constrained Extrema and Lagrange Multipliers | 约束极值与拉格朗日乘数法

of Lagrange multipliers states that at an extremum, the gradients of f and g are parallel.

有时我们需要在约束 g(x,y) = 0 下最大化或最小化 f(x,y)。拉格朗日乘数法指出,在极值点处,f 和 g 的梯度平行。

∇f = λ∇g, combined with g(x,y) = 0

y = 1. The equations y = λ, x = λ, and x + y = 1 give x = y = 1/2, so the maximum is 1/4.

举一个简单例子:在 x + y = 1 的约束下最大化 f(x,y) = xy。方程 y = λ,x = λ,以及 x + y = 1 给出 x = y = 1/2,因此最大值为 1/4。


9. Common Pitfalls and Exam Tips | 常见错误与考试提示


10. Summary | 总结

Stationary points of a multivariable function are found by solving ∇ r methods are required. Global extrema on closed regions demand a check of both interior stationary points and the boundary. The ideas extend naturally to constrained optimisation through Lagrange multipliers.

多元函数的驻点通过求解 ∇f = 0 得到。局部最大值、局部最小值和鞍点随后用判别式 D = fₓₓ fᵧᵧ − (fₓᵧ)² 来分类。当 D = 0 时,需要直接分析或使用其他方法。闭区域上的全局极值要求同时检查区域内部的驻点和边界。这些思想通过拉格朗日乘数法可以自然推广到约束优化问题。


Published by TutorHao | IB Mathematics Revision Series | aleveler.com

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