从几何上看，这意味着目标函数的等值面 $f(x, y, z)$ 与约束曲面相切 $g(x, y, z) = 0$ 于驻点处。

这可以简洁地总结为 $\phi(x, y, z) = f(x, y, z) - \lambda g(x, y, z)$ 。那么 $f(x, y, z)$ 将在约束条件 $g(x, y, z) = 0$ 下取得驻点，当

$$\frac{\partial \phi}{\partial x} = \frac{\partial \phi}{\partial y} = \frac{\partial \phi}{\partial z} = 0 \quad \text{and} \quad g(x, y, z) = 0 \quad (9.41)$$

这给出了四个方程来确定 $(x, y, z; \lambda)$ （驻点处）。标量乘子 $\lambda$ 称为 拉格朗日乘子，而函数 $\phi(x, y, z)$ 称为 辅助函数。

例 9.42 用拉格朗日乘子法重做例 9.41。

解 这里我们需要求函数的极值

$$f(x, y) = 2x^2 + 3y^2$$

在约束条件下

$$g(x, y) = 2x + y - 1 = 0$$

辅助函数为

$$\begin{aligned}\phi(x, y, z) &= f(x, y) - \lambda g(x, y) \ &= 2x^2 + 3y^2 - \lambda(2x + y - 1)\end{aligned}$$

并且我们发现 $f(x, y)$ 的条件极值由下式给出

$$\frac{\partial \phi}{\partial x} = \frac{\partial \phi}{\partial y} = 0, \quad g(x, y) = 0$$

即

$$\frac{\partial \phi}{\partial x} = 4x - 2\lambda = 0 \quad (9.42)$$

$$\frac{\partial \phi}{\partial y} = 6y - \lambda = 0 \quad (9.43)$$

$$g(x, y) = 2x + y - 1 = 0 \tag{9.44}$$

求解 (9.42)–(9.44) 得

$$\lambda = \frac{6}{7}, \quad x = \frac{3}{7} \quad \text{and} \quad y = \frac{1}{7}.$$

因此 $f(x, y)$ 的条件极值为 $\frac{3}{7}$ 且在 $x = \frac{3}{7}$ , $y = \frac{1}{7}$ 处取得。直线 $y = \frac{1}{7}$ 位于 $x = \frac{3}{7}$ ，因此 $f(\frac{3}{7}, \frac{1}{7}) = \frac{3}{7}$ 。解为 $(It is clear from the level curves of $f(x, y)$ , shown in Figure 9.21, that the function has a minimum at $(\frac{3}{7}, \frac{1}{7})$ . In general, however, to determine the nature of the conditional stationary point, we have to resort to Taylor's theorem and consider the sign of the difference $f(x + h, y + k) - f(x, y)$ . Taking a point near $(\frac{3}{7}, \frac{1}{7})$ , say $(\frac{3}{7} + h, \frac{1}{7} + k)$ ,$