Section03_多元函数微分学的应用

求极值

对于一元 $y=f(x)$，求极值

1. 找出 $x\in \mathbb{D}$
2. 找出 $f'(x)\begin{cases}=0\\\nexists\end{cases}$ 对应的 $x$
3. 判别法
   1. Th1
   2. Th2

定义

• 对于$z = f(x,y), (x,y)\in\mathbb{D}, M_{0}(x_{0},y_{0}) \in\mathbb{D}$，若 $\exists\ \delta > 0$， 但 $0 < \sqrt{(x-x_{0})^{2} + (y-y_{0})^{2}}<\delta$ 时
  • $f(x,y) < f(x_{0},y_{0})$ 则 $(x_{0},y_{0})$ 为极大点，$f(x_{0},y_{0})$ 为极大值
  • $f(x,y) > f(x_{0},y_{0})$ 则 $(x_{0},y_{0})$ 为极小点，$f(x_{0},y_{0})$ 为极小值

极值问题

二元函数的无条件极值

$\color{#D0104C}{z=f(x,y),(x,y)\in\mathbb{D}\quad (\mathbb{D}\text{为开区域})}$

步骤

1. 求出 $\begin{cases}\displaystyle\frac{\partial z}{\partial x} = 0\\\displaystyle\frac{\partial z}{\partial y}=0\end{cases}$ 对应的 $\begin{cases}\displaystyle x_{0},\cdots\\ \displaystyle y_{0},\cdots\end{cases}$
2. 设 $(x_{0},y_{0})$ 为 $f(x,y)$ 的一个驻点
   • $\displaystyle A = \left.\frac{\partial^{2}{z}}{\partial x^{2}}\right\vert_{(x_{0},y_{0})}$
   • $\displaystyle B = \left.\frac{\partial^{2}{z}}{\partial x\partial y}\right\vert_{(x_{0},y_{0})}$
   • $\displaystyle C = \left.\frac{\partial^{2}{z}}{\partial y^{2}}\right\vert_{(x_{0},y_{0})}$
3. 判断 $$AC-B^{2} \begin{cases} < 0 \quad (x_{0},y_{0})\text{不为极值点} \\ > 0 \quad \begin{cases} A < 0 & (x_{0},y_{0}) \text{为极大点}\\ A > 0 & (x_{0},y_{0}) \text{为极小点}\\ \end{cases} \end{cases}$$

例题

1. $z=f(x,y) =x^{3} - 3x^{2}-9x+y^{2}+2y - 2$，求极值

$$\begin{array}{ll} & \begin{cases} \displaystyle \frac{\partial{z}}{\partial{x}} = 3x^{2} - 6x - 9 = 0 \\ \displaystyle \frac{\partial{z}}{\partial{y}} = 2y + 2 = 0 \\ \end{cases} \Rightarrow \begin{cases} x = -1,3 \\ y = -1 \end{cases} \\ & \text{对}(-1,-1) \begin{cases} \displaystyle A = \frac{\partial^{2}{z}}{\partial{x^{2}}} = -12 \\ \displaystyle B = \frac{\partial^{2}{z}}{\partial{x}\partial{y}} = 0 \\ \displaystyle C = \frac{\partial^{2}{z}}{\partial{y^{2}}} = 2 \\ \end{cases}\\ \therefore & AC - B^{2} < 0, (-1,-1) \text{不为极值点} \\ & \text{对}(3,-1) \begin{cases} \displaystyle A = \frac{\partial^{2}{z}}{\partial{x^{2}}} = 12 \\ \displaystyle B = \frac{\partial^{2}{z}}{\partial{x}\partial{y}} = 0 \\ \displaystyle C = \frac{\partial^{2}{z}}{\partial{y^{2}}} = 2 \\ \end{cases}\\ \therefore & AC - B^{2} > 0, A > 0 (3,-1) \text{为极小值点} \\ & \text{极小值为} f(3,-1) = -30 \end{array}$$

多元函数条件极值-Lagrange乘数法

步骤

1. Case 1 $z= f(x,y)\quad \text{s.t.}\ \phi(x,y) = 0$
   1. $F=f(x,y)+\lambda \phi(x,y) = 0$
   2. 由 $\begin{cases}F_{x} = f_{x} + \lambda \phi_{x} = 0\\F_{y} = f_{y} + \lambda\phi_{y} = 0\\F_{\lambda} = \phi(x,y) = 0\end{cases}$ ==> $\begin{cases}x=?\\y=?\end{cases}$
   3. 一一代入进行比较

例题

1. $u=u(x,y)$ 满足：a. $du=2xdx-2ydy$；b. $u(0,0) = 2$ 求
   1. $u(x,y)$ $$\begin{array}{ll} & \text{Method 1} \\ \because & du = 2xdx- 2y dy \\ \therefore & du = d(x^{2} - y^{2}) \\ \therefore & u(x,y) = x^{2} - y^{2} + C \\ \because & u(0,0) = 2 \\ \therefore & u(x,y) = x^{2}-y^{2}+2 \\\\ & \color{#D0104C}{\text{Method 2}} \\ \because & du = 2xdx - 2y dy \\ \therefore & \frac{\partial{z}}{\partial{x}} = 2x \\ \therefore & u = x^{2} - \phi(y) \\ \because & \frac{\partial{z}}{\partial{y}} = \phi'(y) = -2y \\ \therefore & \phi(y) = y^{2} + C \\ \therefore & u(x,y) = x^{2} - y^{2} + C\\ \because & u(0,0) = 2 \\ \therefore & u(x,y) = x^{2}-y^{2}+2 \\ \end{array}$$
   2. $u(x,y)$ 在 $x^{2} + 4y^{2}\le 4$ 上的 $m,M$ $$\begin{array}{ll} & \text{对于} x^{2} + 4y^{2} < 4 \\ \because & \begin{cases} \frac{\partial{u}}{\partial{x}} = 2x = 0 \\ \frac{\partial{u}}{\partial{y}} = 2y = 0 \\ \end{cases} \Rightarrow \begin{cases} x=0 \\ y=0 \end{cases} \\ \therefore & u(0,0) = 2 \\ & \text{对于} x^{2} + 4y^{2} = 4 \\ \therefore & F = x^{2} - y^{2} +2 + \lambda(x^{2} + 4y^{2} - 4) \\ & \begin{cases} F_{x} = 2x +2\lambda x = 2x(1+\lambda) = 0 \\ F_{y} = -2y +8\lambda y = 2y(4\lambda-1) = 0\\ F_{\lambda} = x^{2} + 4y^{2} -4 =0 \end{cases} \\ \therefore & \begin{cases} \lambda = -1 \\ y=0 \\ x = \pm 2 \end{cases} \begin{cases} \lambda = \frac{1}{4} \\ x=0 \\ y = \pm 1 \end{cases} \\ \because & u(\pm 2, 0) = 6 \\ & u(0, \pm 1) = 1 \\ \because & u(0,0) = 2 \\ \therefore & m = u(0,\pm 1) = 1, \\& M = u(\pm 2,0)=6 \\ \end{array}$$