高数强化NO12|多元函数的极值|无条件极值|条件极值|有界闭区域上的最值

无条件极值

对概念和定理的考查

$$\begin{array}{rl}& \text{解: }f(x,y)\text{在有界闭区域上连续，则一定存在最大值和最小值.}\\ & \\ & \text{若不在边界上取到，就在区域内取到. 若不在区域内取到，则一定在边界取到}\\ & \\ & AC-B^2=\frac{{\partial }^{2}u}{\partial x^2}\cdot \frac{{\partial }^{2}u}{\partial y^2}-{\left(\frac{{\partial }^{2}u}{\partial x\partial y}\right)}^2=-{\left(\frac{{\partial }^{2}u}{\partial x^2}\right)}^2-{\left(\frac{{\partial }^{2}u}{\partial x\partial y}\right)}^2<0,\\ & \\ & \text{则在区域内不存在极值，一定不存在最值.}\\ & \\ & \text{故最大值和最小值一定在边界上取得. 选}(A)\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{当题目中给出函数存在二阶连续导数时，用充分条件判断某点是否为极值点，}\\ & 当题目中出现抽象函数且未知导数存在性，考虑用定义判断某点是否为极值点。\end{array}$$

$$\begin{array}{rl}& \text{法1: 特殊值（蒙猜）}\\ & \text{令}f(x,y)=xy+(x^2+y^2{)}^2\\ & f(x,y)\text{在}(0,0)\text{的一个非常小邻域}\\ & f(x,y)-xy>0,xy\text{在}(0,0)\text{的一个非常小邻域}\\ & \text{可正可负，}f(x,y)\text{的符号只要可正可负即可.}\\ & f(0,0)=0,\text{故}f(x,y)\text{在}(0,0)\text{处不取极值.}\\ & \text{选}(A)(\text{这种题一般不会选}D)\end{array}$$
$$\begin{array}{rl}& \text{法2: 由极限与无穷小的关系}\\ & 0=\underset{\begin{array}{c}x\to 0\\ y\to 0\end{array}}{\lim }\frac{f(x,y)-xy}{(x^2+y^2{)}^2}-1=\underset{\begin{array}{c}x\to 0\\ y\to 0\end{array}}{\lim }\frac{f(x,y)-xy-(x^2+y^2{)}^2}{(x^2+y^2{)}^2}=0\\ & \\ & f(x,y)=xy+(x^2+y^2{)}^2+o\left((x^2+y^2{)}^2\right)\\ & \\ & \text{在}(0,0)\text{的充分小邻域内, 取}y=x,f(x,x)=x^2+(x^2+x^2{)}^2+o(x^4)>0\\ & \text{取}y=-x,f(x,-x)=-x^2+(x^2+x^2{)}^2+o(x^4)<0\\ & \\ & \text{故}(0,0)\text{不是极值点.}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{本题的难点在于将一元函数极限与无穷小的关系搬到二元函数中运用，在}(0,0)\\ & \text{足够小的去心邻域内，找到两条不同路径，一条路径使 }f(x,y)<f(0,0)\text{成立，一条路}\\ & \text{径使}f(x,y)>f(0,0)\text{成立。}\end{array}$$

极值点和极值的计算

$$\begin{array}{rl}& \text{解: 记方程}{x}^2-6xy+10y^2-2yz-z^2+18=0\text{为}①\\ & \\ & \text{在}①\text{两边同时对}x\text{求偏导, 所得: }2x-6y-2y\frac{\partial z}{\partial x}-2z\frac{\partial z}{\partial x}=0②\\ & \\ & \text{在}①\text{两边同时对}y\text{求偏导, 所得: }-6x+20y-2z-2y\frac{\partial z}{\partial y}-2z\frac{\partial z}{\partial y}=0③\\ & \\ & \text{令}\frac{\partial z}{\partial x}=0,\frac{\partial z}{\partial y}=0,\text{代入可得: }\{\begin{array}{l}x=3y\\ y=z\end{array}\text{代入}①\\ & \\ & 9y^2-18y^2+10y^2-2y^2-y^2+18=0,y=\pm 3\\ & \\ & \text{即}\{\begin{array}{l}x=9\\ y=3\\ z=3\end{array}\text{或}\{\begin{array}{l}x=-9\\ y=-3\\ z=-3\end{array}\end{array}$$
$$\begin{array}{rl}& \text{在}②\text{两边同时对}x\text{求导: }2-2y\left(\frac{{\partial }^{2}z}{\partial x^2}\right)-2{\left(\frac{\partial z}{\partial x}\right)}^2-2z\left(\frac{{\partial }^{2}z}{\partial x^2}\right)=0④\\ & \\ & \text{在}②\text{两边同时对}y\text{求导: }-6-2\frac{\partial z}{\partial x}-2y\frac{{\partial }^{2}z}{\partial x\partial y}-2\frac{\partial z}{\partial y}\cdot \frac{\partial z}{\partial x}-2z\frac{{\partial }^{2}z}{\partial x\partial y}=0⑤\\ & \\ & \text{在}③\text{两边同时对}y\text{求导: }20-2\frac{\partial z}{\partial y}-2\frac{\partial z}{\partial y}-2y\frac{{\partial }^{2}z}{\partial y^2}-2{\left(\frac{\partial z}{\partial y}\right)}^2-2z\frac{{\partial }^{2}z}{\partial y^2}=0⑥\\ & \\ & \text{当}x=9,y=3,z=3,\frac{\partial z}{\partial x}{|}_{(9,3,3)}=0,\frac{\partial z}{\partial y}{|}_{(9,3,3)}=0\text{时:}\\ & AC-B^2=\frac{1}{6}\cdot \frac{5}{3}-{\left(-\frac{1}{2}\right)}^2=\frac{1}{36}>0,\text{又}A=\frac{1}{6}>0,\text{故}(9,3)\text{为}f(x,y)\text{的极小值点,}\\ & \text{极小值为}f(9,3)=3.\\ & \\ & \text{当}x=-9,y=-3,z=-3,\frac{\partial z}{\partial x}{|}_{(-9,-3,-3)}=0,\frac{\partial z}{\partial y}{|}_{(-9,-3,-3)}=0\text{时:}\\ & AC-B^2=\left(-\frac{1}{6}\right)\cdot \left(-\frac{5}{3}\right)-{\left(\frac{1}{2}\right)}^2=\frac{1}{36}>0,\text{又}A=-\frac{1}{6}<0,\text{故}(-9,-3)\text{为}f(x,y)\text{的极大值点,}\\ & \text{极大值为}f(-9,-3)=-3.\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{无条件极值的求解步骤：}\\ & (1)\text{计算一阶偏导数，用必要条件找到所有驻点；}\\ & (2)\text{计算二阶偏导数，对每个驻点逐一用充分条件去判断。}\\ & \\ & 2.\text{本题的函数形式为隐函数，基本解题思路类似，区别在于求导时需要用到隐函数求导的方法。}\\ & \\ & 3.\text{注意由偏导数化简得出}x,y,z\text{的关系后，需要代入原方程求解。}\end{array}$$

$$\begin{array}{rl}& \text{解: }{f}_x^{\prime }(x,y)=\int f_{xy}^{\prime \prime }(x,y)dy=\int 2(y+1)e^{x}dy=e^x(y+1{)}^2+C_1(x)\\ & \\ & f_x^{\prime }(x,0)=e^x+C_1(x)=(x+1)e^x⟹C_1(x)=x{e}^x\\ & \\ & f(x,y)=\int f_x^{\prime }(x,y)dx=\int \left[e^x(y+1{)}^2+x{e}^x\right]dx=e^x(y+1{)}^2+(x-1)e^x+C_2(y)\\ & \\ & f(0,y)=y^2+2y=(y+1{)}^2-1+C_2(y)⟹C_2(y)=0\\ & \\ & f(x,y)=e^x(y+1{)}^2+(x-1)e^x\\ & \\ & \text{令}\{\begin{array}{l}{f}_x^{\prime }=e^x(y+1{)}^2+x{e}^x=0\\ f_y^{\prime }=2e^x(y+1)=0\end{array}⟹\{\begin{array}{l}x=0\\ y=-1\end{array}\\ & \\ & f_{xx}^{\prime \prime }=2e^x(y+1)+(x+1)e^x,f_{xy}^{\prime \prime }=2e^x(y+1),f_{yy}^{\prime \prime }=2e^x\\ & \\ & A=f_{xx}^{\prime \prime }(0,-1)=1,B=f_{xy}^{\prime \prime }(0,-1)=0,C=f_{yy}^{\prime \prime }(0,-1)=2\\ & \\ & AC-B^2=2-0=2>0,\text{又}A>0,\text{故极小值为}f(0,-1)=-1\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{本题没有直接给出函数表达式，需要先根据题设条件求出}\\ & f(x,y)\text{的具体表达式，再计算极值点和极值。}\\ & \\ & 2.\text{二阶混合偏导数积分时，需要注意：}{f}_{xy}^{\prime \prime }\text{对}x\text{积分，得到的是}{f}_y^{\prime }+C(y)\text{；}\\ & f_{xy}^{\prime \prime }\text{对}y\text{积分，得到的是}{f}_x^{\prime }+C(x)\text{。}\end{array}$$

条件极值

对定理内容的考查

$$\begin{array}{rl}& \text{解: 设}F(x,y,\lambda )=f(x,y)+\lambda \phi (x,y)\\ & \\ & \{\begin{array}{l}{F}_x^{\prime }=f_x^{\prime }+\lambda {\phi }_x^{\prime }=0\\ F_y^{\prime }=f_y^{\prime }+\lambda {\phi }_y^{\prime }=0\\ F_{\lambda }^{\prime }=\phi (x,y)=0\end{array}\\ & \\ & \text{若}{f}_x^{\prime }=0,\lambda {\phi }_x^{\prime }=0,\text{即}\lambda =0\text{或}{\phi }_x^{\prime }=0\\ & \text{得不到}{f}_y^{\prime }(x_0,y_0)=0\\ & \\ & \text{若}{f}_x^{\prime }\ne 0,\lambda {\phi }_x^{\prime }\ne 0,\lambda \ne 0\text{且}{\phi }_x^{\prime }(x_0,y_0)\ne 0\\ & \text{题意可知}{\phi }_y^{\prime }\ne 0,\lambda {\phi }_y^{\prime }\ne 0,⟹f_y^{\prime }\ne 0\\ & \\ & \text{选}(D)\end{array}$$

最值的计算

$$\begin{array}{rl}& \text{解: 令}F(x,y,z,\lambda ,\mu )=x^2+y^2+z^2+\lambda (z-x^2-y^2)+\mu (x+y+z-4)\\ & \\ & \{\begin{array}{l}{F}_x^{\prime }=2x-2\lambda x+\mu =0①\\ F_y^{\prime }=2y-2\lambda y+\mu =0②\\ F_z^{\prime }=2z+\lambda +\mu =0③\\ F_{\lambda }^{\prime }=z-x^2-y^2=0④\\ F_{\mu }^{\prime }=x+y+z-4=0⑤\end{array}\\ & \\ & \text{解方程组可得}\{\begin{array}{l}{x}_1=-2\\ y_1=-2\\ z_1=8\end{array}\text{或}\{\begin{array}{l}{x}_2=1\\ y_2=1\\ z_2=2\end{array}\\ & \\ & u(-2,-2,8)=72,u(1,1,2)=6\\ & 故最大值为72，最小值为6\end{array}$$
$$\begin{array}{rl}& \text{计算过程：分析方程组为}①\text{和}②\\ & \text{用}①-②:\\ & (1-\lambda )(y-x)=0\\ & \\ & \text{若}\lambda =1,\text{由}①\text{可得}\mu =0\\ & \text{由}③\text{可得}z=-\frac{1}{2},\text{代入}④\text{可知}④\text{为矛盾方程}\\ & \\ & \text{故}\lambda \ne 1.\\ & \text{则}y=x\\ & \text{由}④\text{可得}z=2x^2,\text{代入}⑤\\ & x+x+2x^2-4=0,x_1=-2,x_2=1\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{计算函数条件极值的基本步骤：}\\ & (1)\text{构造拉格朗日函数；}\\ & (2)\text{求拉格朗日函数的驻点，解方程组；}\\ & (3)\text{根据实际条件判断所求出的点是极大值点和极小值点。}\\ & \\ & 2.\text{拉格朗日求驻点，解方程时计算一般比较大，可先从次数较低的方程中消去}\lambda \text{和}\mu ,\\ & \text{找到变量之间的关系，然后代入次数较高的方程，进而求解。}\end{array}$$

有界闭区域上的最值

$$\begin{array}{rl}& \text{解: 由}dz=2xdx-2ydy\text{可以读出, }\frac{\partial f}{\partial x}=2x,\frac{\partial f}{\partial y}=-2y\\ & \\ & f(x,y)=\int 2xdx=x^2+C(y),\frac{\partial f}{\partial y}=C^{\prime }(y)=-2y,\text{故}C(y)=-y^2+C.\\ & f(x,y)=x^2-y^2+C,f(1,1)=2,\text{故}f(x,y)=x^2-y^2+2.\\ & \\ & \text{先求区域内}\\ & \{\begin{array}{l}{f}_x^{\prime }=2x=0\\ f_y^{\prime }=-2y=0\end{array},\text{解得唯一的驻点为}(0,0).\\ & \\ & A=f_{xx}^{\prime \prime }=2,B=f_{xy}^{\prime \prime }=0,C=f_{yy}^{\prime \prime }=-2,AC-B^2{|}_{(0,0)}=-4<0,\text{则}(0,0)\text{不是极值,}\\ & \text{故}(0,0)\text{一定不是最值.}\\ & f(0,0)=2\\ & \\ & \text{再求区域边界, 由边界方程可知: }{y}^2=4-4x^2.\\ & \text{则}z=f(x,y)=x^2-y^2+2=x^2-(4-4x^2)+2=5x^2-2,x\in [-1,1]\\ & \\ & z_x^{\prime }=10x=0,x=0,z(0)=-2,z(-1)=3,z(1)=3.\\ & \\ & \text{故最大值为}3,\text{最小值为}-2.\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{计算二元函数在有界闭区域上最值的基本步骤：}\\ & (1)\text{求出函数在区域内部所有的驻点；}\\ & (2)\text{求出函数在区域边界上最值；}\\ & (3)\text{找出函数边界所有的分界点。}\\ & \text{上述所有点中，函数值最大的点即最大值点，函数值最小的点即最小值点。}\\ & \\ & 2.\text{求函数在区域边界上最值，相当于有约束条件下计算条件极值，可构造拉格朗日函数}\\ & \text{求最值，但一般情况下边界上的条件极值比较简单，不需要构造拉格朗日函数，直接将}\\ & \text{约束条件代入目标函数计算最值即可。}\end{array}$$