多元微分学

概念

二重极限

$$\begin{array}{rl}& 若对\forall ϵ>0,\exists \delta >0,当0<\sqrt{(x-x_0{)}^2+(y-y_0{)}^2}<\delta 时，\\ & 恒有|f(x,y)-A|<ϵ,则称A是f(x,y)在(x_0,y_0)点的极限\\ [注](1)& \{\begin{array}{l}一元极限中x\to x_0有且仅有两种方式\\ 二元极限中有无穷任意多种方式\end{array}\\ (2)& 若有两条不同路径（如直线y=kx，抛物线x=y^2）使极限\underset{x\to 0,y\to 0}{\lim }f(x,y)值不相等\\ & 或某一条路径使极限\underset{x\to 0,y\to 0}{\lim }f(x,y)值不存在，则说明二重极限\underset{x\to 0,y\to 0}{\lim }f(x,y)不存在\\ (3)& 主要方法：化成一元极限、等价代换、无穷小乘有界、夹逼准则\\ (4)& 二重极限保持了一元极限的各种性质，如唯一性、局部有界性、局部保号性及运算性质\\ (5)& 所求极限得二元函数f(x,y)如果使齐次有理式函数，即分子、分母分别均是齐次有理函数，\\ & 考察(x,y)\to (0,0)时得极限，可用下述命题：\\ & 设f(x,y)=\frac{P(x,y)}{Q(x,y)}=\frac{m次}{n次},其中分子分母是互质多项式，则\\ & 1.当m>n时，若方程Q(1,y)=0与Q(x,1)=0均无实根，则\underset{x\to 0,y\to 0}{\lim }f(x,y)=0;\\ & 若方程Q(1,y)=0与Q(x,1)=0有实根，则\underset{x\to 0,y\to 0}{\lim }f(x,y)不存在\\ & 2.当m\le n时，\underset{x\to 0,y\to 0}{\lim }f(x,y)不存在\end{array}$$

$$\begin{array}{rl}& [注3相关]\\ 1.& \underset{x\to 0,y\to 0}{\lim }\frac{\sqrt{x^2+y^2}-\sin \sqrt{x^2+y^2}}{(\sqrt{x^2+y^2}{)}^3}\\ & 令\sqrt{x^2+y^2}=t,则I=\underset{t\to 0}{\lim }\frac{t-\sin t}{t^3}=\frac{1}{6}\\ 2.& \underset{x\to 0,y\to 0}{\lim }\frac{\sin xy}{y}\\ & =\underset{x\to 0,y\to 0}{\lim }\frac{xy}{y}=\underset{x\to 0,y\to 0}{\lim }x=0\\ 3.& \underset{x\to 0,y\to 0}{\lim }\frac{x{y}^2}{x^2+y^2}=\underset{x\to 0,y\to 0}{\lim }x\cdot \frac{y^2}{x^2+y^2}=0\\ & [注5相关]\\ 1.& \underset{x\to 0,y\to 0}{\lim }\frac{x^3+y^3}{x^2+y^2}\\ & m=3>n=2,且Q(1,y)=1+y^2=0和Q(x,1)=x^2+1=0无实根⟹I=0\\ 2.& \underset{x\to 0,y\to 0}{\lim }\frac{xy}{x+y}\\ & m=2>n=1,且Q(1,y)=1+y=0有实根⟹不\exists \\ 3.& \underset{x\to 0,y\to 0}{\lim }\frac{x+y}{x-y}\\ & m=1=n=1⟹不\exists \\ 4.& \underset{x\to 0,y\to 0}{\lim }\frac{x^2+y^2}{x^3+y^3}\\ & m=2<n=3⟹不\exists \\ & [注2相关]\\ & 证明\underset{x\to 0,y\to 0}{\lim }\frac{x^{2}y}{x^4+y^2}不存在\\ & I\underrightarrow{y=kb}=\underset{x\to 0}{\lim }\frac{k{x}^3}{x^4+k^2{x}^2}=\underset{x\to 0}{\lim }\frac{kx}{x^2+k^2}=0\\ & I\underrightarrow{y=x^2}\underset{x\to 0}{\lim }\frac{x^4}{x^4+x^4}=\frac{1}{2}\\ & \therefore 二元极限不存在\end{array}$$

连续

$$\begin{array}{rl}& 若\underset{x\to 0,y\to 0}{\lim }f(x,y)=f(x_0,y_0),则称f(x,y)在点(x_0,y_0)处连续\\ & [注]不讨论间断点\end{array}$$

偏导数

$$\begin{array}{rl}& z=f(x,y)在(x_0,y_0)处的偏导数\\ & f_x^{\prime }(x_0,y_0)=\underset{\Delta x\to 0}{\lim }\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}=\underset{x\to x_0}{\lim }\frac{f(x,y_0)-f(x_0,y_0)}{x-x_0}\\ & f_y^{\prime }(x_0,y_0)=\underset{\Delta y\to 0}{\lim }\frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y}=\underset{y\to y_0}{\lim }\frac{f(x_0,y)-f(x_0,y_0)}{y-y_0}\\ & 1.偏导数实际上就是对应一元函数得导数，如:\\ & f_x^{\prime }(x_0,y_0)={\phi }^{\prime }(x){|}_{x=x_0}=[f(x,y_0){]}^{\prime }{|}_{x\to x_0}\\ & f_y^{\prime }(x_0,y_0)={\phi }^{\prime }(y){|}_{y=y_0}=[f(x_0,y){]}^{\prime }{|}_{y\to y_0}\\ & 2.求f(x,y)的偏导数只需先把其中一个变量视为常数即可，\\ & 如求f_x^{\prime }(x,y)时，把f(x,y)中的y先视为常数，y偏导同理。\end{array}$$

$$\begin{array}{rl}1.& 设f(x,y)=x^2+(y-1)\arcsin \sqrt{\frac{y}{x}},则\frac{\partial f}{\partial x}{|}_{(2,1)}=\\ & f(x,1)=x^2⟹\frac{\partial f}{\partial x}{|}_{(2,1)}=(x^2{)}^{\prime }{|}_{x=2}=4\\ 2.& 设f(x,y)=\frac{e^x}{x-y},则\underline{}\\ & f_x^{\prime }=\frac{e^x\cdot (x-y)-e^x\cdot (1-0)}{(x-y{)}^2}\\ & f_y^{\prime }=\frac{0-e^x\cdot (0-1)}{(x-y{)}^2}\\ & ⟹f_x^{\prime }+f_y^{\prime }=\frac{e^x}{x-y}=f\\ 3.& 设f(x,y)=e^{x+y}[x^{\frac{1}{3}}(y-1{)}^{\frac{1}{3}}+y^{\frac{1}{3}}(x-1{)}^{\frac{2}{3}}],则在点(0,1)处的两个偏导数f_x^{\prime }(0,1)=\underline{},f_y^{\prime }(0,1)=\underline{}\\ & 令y=1⟹f(x,1)=e^{x+1}(x-1{)}^{\frac{2}{3}}\\ & ⟹f_x^{\prime }(x,1)=e^{x+1}(x-1{)}^{\frac{2}{3}}+e^{x+1}\frac{2}{3}(x-1{)}^{-\frac{1}{3}}\\ & 令x=0⟹f_x^{\prime }(0,1)=e+e\cdot (-\frac{2}{3})=\frac{e}{3}\\ & f_y^{\prime }(x_0,y_0)=\underset{\Delta y\to 0}{\lim }\frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y}\\ & 令x=0⟹f(0,y)=e^y\cdot y^{\frac{1}{3}}⟹f_y^{\prime }(0,y)=e^y\cdot y^{\frac{1}{3}}+e^y\cdot \frac{1}{3}{y}^{-\frac{2}{3}}\\ & 令y=1,f_y^{\prime }(0,1)=e+e\cdot \frac{1}{3}=\frac{4}{3}e\end{array}$$

可微

$$\begin{array}{rl}& 判定二元函数f(x,y)在点(x_0,y_0)是否可微的方法\\ & 先求f_x^{\prime }(x_0,y_0)与f_y^{\prime }(x_0,y_0),若有一个不存在，则直接不可微；\\ & 若都存在，则检查\underset{x\to 0,y\to 0}{\lim }\frac{f(x,y)-f(x_0,y_0)-f_x^{\prime }(x_0,y_0)(x-x_0)-f_y^{\prime }(x_0,y_0)(y-y_0)}{\sqrt{(x-x_0{)}^2+（y-y_0{)}^2}}=0?\\ & 若等于0，则可微，若不等于0或不存在，则不可微。\end{array}$$

$$\begin{array}{rl}1.& 讨论f(x,y)=\{\begin{array}{l}\frac{x^{2}y}{x^2+y^2},(x,y)̸=(0,0)\\ 0,(x,y)=(0,0)\end{array}在(0,0)点可微性\\ & f_x^{\prime }(0,0)=\underset{x\to 0}{\lim }\frac{f(x,0)-f(0,0)}{x}=\underset{x\to 0}{\lim }\frac{0-0}{x}=0(\exists )\\ & 同理f_y^{\prime }(0,0)=0(\exists )\\ & \underset{x\to 0,y\to 0}{\lim }\frac{f(x,y)-f(0,0)-0}{\sqrt{x^2+y^2}}=\underset{x\to 0,y\to 0}{\lim }\frac{x^{2}y}{(x^2+y^2{)}^{3/2}}\\ & (m=3=n=3)故f(x,y)在(0,0)处不可微\end{array}$$

概念之间的关系

两个偏导数连续→二元函数可微→二元函数连续→极限存在 且 二元函数可偏导

$$\begin{array}{rl}1.& 设f(x,y)具有一阶连续偏导数，且对任意的(x,y)都有\frac{\partial f(x.y)}{\partial x}>0,\frac{\partial f(x.y)}{\partial y}<0,则\\ & A.f(0,0)>f(1,1)B.f(0,0)<f(1,1)C.f(0,1)>f(1,0)D.f(0,1)<f(1,0)\\ & 由题意，得对x单调递增；对y单调递减\\ & 画图可得D\\ 2.& 设f(x,y)=e^{\sqrt{x^2+y^4}},则\\ & f(x,0)=e^{|x|}在x=0处不可导⟹\frac{\partial f}{\partial x}{|}_{(0,0)}不\exists \\ & f(0,y)=e^{y^2}在y=0处可导⟹\frac{\partial f}{\partial y}{|}_{(0,0)}\exists \\ 3.& 设f(x,y)=\{\begin{array}{l}\frac{x^2{y}^2}{x^2+y^2},(x,y)̸=(0,0)\\ 0,(x,y)=(0,0)\end{array},讨论f(x,y)在点(0,0)的连续性、可导性及可微性\\ & \underset{x\to 0,y\to 0}{\lim }f(x,y)=\underset{x\to 0,y\to 0}{\lim }\frac{x^2{y}^2}{x^2+y^2}=0=f(0,0),故连续\\ & f_x^{\prime }(0,0)=\underset{x\to 0}{\lim }\frac{f(x,0)-f(0,0)}{x}=\underset{x\to 0}{\lim }\frac{0-0}{x}=0\exists \\ & 由对称性可知，f_y^{\prime }(0,0)=0\exists \\ & \underset{x\to 0,y\to 0}{\lim }\frac{f(x,y)-f(0,0)-0}{\sqrt{x^{+}{y}^2}}=\underset{x\to 0,y\to 0}{\lim }\frac{x^2{y}^2}{(x^2+y^2{)}^{3/2}}=0\\ & ⟹f(x,y)在(0,0)处可微\\ 4.& 已知(ax{y}^3-y^2\cos x)dx+(1+by\sin x+3x^2{y}^2)dy为某一函数u(x,y)的全微分，则\\ & \frac{\partial u}{\partial x}=ax{y}^3-y^2\cos x\\ & \frac{\partial u}{\partial y}=1+by\sin x+3x^2{y}^2\\ & ⟹\frac{{\partial }^{2}u}{\partial x\partial y}=3ax{y}^2-2y\cos x\\ & ⟹\frac{{\partial }^{2}u}{\partial y\partial x}=by\cos x+6x{y}^2\\ & ⟹a=2,b=-2\end{array}$$

计算

链式求导规则+高阶偏导复合结构不变

$$\begin{array}{rl}1.& 设z=f(u,v,x),u=u(x,y),v=v(y)都是可微函数，求\frac{\partial z}{\partial x}和\frac{\partial z}{\partial y}\\ & \frac{\partial z}{\partial x}=f_1^{\prime }\cdot \frac{\partial u}{\partial x}+f_3^{\prime }\cdot 1\\ & \frac{\partial z}{\partial y}=f_1^{\prime }\frac{\partial u}{\partial y}+f_2^{\prime }\cdot \frac{dv}{dy}\\ & [注]u是二元\to \partial ,v是一元\to d\\ 2.& 设z=f(e^x\sin y,x^2+y^2),其中f具有二阶连续偏导数，求\frac{{\partial }^{2}z}{\partial x\partial y}\\ & \frac{\partial z}{\partial x}=f_1^{\prime }\cdot e^x\sin y+f_2^{\prime }\cdot 2x\\ & \frac{{\partial }^{2}z}{\partial x\partial y}=(f_{11}^{\prime \prime }\cdot e^x\cos y+f_{12}^{\prime \prime }\cdot 2y)\cdot e^x\sin y+f_1^{\prime }\cdot e^x\cos y+2x(f_{21}^{\prime \prime }\cdot e^x\cos y+f_{22}^{\prime \prime }\cdot 2y)\\ & =e^{2x}\sin y\cos y\cdot f_{11}^{\prime \prime }+(2y{e}^x\sin y+2x{e}^x\cos y)\cdot f_{12}^{\prime \prime }+e^x\cos y{f}_1^{\prime }+4xy{f}_{22}^{\prime \prime }\\ 3.& 设z=f(u,v,x),u=x{e}^y,其中f具有二阶连续偏导数，求\frac{{\partial }^{2}z}{\partial x\partial y}\\ & \frac{\partial z}{\partial x}=f_1^{\prime }\cdot e^y+f_2^{\prime }\cdot 1\\ & \frac{{\partial }^{2}z}{\partial x\partial y}=(f_{11}^{\prime \prime }\cdot x{e}^y+f_{13}^{\prime \prime }\cdot 1)\cdot e^y+f_1^{\prime }\cdot e^y+(f_{21}^{\prime \prime }\cdot x{e}^y+f_{23}^{\prime \prime }\cdot 1)\\ 4.& 设z=f(2x-y)+g(x,xy),其中f二阶可导，g具有二阶连续偏导数，求\frac{{\partial }^{2}z}{\partial x\partial y}\\ & \frac{\partial z}{\partial x}=f^{\prime }\cdot 2+g_1^{\prime }\cdot 1+g_2^{\prime }\cdot y\\ & \frac{{\partial }^{2}z}{\partial x\partial y}=2f^{\prime \prime }\cdot (-1)+g_{12}^{\prime \prime }\cdot x+g_{22}^{\prime \prime }\cdot x\cdot y+g_2^{\prime }\cdot 1\\ 5.& 设F(u,v)二阶偏导连续，并设z=F(\frac{y}{x},x^2+y^2),求\frac{{\partial }^{2}z}{\partial x\partial y}\\ & \frac{\partial z}{\partial x}=F_1^{\prime }(-\frac{y}{x^2})+F_2^{\prime }\cdot 2x\\ & \frac{{\partial }^{2}z}{\partial x\partial y}=\frac{\partial (\frac{\partial z}{\partial x})}{\partial y}=\frac{\partial (F_1^{\prime }\cdot (-\frac{y}{x^2}))}{\partial y}+\frac{\partial (F_2^{\prime }\cdot 2x)}{\partial y}\\ & =\frac{\partial F_1^{\prime }}{\partial y}(-\frac{y}{x^2})+F_1^{\prime }(-\frac{1}{x^2})+2x\frac{\partial F_2^{\prime }}{\partial y}\\ & =(F_{11}^{\prime \prime }(\frac{1}{x})+F_{12}^{\prime \prime }2y)(-\frac{y}{x^2})+F_1^{\prime }(-\frac{1}{x^2})+2x(F_{21}^{\prime \prime }(\frac{1}{x})+F_{22}^{\prime \prime }2y)\end{array}$$

隐函数求导

$$\begin{array}{rl}& F(x,y,z)=0⟹z=z(x,y)⟹F(x,y,z(x,y))=0\end{array}$$

$$\begin{array}{rl}& \frac{\partial F}{\partial x}=F_x^{\prime }=F_1^{\prime }+F_z^{\prime }\frac{\partial z}{\partial x}=0⟹\{\begin{array}{l}\frac{\partial z}{\partial x}=-\frac{F_x^{\prime }}{F_z^{\prime }}\\ \frac{\partial z}{\partial y}=-\frac{F_y^{\prime }}{F_z^{\prime }}\end{array}\end{array}$$

$$\begin{array}{rl}1.& 设z=z(x,y)由方程\ln z+e^{z-1}=xy确定，则\frac{\partial z}{\partial x}{|}_{(2,\frac{1}{2})}\\ 方法一.& 由\ln z+e^{z-1}=xy⟹\frac{1}{z}\cdot z_x^{\prime }+e^{z-1}\cdot z_x^{\prime }=y\\ & 由x=2,y=\frac{1}{2},代入前者，则z=1;代入后者，则z_x^{\prime }=\frac{1}{4}\\ 方法二.& 令F(x,y,z)=\ln z+e^{z-1}-xy=o\\ & ⟹\frac{\partial z}{\partial x}=-\frac{F_x^{\prime }}{F_z^{\prime }}=-\frac{-y}{\frac{1}{z}+e^{z-1}}{|}_{(2,\frac{1}{x},1)}=\frac{1}{4}\\ 2.& 设z=z(x,y)由方程F(x+\frac{z}{y},y+\frac{z}{x})=0确定，其中F由连续偏导数，求x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}\\ 方法一.& F_1^{\prime }\cdot (1+\frac{1}{y}\frac{\partial z}{\partial x})+F_2^{\prime }\cdot \frac{\frac{\partial z}{\partial x}\cdot x-z}{x^2}=0(对x)\\ & F_1^{\prime }\cdot \frac{\frac{\partial z}{\partial y}\cdot y-z}{y^2}+F_2^{\prime }\cdot (1+\frac{1}{x}\cdot \frac{\partial z}{\partial y})=0(对y)\\ & ⟹x\cdot \frac{\partial z}{\partial x}+y\cdot \frac{\partial z}{\partial y}=z-xy\\ 方法二.& \frac{\partial z}{\partial x}=-\frac{F_x^{\prime }}{F_z^{\prime }}=-\frac{F_1^{\prime }+F_2^{\prime }(-\frac{z}{x^2})}{F_1^{\prime }(\frac{1}{y})+F_2^{\prime }(\frac{1}{x})}\\ & \frac{\partial z}{\partial y}=-\frac{F_y^{\prime }}{F_z^{\prime }}=-\frac{F_1^{\prime }(-\frac{z}{y^2})+F_2^{\prime }}{F_1^{\prime }(\frac{1}{y})+F_2^{\prime }(\frac{1}{x})}\\ & I=z-xy\\ 3.& 设\{\begin{array}{l}u=f(x-ut,y-ut,z-ut)\\ g(x,y,z)\end{array},求\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}\\ & [分析]一般有几个方程，就有几个因变量，其余的字母都是自变量\\ & \frac{\partial u}{\partial x}=f_1^{\prime }\cdot (1-\frac{\partial u}{\partial x}t)+f_2^{\prime }\cdot (-\frac{\partial u}{\partial x}t)+f_3^{\prime }\cdot (\frac{\partial z}{\partial x}-\frac{\partial u}{\partial x}t)\\ & g_1^{\prime }\cdot 1+g_2^{\prime }\cdot 0+g_3^{\prime }\cdot \frac{\partial z}{\partial x}=0\\ & 解得\frac{\partial u}{\partial x}=\frac{f_1^{\prime }{g}_3^{\prime }-f_3^{\prime }{g}_1^{\prime }}{g_3^{\prime }[1+t(f_1^{\prime }+f_2^{\prime }+f_3^{\prime })]}\\ & 对y求偏导数同样可得\frac{\partial u}{\partial y}=\frac{f_2^{\prime }{g}_3^{\prime }-f_3^{\prime }{g}_2^{\prime }}{g_3^{\prime }[1+t(f_1^{\prime }+f_2^{\prime }+f_3^{\prime })]}\end{array}$$

应用

无条件极值

$$\begin{array}{rl}1.& 必要条件\{\begin{array}{l}{f}_x^{\prime }(x_0,y_0)=0\\ f_y^{\prime }(x_0,y_0)=0\end{array}\\ 2.& 充分条件\{\begin{array}{l}A>0且AC-B^2>0⟹极小\\ A<0且AC-B^2>0⟹极大\\ AC-B^2<0⟹非极值点\end{array}\\ [注]& 记A=f_{xx}^{\prime \prime }(x_0,y_0),B=f_{xy}^{\prime \prime }(x_0,y_0),C=f_{yy}^{\prime \prime }(x_0,y_0)\end{array}$$

条件极值

$$\begin{array}{rl}1.& 构造拉格朗日函数F(x,y,\lambda )=f(x,y)+\lambda \phi (x,y),\\ 2.& 令\{\begin{array}{l}{F}_x^{\prime }=f_x^{\prime }(x,y)+\lambda {\phi }_x^{\prime }(x,y)=0\\ F_y^{\prime }=f_y^{\prime }(x,y)+\lambda {\phi }_y^{\prime }(x,y)=0\\ F_{\lambda }^{\prime }=\phi (x,y)=0\end{array}\\ 3.& 比较上述各函数值的大小，最大的为最大值，最小的为最小值\end{array}$$

$$\begin{array}{rl}1.& 设z=z(x,y)由方程(x^2+y^2)z+\ln z+2(x+y+1)=0确定，求z=z(x,y)的极值\\ & 由方程得2xz+(x^2+y^2)z_x^{\prime }+\frac{1}{z}{z}_x^{\prime }+2=0\\ & ⟹2yz+(x^2+y^2)z_y^{\prime }+\frac{1}{z}{z}_y^{\prime }+2=0\\ & 令z_x^{\prime }=0,z_y^{\prime }=0,则x=y=-\frac{1}{z}代入第一个式子\\ & 得：\frac{2}{z}+\ln z-\frac{4}{z}+2=0⟹z=1\\ & \therefore x=y=-1,z=1\\ & ⟹2z+2x{z}_x^{\prime }+2x{z}_x^{\prime }+(x^2+y^2)z_{xx}^{\prime \prime }+\frac{z_{xx}^{\prime \prime }\cdot z-z_x^{\prime 2}}{z^2}=0\\ & ⟹2x{z}_y^{\prime }+2y{z}_x^{\prime }+(x^2+y^2)z_{xy}^{\prime \prime }+\frac{z_{xy}^{\prime \prime }\cdot z-z_x^{\prime }\cdot z_y^{\prime }}{z^2}=0\\ & A=z_{xx}^{\prime \prime }(-1,-1)=-\frac{2}{3}=c(对称性)\\ & B=z_{xy}^{\prime \prime }(-1,-1)=0,由B^2-AC<0且A<0\\ & 故z(-1,-1)=1极大值\\ 2.& 求函数u=x^2+y^2+z^2在条件z=x^2+y^2及x+y+z=4下的最大值与最小值\\ & 令F(x,y,z,\lambda ,\mu )=x^2+y^2+z^2+\lambda (x^2+y^2-z)+\mu (x+y+z-4)\\ 得& \{\begin{array}{l}{F}_x^{\prime }=2x+2\lambda x+\mu =0\\ F_y^{\prime }=2y+2xy+\mu =0\\ F_z^{\prime }=2z-\lambda +\mu =0\\ F_{\lambda }^{\prime }=x^2+y^2-z=0\\ F_{\mu }^{\prime }=x+y+z-4=0\end{array}解得：P_1(1,1,2),P_2(-2,-2,8)\\ & 由u(P_1)=6为最小值,且u(P_2)=72为最大值\\ 3.& 求u=\sqrt{x^2+y^2+z^2}在(x-y{)}^2-z^2=1条件下的最小值\\ & 令F(x,y,z,\lambda )=x^2+y^2+z^2+\lambda ((x-y{)}^2-z^2-1)\\ & \{\begin{array}{l}{F}_x^{\prime }=2x+2\lambda (x-y)=0\\ F_y^{\prime }=2y-2\lambda (x-y)=0\\ F_z^{\prime }=2z-2\lambda z=0\\ F_{\lambda }^{\prime }=(x-y{)}^2-z^2-1=0\end{array}解得：P_1(-\frac{1}{2},\frac{1}{2},0),P_2(\frac{1}{2},-\frac{1}{2},0)\\ & 由u(P_1)=u(P_2)=\frac{\sqrt{2}}{2}\\ 4.& 求f(x,y)=x^2-y^2+2在椭圆域D:x^2+\frac{y^2}{4}\le 1上的最大值与最小值\\ & 内部\to f(x,y),由f_x^{\prime }=2x=0,f_y^{\prime }=-2y=0\\ & 边界\to F(x,y,\lambda )\\ & F(x,y,\lambda )=x^2-y^2+\lambda (x^2+\frac{y^2}{4}-1)\\ & 由F_x^{\prime }=2x+2\lambda x=0,F_y^{\prime }=-2y+\frac{\lambda }{2}y=0\\ & F_x^{\prime }=x^2+\frac{y^2}{4}-1=0\\ & f(0,0)=2,f(+-1,0=3)最大,f(0,+-2)=-2最小\\ 5.& 求f(x,y)=x+xy-x^2-y^2在闭区域D:0\le x\le 1,0\le y\le 2上得最大值与最小值\\ & 内部\to f(x,y),由f_x^{\prime }=1+y-2x=0,f_y^{\prime }=x-2y=0\\ & 边界(代入法)L_1:y=0(0\le x\le 1)⟹f(x,0)=x-x^2=\phi (x)⟹(\frac{1}{2},0),(0,0),(1,0)\\ & L_2:x=1(0<y<2)⟹f(1,y)=y-y^2⟹(1,\frac{1}{2})\\ & L_3:y=2(0\le x\le 1)⟹f(x,2)=3x-x^2-4⟹(0,2),(1,2)\\ & L_4:x=0(0<y<2)⟹f(0,y)=-y^2\\ & 比较得最大值为\frac{1}{3}，最小值为-4\end{array}$$