定理4.3（极值的必要条件）#

设函数$f$在点$X_{0}$的一阶偏导数都存在，若$f$在$X_{0}$取极值，则$\nabla f(X_{0})=0$
选$\Phi(t)=f(x_{0}+tk,y_{0}+th)$需要可微
满足$\nabla f(X_{0})=0$的点$X_{0}$称为驻点
极值点一定是驻点（函数一阶偏导数存在时），反过来不成立，例如鞍点
由$Taylor$公式
$f(X_{0}+\Delta X)-f(X_{0})=\langle\nabla f(X_{0}),\Delta X\rangle+\frac{1}{2}(\Delta X)^{T}H_{f}(X_{0})\Delta X+o(||\Delta X||^{2})$

若$X_{0}$是驻点，则
$f(X_{0}+\Delta X)-f(X_{0})=\frac{1}{2}(\Delta X)^{T}H_{f}(X_{0})\Delta X+o(||\Delta X||^{2})$

定理4.4（极值充分条件）设$f(X)$在$U(X_{0})$内二阶偏导数连续，$\nabla f(X_{0})=0$且$H_{f}(X_{0})$为$f$在点$X_{0}$的$Hesse$矩阵$(\frac{\partial^{2}f}{\partial X_{i}\partial X_{j}})(X_{0})_{n\times n}$,#

若$H_{f}(X_{0})$是正定（负定）矩阵，则$f(X_{0})$为极小（大）值，
因为$H_{f}(X_{0})$是实对称矩阵，存在实正交矩阵$Q$，使得
$Q^{T}H_{f}(X_{0})Q=diag\{\lambda_{1},...,\lambda_{n}\}$
其中$\lambda_{1},...,\lambda_{n}$为$H_{f}(X_{0})$的实特征值，令$\Delta X=Q\cdot y$，则

对二元函数$f(x,y)$,记
$A=f_{xx}(X_{0}),B=f_{xy}(X_{0}),C=f_{yy}(X_{0}),H_{f}(X_{0})=\begin{pmatrix}A&B\\B&C\end{pmatrix}$

$f(x,y)=x^{2},(0,0)$是极小值点，但$A=2,B=C=0$
$g(x,y)=x^{2}+y,(0,0)$不是极值点，但$A=2,B=C=0$

例4.2 求$f(x,y)=x^{3}+y^{3}+3xy$的极值#

解：
先求驻点
$f_{x}=3x^{2}+3y=0$
$f_{y}=3y^{2}+3x=0$
则$P_{1}(0,0),P_{2}(-1,-1)$为两个驻点
再求二阶偏导数
$f_{xx}=6x$
$f_{xy}=3$
$f_{yy}=6y$

于是
$H_f(P_{1})=\begin{pmatrix}0&3\\3&0\end{pmatrix},H_f(P_{2})=\begin{pmatrix}-6&3\\3&-6\end{pmatrix}$
因为$H_{f}(P_{1})$不定，$H_f(P_{2})$负定，则$P_{1}$不是极值点，$P_{2}$为极大值点，$f(P_{2})=1$为极大值

例4.3 $f(x,y)=2x^{2}-3xy^{2}+y^{4}$的极值点#

解：
$f_{x}=4x-3y^{2}=0$
$f_{y}=-6xy+4y^{3}=0$
驻点$P(0,0)$
$f_{xx}=4$
$f_{xy}=-6y$
$f_{yy}=-6x+12y^{2}$
$f_{xx}(0,0)=4,f_{xy}(0,0)=0,f_{yy}(0,0)=0$
此时$AC-B^{2}=0$
有因为$f(x,y)-f(0,0)=(2x-y^{2})(x-y^{2})$
在$x>y^{2},2x<y^{2}$时大于零，$\frac{1}{2}y^{2}<x<y^{2}$时小于零
因此，$f(0,0)=0$不是极值

ex.2 方程$2x^{2}+y^{2}+z^{2}+2xy-2x-2y-4z+4=0(*)$所确定的隐函数$z=Z(x,y)$的极值#

（下面的求导是我自己算的，解法在解后面）
$z_{x}=4x+2z\frac{\partial z}{\partial x}+2y-2-4\frac{\partial z}{\partial x}=4x+2y-2=0$
$z_{y}=2y+2z\frac{\partial z}{\partial y}+2x-2-4\frac{\partial z}{\partial y}=2y+2x-2=0$
$z_{xx}=4+2(\frac{\partial z}{\partial x})^{2}+(2z-4)\frac{\partial^{2}z}{\partial x^{2}}$
$z_{xy}=2\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}+2-4\frac{\partial^{2}z}{\partial x\partial y}$
$z_{yy}=2+2(\frac{\partial z}{\partial x})^{2}+(2z-4)\frac{\partial^{2}z}{\partial x^{2}}$
解：
对$(*)$取微分

先求驻点，令$dz=0$
$(4x+2y-2)dx+(2y-2x-2)dy=0$
由于$dx,dy$的任意性
$4x+2y-2=0,2y-2x-2=0$
则$x=0,y=1$带入$(*)$得$z^{2}-4z+3=0$
则$z=1$或$z=3$，说明$(*)$在$(0,1,1),(0,1,3)$附近分别确定一个隐函数$z=Z_{1}(x,y),z=Z_{2}(x,y)$ 对$(*2)$取微分，得

则
$d^{2}z=\frac{1}{2-z}(2(dx)^{2}+2dxdy+(dy)^{2})$
当$x=0,y=1,z=1$时，$d^{2}z=2(dx)^{2}+2dxdy+(dy)^{2}$
$A=2,B=2,C=1,H_{z_{1}}(0,1)=1,Z_{1}(0,1)=1$为极小值
当$x=0,y=1,z=3$时，$d^{2}z=-2(dx)^{2}-2dxdy-(dy)^{2}$
$A=-2,B=-2,C=-1,H_{z_{2}}(0,1)=1,Z_{1}(0,1)=3$为极大值

ex.3证明：圆的外切三角形中，正三角形的面积最小#

Proof
假设圆的半径为$a$,则三角形面积为
$S=a^{2}(tan(\frac{\alpha}{2})+tan(\frac{\beta}{2})+tan(\frac{\gamma}{2}))=a^{2}(tan(\frac{\alpha}{2})+tan(\frac{\beta}{2})-tan(\frac{\alpha+\beta}{2}))$
$\alpha+\beta+\gamma=2\pi,D:0<\alpha,\beta\pi,\alpha+\beta>\pi$
即$\triangle ABC$,不含三边
先求驻点
$S_{\alpha}=a^{2}(sec^{2}(\frac{\alpha}{2})-sec^{2}(\frac{\alpha+\beta}{2}))=0$
$S_{\beta}=a^{2}(sec^{2}(\frac{\beta}{2})-sec^{2}(\frac{\alpha+\beta}{2}))=0$
$cos(\frac{\alpha}{2})=cos(\frac{\beta}{2})=|cos(\frac{\alpha+\beta}{2})|$
有唯一解$\alpha=\beta=\frac{2}{3}\pi$
$A=S_{\alpha\alpha}=4\sqrt{3}a^{2}$
$B=S_{\alpha\beta}=2\sqrt{3}a^{2}$
$B=S_{\alpha\beta}=4\sqrt{3}a^{2}$
$A>0,AC-B^{2}>0$
因此$S$在$(\frac{2}{3}\pi,\frac{2}{3}\pi)$取到极小值$3\sqrt{3}a^{2}$
下面证明：$min_{D}S=3\sqrt{3}a^{2}$
其一，$S$在$D$上连续，但$D$不是有解闭域，$S$在$D$上处处有偏导数
其二，$S$在$\triangle ABC$的边界附近的值远大于$3\sqrt{3}a^{2}$
当$\alpha+\beta->\pi+0$时，$S>\frac{a^{2}}{2}(-tan(\frac{\alpha+\beta}{2}))$趋向正无穷
当$\alpha->\pi-0$时，$S>\frac{a^{2}}{2}tan(\frac{\alpha}{2})$趋向正无穷
$\beta$类似
其三，直线$\alpha+\beta=\pi-\delta,\alpha=\pi-\delta,\beta=\pi-\delta$围成$\triangle A_{1}B_{1}C_{1}$
则$D=\triangle A_{1}B_{1}C_{1}\cup(int\triangle ABC-\triangle A_{1}B_{1}C_{1})=D_{1}\cup(D-D_{1})$
其中$D_{1}=\triangle A_{1}B_{1}C_{1}$时有界闭域且存在$\delta>0$使得$S|_{D-D_{1}}>6\sqrt{3}a^{2}$
于是$min_{D}S=min_{D_{1}}S$
在$D_{1}$的内点$(\frac{2}{3}\pi,\frac{2}{3}\pi)$取到
因此$min_{D}=S(\frac{2}{3}\pi,\frac{2}{3}\pi)=3\sqrt{3}a^{2}$