Question

x^2+y^2+z^2-2x-2y-4z-10=0确定的隐函数z=f(x,y)有两个极小值吗

Answer 1

对x求偏导：2x+2zZ'x-2-4Z'x=0

得：Z'x=(1-x)/(z-2)

同理得Z'y=(1-y)/(z-2)

由Z'x=0,Z'y=0,得x=1,y=1

将其代入原方程得：1+1+z²-2-2-4z-10=0

得：z²-4z-12=0

(z+2)(z-6)=0

z=-2,6

Z"xx=[-(z-2)-(1-x)Z'x]/(z-2)²

Z"xy=-(1-x)/(z-2)²

Z"yy=[-(z-2)-(1-y)Z'y]/(z-2)²

当x=1,y=1,z=-2时,A=Z"xx=-1/(z-2)=1/4,B=Z"xy=0,C=Z"yy=-1/(z-2)=1/4; AC-B²=1/16>0,A>0,因此这是极小值点；

当x=1,y=1,z=6时,A=Z"xx=-1/(z-2)=-1/4,B=Z"xy=0,C=Z"yy=-1/(z-2)=-1/4;AC-B²=1/16>0,A