求Z=x^3-y^3-3xy的极值和极值点
1个回答

Z=f(x,y)=x^3-y^3-3xy

分别对x,y求偏导:

fx=3x^2-3y

fy=-3y^2-3x.

令fx=0,fy=0,可得x=0,y=0,或x=-1,y=1这2个驻点.

然后求二阶偏导:

fxx=6x,fxy=-3,fyy=-6y.

x=0,y=0时,fxx=0,fxy=-3,fyy=0,(-3)^2-0>0,所以(0,0)不是极值点;

x=-1,y=1时,fxx=-6,fxy=-3,fyy=-6,(-3)^2-(-6)^2