请问:怎样用maple求二元函数的极值?答:2,2](f)(x0,y0);Delta:=A*C-B^2;f(x0,y0);第四个驻点:f:=(x,y)->x^3-y^3+3*x^2+3*y^2-9*x;x0:=-3:y0:=2:A:=D[1,1](f)(x0,y0);B:=D[1,2](f)(x0,y0);C:=D[2,2](f)(x0,y0);Delta:=A*C-B^2;f(x0,y0);函数的图形:函数的等值线:
二元函数z=x^3+y^3-3x^2-3y^2的极小值点怎么求啊答:zx=3x²-6x=0 x1=0,x2=2 zy=3y²-6y=0 y1=0,y2=2 驻点(0,0)(0,2),(2,0),(2,2)zxx=6x-6,zxy=0,zyy=6y-6 (0,0)AC-B²=36>0,A=-6<0,所以 取极大值f(0,0)=0 (0,2)AC-B²<0 无极值 (2,0)AC-B²<0 无极值;(2,2)AC-B...