Question

看图求解；；；；；；

Answer 1

z=e^(x-y) (x²-2y²)
z'x=e^(x-y) (x²-2y²) + e^(x-y) *2x=e^(x-y) (x²+2x-2y²)
z'y=-e^(x-y) (x²-2y²) + e^(x-y) *(-4y)=-e^(x-y) (x²+4y-2y²)
令z'x=0和z'y=0
x²+2x-2y²=0
x²+4y-2y²=0
得驻点(0，0)，(-4，-2)
z''xx=e^(x-y) (x²+2x-2y²)+e^(x-y) (2x+2)=e^(x-y) (x²+4x-2y²+2)=A
z''xy=-e^(x-y) (x²+2x-2y²) + e^(x-y) (-4y)=-e^(x-y) (x²+2x+4y-2y²)=B
z''yy=-[-e^(x-y) (x²+4y-2y²) + e^(x-y) (4-4y)]=e^(x-y) (x²+8y-2y²-4)=C
当驻点为(0，0)时
A=2
B=0
C=-4
AC-B²＜0，不是极值
当驻点为(-4，-2)时，
A=-6
B=-8
C=-12
AC-B²＞0，有极值，且A＜0，有极大值
极大值为z(-4,-2)=8e^(-2)