Question

高数求解求由方程 2x^2+2y^2+2z^2+8yz-z+8=0 所确定的函数z=z(x,y)的极值

Answer 1

（1）一阶偏导 Zx(x,y)=(-4x)/(4z+8y-1)

Zy(x,y)=(-4y-8z)/(4z+8y-1)

(2)二阶偏导 Zxx(x,y)=[-4(4z+8y-1)^2-64x^2]/(4z+8y-1)^3

Zxy(x,y)=[48y-8]/(4z+8y-1)^3

Zyy(x,y)=(192z^2+32z+32y-192y^2-4)/(4z+8y-1)^3

(3)由一阶偏导,且设z=k.可得驻点（0,-2k,k） 且 k不等于-1/12

（4）将驻点带入二阶偏导 Zxx=4/(12k+1)=A

Zxy=8/(12k+1)^2=B

Zyy=(32k+4-576k^2)/(12k+1)^3=C

(5)AC-B^2=[4(32k+4-576k^2)-8^2]/(12k+1)^4

=[-(48k-4/3)^2-48+16/9]/(12k+1)^4