f=x-y+2z+h(x^2+y^2+2z^2-16)
df/dx=1+2hx=0;x=-1/(2h)
df/dy=-1+2hy=0;y=1/(2h)
df/dz=2+4hz=0;z=-1/(2h)
(-1/2h)^2+(1/2h)^2+2(1/2h)^2-16=0
h=+1/4;h=-1/4
x=-1/8;y=1/8;z=-1/8;u=-1/8-1/8-2*1/8=-4/8;極小
x=+1/8;y=-1/8;z=1/8;u=1/8+1/8+2/8=+4/8;極大
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但如果我們用JACOBIAN的行列式來做;2X+2Y=0;-4Z-4Y=0;解之;X=-Y;Z=-Y;解之;得Y=+2;-2;代入;U=0為極大;U=-8為極小;答案不是唯一的;為甚麼?
函數(shù)F(X,Y,Z,...)極大極小時(shí);必需條件為偏微分DF/DX=DF/DY=DF/DZ=...=0
如果是F(X,Y);A=F11,B=F12,C=F22,1對(duì)X微分;2對(duì)Y微分;則
B^2-AC