z=x^2+y^2
x+y+z=1
橢圓方程為(x+1/2)^2+(y+1/2)^2=3/2
z=1-x-y
原點到這橢圓上點的距離r=根號{x^2+y^2+z^2}
極值點坐標(biāo)滿足dr/dx=0
dr/dx=[2x+2y*dy/dx+2z*dz/dx]/2r
=x+y*dy/dx+(1-x-y)*(-1-dy/dx)
=(2x+y-1)+(x+2y-1)*dy/dx
對橢圓方程求導(dǎo)2*(x+1/2)+2*(y+1/2)*dy/dx=0
dy/dx=-(2x+1)/(2y+1)
dr/dx=(2x+y-1)-(x+2y-1)*(2x+1)/(2y+1)
=(2x+2y-3)*(y-x)/(2y+1)
dr/dx=0,=> (2x+2y-3)*(y-x)=0
x=y=+(-)根號3/2-1/2 ; x+y=3/2>1（舍去）
r=根號{x^2+y^2+z^2}=根號{2x^2+4y^2}=根號{(11+(-)6*根號3)/2}
r(min)=根號{(11-6*根號3)/2}
r(max)=根號{(11+6*根號3)/2}