求自原點到曲面x^2y-z^2+9=0之最短距離?

求自原點到曲面x^2y-z^2+9=0之最短距離?F(x,y,z)=√(x^2+y^2+z^2)+λ(yx^2-z^2+9)Fx=x/√(x^2+y^2+z^2) + 2xyλ=0 => λ=-1/2y√(x^2+y^2+z^2)...(1)Fy=y/√(x^2+y^2+z^2) + λx^2=0 => λ=-y/2x^2*√(x^2+y^2+z^2)...(2)Fz=z/√(x^2+y^2+z^2) - 2λz=0 => λ=1/2√(x^2+y^2+z^2)...(3)Eq.(1) = Eq.(3): y=-1Eq.(1) = Eq.(2): x^2=y^2 => x=+-y=-+1原式: z=+-√(yx^2+9)=+-√8=> min=F(-+1,-1,+-√8)=√(1+1+8)=√10

2014-12-28 15:05:12 補充：
min=F(-+1,-1,+-√8) 修改為:

min=√(1+1+8)=√10

最短距離 ＝ 3/2*(48)^(1/6) ≈ 2.8596