1.Total differential vanishes.
Ans:
Ellipsoid: z^2=(4-x^2-y^2+xy)/2
Distance from O(0,0,0) to P(x,y,z):
F(x,y)=√(x^2+y^2+z^2)
=√[(2x^2+2y^2+4-x^2-y^2+xy)/2]
=√[(4+x^2+y^2+xy)/2]
Fx=(2x+y)/√[2(4+x^2+y^2+xy)]=0 => 2x+y=0
Fy=(x+2y)/√[2(4+x^2+y^2+xy)]=0 => x+2y=0
The both equations: x=y=0
Thus z^2=(4-x^2-y^2+xy)/2=2 => z=+-√2
Fxx=Fyy=2>0 => min
So P(x,y,z)=(0,0,+-√2).....ans
2.Lagrange multipliers.
Ans: L=Lamda
F(x,y,z)=√(x^2+y^2+z^2)+L*(x^2-xy+y^2+2z^2-4)
Fx=x/√(x^2+y^2+z^2)+(2x-y)*L=0
Fy=y/√(x^2+y^2+z^2)+(2y-x)*L=0
Fz=z/√(x^2+y^2+z^2)+4z*L=0
Partial(F)/Partial(L)=x^2-xy+y^2+2z^2-4=0
The former 3 equations:
L*√(x^2+y^2+z^2)=x/(y-2x)=y/(x-y)=-1/4
=> 2x+y=x+2y=0 => x=y=0
The last equation: x^2-xy+y^2+2z^2-4=0 => z=+-√2
So P(x,y,z)=(0,0,+-√2).....ans
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