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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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