Can Lagrange's Undetermined Multiplier Have Imaginary Points?

[KaGa]

Hey guys, new here and this is my first post. Wondering if anyone could help me.
So I've encountered a problem on Lagrange's undetermined multiplier. Usually i have no problem with these, but this one caught me off a little.

g(x,y) = x^2 + y^2 - 4xy - 6 = 0
Find the points closest to the origin.
With this in mind:
f(x,y) = x^2 + y^2
Using the formula:
d(f + λg) = 0
Let (f + g) = F
F_x = 2x + 2λx -λ4y
F_y = 2y + 2λy -λ4x
By inspection you can see x = y, so into g(x,y) and...
-2x^2-6=0
x^2 = -3

∴ x = √-3

I haven't encountered an imaginary point yet in this type of question, and since i can't quite make sense of it in my mind, i was wondering if anyone could help me? Have I made an error somewhere, or can you have imaginary points closest to the origin?

Cheers.

 

[AlephZero]

By inspection you can see x = y

That's your error. Check the signs!

 

[KaGa]

That's your error. Check the signs!

F_x = 2x + 2λx -λ4y
F_y = 2y + 2λy -λ4x

I only say by inspection to save writing out the rest of the calculation but I did try it.

2x + λ(2x - 4y) = 0
2y + λ(2y - 4x) = 0

λ= -2x/(2x - 4y) = -2y/(2y - 4x)
[-2x/(2x - 4y) = -2y/(2y - 4x)] *-1
2x/(2x - 4y) = 2y/(2y - 4x)
[2x/(2x - 4y) = 2y/(2y - 4x)]/2
x/(2x - 4y) = y/(2y - 4x)
rearrange:
x(2y - 4x) = y(2x - 4y)
2xy -4x^2 = 2xy -4y^2
-4x^2=-4y^2
y^2 = x^2
y = x.

Unless I'm making a basic floor I'm not seeing.

 

[AlephZero]

$y = x$ is not the only solution of $y^2 = x^2$.

 

[blue_raver22]

I can say that Lagrange's undetermined multiplier method can indeed result in imaginary points. This is because the method involves finding the critical points of a function, which can include complex numbers. In your example, the point closest to the origin may indeed be an imaginary point, which would mean that the minimum distance to the origin is a complex number. This is not uncommon in optimization problems, and it simply means that the solution lies on the imaginary axis rather than the real axis. It is important to remember that imaginary numbers are a valid part of mathematics and can have real-world applications. So while it may seem unfamiliar, it is a valid result in this case. I would recommend double-checking your calculations and if you are still unsure, seeking further assistance from a mathematics expert.