Can Lagrange's Undetermined Multiplier Have Imaginary Points?

  • Thread starter KaGa
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
AlephZero
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By inspection you can see x = y
That's your error. Check the signs!
 
  • #3
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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.
 
  • #4
AlephZero
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##y = x## is not the only solution of ##y^2 = x^2##.
 

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