Lagrange Multiplier Question

  • #1
ElijahRockers
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Homework Statement



f(x,y) = y2-x2, g(x,y) = x2/4 +y2=9

Homework Equations



[itex]\nabla f = \lambda \nabla g[/itex]

[itex]-2x = \lambda \frac{x}{2}[/itex]
[itex]2y = 2\lambda y[/itex]
[itex]\frac{1}{4} x^2 + y^2 = 9[/itex]

The Attempt at a Solution



I arrived at the three equations above. So according to the first equation, lambda can equal -4. According to the second equation, it can equal 1. After this, I am algebraically lost. The x's and y's cancel themselves out from the first two equations. What does this mean?
 

Answers and Replies

  • #2
Dick
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−2x=λx/2 means EITHER λ=(-4) OR x=0. You have to check both options.
 
  • #3
Ray Vickson
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Homework Statement



f(x,y) = y2-x2, g(x,y) = x2/4 +y2=9

Homework Equations



[itex]\nabla f = \lambda \nabla g[/itex]

[itex]-2x = \lambda \frac{x}{2}[/itex]
[itex]2y = 2\lambda y[/itex]
[itex]\frac{1}{4} x^2 + y^2 = 9[/itex]

The Attempt at a Solution



I arrived at the three equations above. So according to the first equation, lambda can equal -4. According to the second equation, it can equal 1. After this, I am algebraically lost. The x's and y's cancel themselves out from the first two equations. What does this mean?

If x ± 0 then λ = -4, so in the second equation you must have y = 0.

RGV
 
Last edited:
  • #4
Dick
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If x ± 0 then λ = 4, so in the second equation you must have y = 0.

RGV

What are you talking about?
 
  • #5
Ray Vickson
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What are you talking about?

I had the typo λ = 4 instead of the correct λ = -4, but that still implies we need y = 0 to satisfy the second equation (which would be 2y = -8y).

RGV
 
  • #6
ElijahRockers
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Ok, well a lambda of -4 makes the other equation untrue. So lambda can not be -4 then, right?
 
  • #7
Dick
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Ok, well a lambda of -4 makes the other equation untrue. So lambda can not be -4 then, right?

Yes, it can. lambda can be -4 if y is zero.
 
  • #8
ElijahRockers
Gold Member
270
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Duh, ok, I think I got it. Thanks again.
 

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