Lagrange Multipliers Question?

  • #1
Lagrange Multipliers Question???

Homework Statement


Find the minimum and maximum values of the function subject to the given constraint.

f (x,y,z) = x^2 - y - z, x^2 - y^2 +z = 0



The Attempt at a Solution



Okay this is what I did:

Gradient f = <2x,-1,-1> Gradient g = <2x,-2y,1>

Gradient f = (Lambda) Gradient g
<2x,-1,-1> = L <2x,-2y,1>

2x = L(2x) L=1
-1=-L(2y)
-1=L(1)

by this Lambda = 1 and -1? How can this be, am I doing something wrong? Thanks for the help.
 

Answers and Replies

  • #2
hotvette
Homework Helper
996
5


I haven't seen this terminology before, but are you sure it isn't Gradient f = - (Lambda) Gradient g?

In any event, look more closely at your first equation. What assumption did you make to obtain the value for L?
 
Last edited:
  • #3


I checked and yes I got the formula right. To get my values for lambda all I did was set the two gradients with g multiplied lambda. This gives L = 1, -1. I dont know how to get the values of x , y , and z from this which is what I need.
 
  • #4
hotvette
Homework Helper
996
5


I'm referring to the following step:

2x = L(2x) L=1

What assumption is implied about x and is the assumption valid?
 
  • #5
558
1


There's a conceptual mistake here:
Gradient f = (Lambda) Gradient g

<2x,-1,-1> = L <2x,-2y,1>

You have a function:

[tex]F(x,y,z)= x^2 - y - z[/tex]

and a vinculum which will be equal to zero:

[tex]G(x,y,z)= x^2 - y^2 +z[/tex]

build the Lagrange formula like this

[tex]H(x,y,z)= F(x,y,z) + \lambda G(x,y,z)[/tex]

Then take the partial derivatives and equal them to zero

[tex]H_x = 0[/tex]

[tex]H_y = 0[/tex]

[tex]H_z = 0[/tex]

[tex]H_{\lambda} = 0[/tex]

In doing all this you have missed one thing.
 
  • #6


Hmm wow I didn't know you could do it that way. But anyway I figured out what I was doing wrong in regards to my method. I was assuming that x, y, and z were not zero. I can't automatically do that.
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,847
966


The two methods, [itex]\nabla F= \lambda \nabla G[/itex] and [itex]\nabla F+ \lambda\nabla G= 0[/itex] are exactly the same. They just change the sign on [itex]\lambda[/itex] which is irrelevant to the answer.

[itex]F= x^2- y- z[/itex] so [itex]\nabla F= 2x\vec{i}- \vec{j}-\vec{k}[/itex]

[itex]G= x^2- y^2+ z[/itex] so [itex]\nabla G= 2x\vec{i}- 2y\vec{j}+ \vec{k}[/itex]

[itex]\nabla F= \lambda\nabla G[/itex] becomes [itex]2x= 2\lambda x[/itex], [itex]-1= -2\lambda y[/itex] and [itex]-1= \lambda[/itex]. It does NOT follow from that that [itex]\lambda[/itex] is "both 1 and -1". In particular, the first equation, [itex]2\lambda x= 2x[/itex], does NOT mean [itex]\lambda= 1[/itex]. It means either [itex]\lambda= 1[/itex] or x= 0.

The last equation tells you that [itex]\lambda= -1[/itex]. Putting that into the second equation, [itex]-2\lambda y= 2y= -1[/itex] so that y= -1/2. Putting [itex]\lambda= -1[/itex] into the first equation, [itex]2x= 2x[/itex] we get, as I said before, x= 0.

Since we must have [itex]G(x, y, z)= x^2- y^2+ z= 0[/itex], x= 0, y= -1/2 give z= 1/4.
 
  • #8
558
1


Yes, Halls of Ivy is right.
Thank you for the comment.
 

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