Lagrange Multipliers Question?

[TheSpaceGuy]

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.

 

[hotvette]

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?

 

[TheSpaceGuy]

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 don't know how to get the values of x , y , and z from this which is what I need.

 

[hotvette]

I'm referring to the following step:

2x = L(2x) L=1

What assumption is implied about x and is the assumption valid?

 

[Quinzio]

There's a conceptual mistake here:

Gradient f = (Lambda) Gradient g

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

You have a function:

F(x,y,z)= x^2 - y - z

and a vinculum which will be equal to zero:

G(x,y,z)= x^2 - y^2 +z

build the Lagrange formula like this

H(x,y,z)= F(x,y,z) + \lambda G(x,y,z)

Then take the partial derivatives and equal them to zero

H_x = 0

H_y = 0

H_z = 0

H_{\lambda} = 0

In doing all this you have missed one thing.

 

[TheSpaceGuy]

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.

 

[HallsofIvy]

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

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

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

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

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

Since we must have G(x, y, z)= x^2- y^2+ z= 0, x= 0, y= -1/2 give z= 1/4.

 

[Quinzio]

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