# Lagrange Multipliers Question?

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>

<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

hotvette
Homework Helper

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:

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.

hotvette
Homework Helper

I'm referring to the following step:

2x = L(2x) L=1

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

There's a conceptual mistake here:

<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.

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
Homework Helper

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.

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