# Lagrange Multipliers Question?

• TheSpaceGuy
In summary, the student tried to solve for the minimum and maximum values of the function subject to the given constraint, but made a mistake in their assumption about x and z being nonzero. They were able to solve for the value of lambda using either the \nabla F= \lambda \nabla G or \nabla F+ \lambda\nabla G formulas, but were not able to automatically solve for x and y.
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:

<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.

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

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.

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.

## What is the concept behind Lagrange multipliers?

The concept behind Lagrange multipliers is to find the maximum or minimum value of a function when there are constraints involved. It involves taking the gradient of the function and setting it equal to the gradient of the constraint multiplied by a new variable, known as the Lagrange multiplier.

## What is the purpose of using Lagrange multipliers?

The purpose of using Lagrange multipliers is to find the optimal solution to a constrained optimization problem. It allows for the consideration of constraints without having to rely on trial and error methods or graphical solutions.

## How do you use Lagrange multipliers to solve a problem?

To use Lagrange multipliers to solve a problem, you must first set up the objective function and the constraints. Then, you take the gradient of the objective function and set it equal to the gradient of the constraints multiplied by the Lagrange multiplier. Solve the resulting system of equations to find the optimal solution.

## What are the advantages of using Lagrange multipliers?

One advantage of using Lagrange multipliers is that it simplifies the process of solving constrained optimization problems. It also allows for the consideration of multiple constraints at once, making it a more efficient method compared to trial and error or graphical solutions.

## What are some real-world applications of Lagrange multipliers?

Lagrange multipliers have various real-world applications, such as in economics, where they are used to find the maximum or minimum profit for a company with limited resources. They are also used in physics to determine the path of a particle with constraints, and in engineering to optimize designs under certain constraints.

Replies
8
Views
1K
Replies
2
Views
984
Replies
6
Views
1K
Replies
16
Views
2K
Replies
9
Views
1K
Replies
10
Views
1K
Replies
18
Views
2K
Replies
8
Views
2K
Replies
6
Views
1K
Replies
4
Views
1K