How do I solve a Lagrange Multiplier problem with a given constraint?

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In summary, the conversation discussed using Lagrange Multipliers to solve a problem involving the function f(x,y)=x^2-y^2 and the constraint x^2+y^2=1. The process involves setting the partial derivatives of f + λg to zero and solving for x and y, while also taking the derivative with respect to λ. The constraint can be written as any constant, but for simplicity, it is often written as zero. The process was applied to another problem, F(X)=x^2y and G(X)=x^2+2y^2=6, to find solutions that satisfy all three equations.
  • #1
Arden1528
We have started to do Lagrange Multi. in my class and my book has a very short section on how to solve these. I was wondering if someone couls help.
The problem is f(x,y)=x^2-y^2 with the constraint x^2+y^2=1.
I have found the partial derv. but I am not sure on what else to do. Any help would be sweet, later.
 
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  • #2
If you have a function f(x,y), and a constraint g(x,y) = 0, then to find the constrained extrema you set the partial derivatives of f + λg to zero, and solve for x and y. Remember to take partial derivatives not only with respect to x and y, but also with respect to λ; otherwise, you won't impose the constraint.
 
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  • #3
In my problem I have the constraint equal to 1, should I adjust it and have it equal to 0?
 
  • #4
Originally posted by Arden1528
In my problem I have the constraint equal to 1, should I adjust it and have it equal to 0?

It's traditional to write the constant as zero, but it really doesn't matter; it will work with any constraint g(x,y) = constant, because the constant goes away after you take the derivatives.
 
  • #5
So then for my problem I would get something like

the partial of X: 2X+(lag. symbol)2X=0
thus getting (Lag. Symbol)= -1
 
  • #6
Originally posted by Arden1528
the partial of X: 2X+(lag. symbol)2X=0
thus getting (Lag. Symbol)= -1

Either that, or x=0.
 
  • #7
Then take that 1 and 0 and plug them into the orig. equation of F(X)
and get (+-1,0) or (0,+-1) or (0,0). Then I have the conditions to make this either a max or min. Value. Is this correct?
 
  • #8
Originally posted by Arden1528
Then take that 1 and 0 and plug them into the orig. equation of F(X)
and get (+-1,0) or (0,+-1) or (0,0).

Almost; (0,0) is not a solution of the constraint. Otherwise, you're right.
 
  • #9
Alright, then for my next problem I have
F(X);x^2y and G(X);x^2+2y^2=6

The partials of x I get
2xy+(Lang.)2x=0, giving me (Lang.)=y, can this be true?
 
  • #10
That's one solution (again, another is x=0), but you'll have to impose the other derivative constraints too, and you'll find that, in that case, they restrict what y can be.
 
  • #11
I have thought about the three equations
Partial x
2xy=(Lang.)2x
Partial y
x^2=(Lang.)4y
and
x^2+2y^2=6
and am looking for numbers that satisfy all equations. So I would get something like...
 

1. What is the Lagrange multiplier method?

The Lagrange multiplier method is a mathematical technique used to find the maximum or minimum value of a function subject to a set of constraints. It involves adding a parameter, known as the Lagrange multiplier, to the objective function and setting its derivative equal to zero.

2. When is the Lagrange multiplier method used?

The Lagrange multiplier method is used when solving optimization problems where the objective function is subject to one or more constraints. It is commonly used in economics, engineering, and physics to find the optimal solution in situations with limited resources or restrictions.

3. How does the Lagrange multiplier method work?

To use the Lagrange multiplier method, the objective function and constraints must be written in the form of equations. The Lagrange multiplier is then added to the objective function and the resulting equation is solved by taking the derivative and setting it equal to zero. This will give the values for the variables that maximize or minimize the function.

4. What are the advantages of using the Lagrange multiplier method?

The Lagrange multiplier method offers several advantages, including its ability to handle multiple constraints and its flexibility in handling different types of functions. It also provides a systematic approach to solving optimization problems, making it easier to understand and apply.

5. Are there any limitations to the Lagrange multiplier method?

One limitation of the Lagrange multiplier method is that it can only be used for optimization problems involving continuous functions. It also requires the constraints to be differentiable, which may not always be the case in real-world scenarios. Additionally, the Lagrange multiplier method may not always give a global optimum, but rather a local optimum.

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