Need help understanding Lagrange multipliers at a more fundamental level.

In summary, Lagrange multipliers can be used to solve for extreme values by setting the gradient of the objective function equal to a multiple of the gradient of the constraint function. This method helps determine whether a critical point is a maximum, minimum, or saddle point. Lagrange multipliers have a significant purpose in the theory of partial differential equations, where they can be used to derive Green's functions.
  • #1
smize
78
1
I understand that for Lagrange multipliers,

[itex] ∇f = λ∇g [/itex]

And that you can use this to solve for extreme values.

I have a set of questions because I don't understand these on a basic level.

1. How do you determine whether it is a max, min, or saddle point, especially when you only get one extreme value/critical point.

2. Why does this work? Could someone help paint a picture or better description of why you can find these critical points using Lagrange multipliers?

3. Is there a more significant purpose for Lagrange multipliers?

You may use any problem where you have either [itex] f(x,y) [/itex] with the constraint [itex] g(x,y) = k [/itex] or with [itex] f(x,y,z) [/itex] with the constraint [itex] g(x,y,z) = k [/itex]

Both would be preferred; The former preferred for a basic understanding, the latter for a more complex example.

Any help would be appreciated, I have a quiz and test over it this week.
 
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  • #2
smize said:
I understand that for Lagrange multipliers,

[itex] ∇f = λ∇g [/itex]

And that you can use this to solve for extreme values.

I have a set of questions because I don't understand these on a basic level.

1. How do you determine whether it is a max, min, or saddle point, especially when you only get one extreme value/critical point.
By comparing values of points close by the critical point.

2. Why does this work? Could someone help paint a picture or better description of why you can find these critical points using Lagrange multipliers?
Suppose you want to maximize some function f(x,y,z) without any constraint. The "gradient" of f, [itex]\nabla f[/itex] is a vector pointing in the direction of fastest increase. So, starting at any given point, calculate [itex]\nabla f[/itex] at that point, and move in its direction. Keep doing that until you can no longer do it. The only reason you could "no longer do it" is because the gradient vector does not have a 'direction'- and that would be when [itex]\nabla f= 0[/itex].

Now, suppose we have the constraint g(x,y,z)= constant, so we are constrained to stay on some surface satisfying that equation. We can still caculate [itex]\nabla f[/itex], but if it does not happen to be tangent to the surface, we can not "move in its direction". What we can do is find its projectin tangent to the surface and move in that direction. We can continue doing that until there is no projection of [itex]\nabla f[/itex] tangent to the surface. That will happen if and only if [itex]\nabla f[/itex] is itself normal to the surface. But, of course, for any surface g(x,y,z)= constant, [itex]\nabla g[/itex] is perpendicular to the surface so we are saying that [itex]\nabla f[/itex] and [itex]\nabla g[/itex] are both perpendicular to the surface- they are parallel and so one is a multiple of the other: [itex]\nabla f= \lambda\nabla g[/itex] for some constant, [itex]\lambda[/itex].

3. Is there a more significant purpose for Lagrange multipliers?
Not sure what you mean by this but in the theory of partial differential equations, "Green's functions" can be derived in a manner similar to lagrange multipliers.

You may use any problem where you have either [itex] f(x,y) [/itex] with the constraint [itex] g(x,y) = k [/itex] or with [itex] f(x,y,z) [/itex] with the constraint [itex] g(x,y,z) = k [/itex]

Both would be preferred; The former preferred for a basic understanding, the latter for a more complex example.

Any help would be appreciated, I have a quiz and test over it this week.
 
  • #3
HallsofIvy said:
By comparing values of points close by the critical point.


Suppose you want to maximize some function f(x,y,z) without any constraint. The "gradient" of f, [itex]\nabla f[/itex] is a vector pointing in the direction of fastest increase. So, starting at any given point, calculate [itex]\nabla f[/itex] at that point, and move in its direction. Keep doing that until you can no longer do it. The only reason you could "no longer do it" is because the gradient vector does not have a 'direction'- and that would be when [itex]\nabla f= 0[/itex].

Now, suppose we have the constraint g(x,y,z)= constant, so we are constrained to stay on some surface satisfying that equation. We can still caculate [itex]\nabla f[/itex], but if it does not happen to be tangent to the surface, we can not "move in its direction". What we can do is find its projectin tangent to the surface and move in that direction. We can continue doing that until there is no projection of [itex]\nabla f[/itex] tangent to the surface. That will happen if and only if [itex]\nabla f[/itex] is itself normal to the surface. But, of course, for any surface g(x,y,z)= constant, [itex]\nabla g[/itex] is perpendicular to the surface so we are saying that [itex]\nabla f[/itex] and [itex]\nabla g[/itex] are both perpendicular to the surface- they are parallel and so one is a multiple of the other: [itex]\nabla f= \lambda\nabla g[/itex] for some constant, [itex]\lambda[/itex].


Not sure what you mean by this but in the theory of partial differential equations, "Green's functions" can be derived in a manner similar to lagrange multipliers.

So, is the max and min found a max and min based on the constraint, and not a regular max/min of [itex] f(x,y) [/itex] or [itex] f(x,y,z) [/itex]?
 
  • #4
Nevermind. I understand now. I spent 2 seconds on the wikipedia page for it, and I finally had that "Oh my God. I get it." moment.
 

1. What are Lagrange multipliers and what do they represent?

Lagrange multipliers are a mathematical tool used in optimization problems to find the maximum or minimum value of a function subject to one or more constraints. They represent the rate of change of the objective function with respect to the constraint variables.

2. How do Lagrange multipliers work?

Lagrange multipliers work by creating a new function, called the Lagrangian, which combines the objective function and the constraints using a set of parameters called multipliers. By taking the partial derivatives of the Lagrangian with respect to both the objective function and the constraints, we can find the optimal values for both.

3. What is the intuition behind Lagrange multipliers?

The intuition behind Lagrange multipliers is that at the optimal solution, the gradient of the objective function must be parallel to the gradient of the constraint function. This means that the slope of the objective function must be the same as the slope of the constraint function at the optimal point.

4. When should I use Lagrange multipliers?

Lagrange multipliers are typically used in situations where we need to optimize a function subject to one or more constraints. This could include problems in economics, physics, engineering, and other fields.

5. Are there any limitations to using Lagrange multipliers?

One limitation of using Lagrange multipliers is that it only works for problems with continuous functions and constraints. Additionally, it may not always be the most efficient method for solving optimization problems, and there may be other techniques or algorithms that are better suited for certain types of problems.

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