Need help understanding Lagrange multipliers at a more fundamental level.

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SUMMARY

The discussion focuses on understanding Lagrange multipliers, specifically the equation ∇f = λ∇g, which is used to find extreme values of functions subject to constraints. Key points include determining whether a critical point is a maximum, minimum, or saddle point by evaluating nearby values, and the geometric interpretation of why Lagrange multipliers work by projecting the gradient of the function onto the constraint surface. Additionally, the discussion touches on the broader applications of Lagrange multipliers in fields such as partial differential equations, particularly in deriving Green's functions.

PREREQUISITES
  • Understanding of gradient vectors and their significance in optimization.
  • Familiarity with constraints in multivariable calculus.
  • Basic knowledge of critical points and their classification (maxima, minima, saddle points).
  • Concept of partial differential equations and their applications.
NEXT STEPS
  • Study the geometric interpretation of gradients and constraints in multivariable calculus.
  • Learn about the classification of critical points in detail.
  • Explore applications of Lagrange multipliers in optimization problems.
  • Investigate the role of Green's functions in partial differential equations.
USEFUL FOR

Students preparing for exams in multivariable calculus, mathematicians interested in optimization techniques, and professionals working with partial differential equations.

smize
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I understand that for Lagrange multipliers,

∇f = λ∇g

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 f(x,y) with the constraint g(x,y) = k or with f(x,y,z) with the constraint g(x,y,z) = k

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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smize said:
I understand that for Lagrange multipliers,

∇f = λ∇g

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, \nabla f is a vector pointing in the direction of fastest increase. So, starting at any given point, calculate \nabla f 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 \nabla f= 0.

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 \nabla f, 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 \nabla f tangent to the surface. That will happen if and only if \nabla f is itself normal to the surface. But, of course, for any surface g(x,y,z)= constant, \nabla g is perpendicular to the surface so we are saying that \nabla f and \nabla g are both perpendicular to the surface- they are parallel and so one is a multiple of the other: \nabla f= \lambda\nabla g for some constant, \lambda.

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 f(x,y) with the constraint g(x,y) = k or with f(x,y,z) with the constraint g(x,y,z) = k

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.
 
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, \nabla f is a vector pointing in the direction of fastest increase. So, starting at any given point, calculate \nabla f 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 \nabla f= 0.

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 \nabla f, 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 \nabla f tangent to the surface. That will happen if and only if \nabla f is itself normal to the surface. But, of course, for any surface g(x,y,z)= constant, \nabla g is perpendicular to the surface so we are saying that \nabla f and \nabla g are both perpendicular to the surface- they are parallel and so one is a multiple of the other: \nabla f= \lambda\nabla g for some constant, \lambda.


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 f(x,y) or f(x,y,z)?
 
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.
 

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