Lagrange Multiplers with Two Constraints

In summary, the trick to finding the extrema of a function with constraints is to add the gradients of the two constraint functions and set it parallel to the function to be maximized. This allows us to find the extrema precisely where the gradients are 0, thus avoiding any mistakes in our calculations due to rounding errors.
  • #1
keemosabi
109
0
Why when doing a Lagrange Multipler with two constraints, why do you add the gradients of the two constriant funcions and set it parallel to the function to be maximized.

http://www.libraryofmath.com/pages/lagrange-multipliers-with-two-parameters/Images/lagrange-multipliers-with-two-parameters_gr_20.gif

If g and h are the two constraint functions, why would you add their gradients?
 
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  • #2
Let's say you are to find the extrema of f(x,y,z) under the constraints g(x,y,z)=0 AND h(x,y,z)=0 (for illustration of the argument I have regarded f,g,h as 3-variable functions, the argument holds equally well for an arbitrarily large number of free variables)

Consider the five-variable function:
[tex]F(x,y,z,\gamma,\mu)=f-\gamma{g}-\mu{h}[/tex]

Note the following:
In the region (x,y,z) where g=h=0, F is identically equal to f. Thus, whatever extrema F might have there, will also be extrema of f!

As with any other function, the critical points of F lies where the total derivative of F equals 0.

This yields, for the partial differentiations of F with respect to (x,y,z) (using [itex]\nabla[/itex] as the (x,y,z)-gradient):
[tex]\nabla{F}=\nabla{f}-\gamma\nabla{g}-\mu\nabla{h}=0[/tex]
The partial differentiations of F with respect to [itex]\gamma[/itex]- and [itex]\mu[/itex] simply yields:
g=0 and h=0.
 
  • #3
Thank you for the reply, but it seems as if something did not come out correctly. Do you think that you could please re-post?
 
  • #4
Yeah, I know, Latex doesn't work!

Here's the gist of it:

For a multivariable function (without constraints), we know that extrema will be where the gradient is 0.

This gives us the trick to find where the extrema must be in the case of constraints!

Now, letting L stand for one of the Lagrangian multiplier, M the other,
consider the function F(x,y,z,L,M)=f(x,y,z)-Lg(x,y,z)-Mh(x,y,z)

Clearly, at the region where g=h=0, F coincides with f, and hence, F's extrema there must equal f's extrema there!

But, by the clever construction of the linear sum of the constraint functions, F's extrema will precisely be located within that region!

The partial derivative of F with respect to L will give the equation g=0 when we require that the gradient of F is to be zero.
Similarly, the partial derivative of F with respect to M will give the equation h=0 when we require that the gradient of F is to be zero.
The partial derivatives of F with respect to x,y and z yields the familiar gradient condition upon f.
 
  • #5
arildno said:
Yeah, I know, Latex doesn't work!

Here's the gist of it:

For a multivariable function (without constraints), we know that extrema will be where the gradient is 0.

This gives us the trick to find where the extrema must be in the case of constraints!

Now, letting L stand for one of the Lagrangian multiplier, M the other,
consider the function F(x,y,z,L,M)=f(x,y,z)-Lg(x,y,z)-Mh(x,y,z)

Clearly, at the region where g=h=0, F coincides with f, and hence, F's extrema there must equal f's extrema there!

But, by the clever construction of the linear sum of the constraint functions, F's extrema will precisely be located within that region!

The partial derivative of F with respect to L will give the equation g=0 when we require that the gradient of F is to be zero.
Similarly, the partial derivative of F with respect to M will give the equation h=0 when we require that the gradient of F is to be zero.
The partial derivatives of F with respect to x,y and z yields the familiar gradient condition upon f.
I think I understand what you're saying. The one thing that I'm wondering is that in my textbook they have a few example problems, so I worked backwards and plugged the answer pairs (x,y,z) into my gradient functions and did not get 0. Shouldn't the gradient be 0 when I have reached an extremum?

Also, is our goal to get the two constraint functions times the Lagrange Multipliers equal to each other, so that they cancel out in the equation? This would leave us with the familiar gradient condition of the gradient of F being equal to the gradient of f.
 

What is the concept of Lagrange Multipliers with Two Constraints?

The concept of Lagrange Multipliers with Two Constraints is a mathematical method used in optimization problems to find the maximum or minimum value of a multivariable function subject to two constraints. It involves finding the values of the independent variables that satisfy both the constraints and the gradient of the function.

Why is the Lagrange Multipliers method useful in optimization problems?

The Lagrange Multipliers method is useful because it allows for the optimization of a function subject to multiple constraints, which may be difficult or impossible to solve using other methods. It also provides a systematic way to find the optimal solution without having to exhaustively search through all possible values.

What are the steps to solve a problem using Lagrange Multipliers with Two Constraints?

The steps to solve a problem using Lagrange Multipliers with Two Constraints are as follows:

  1. Formulate the objective function and the two constraints.
  2. Write the Lagrangian function by adding the products of the Lagrange multipliers and the constraints to the objective function.
  3. Find the partial derivatives of the Lagrangian function with respect to each of the variables and set them equal to zero.
  4. Solve the resulting system of equations to find the values of the variables and the Lagrange multipliers.
  5. Check the solutions to ensure they satisfy the constraints and use the values to find the optimal solution to the original problem.

What are some real-world applications of Lagrange Multipliers with Two Constraints?

Lagrange Multipliers with Two Constraints have various real-world applications, including but not limited to:

  • Maximizing profit in a production process while minimizing costs and satisfying resource limitations.
  • Optimizing the shape of a structure to withstand external forces while minimizing material usage and satisfying design constraints.
  • Maximizing the utility of a consumer while minimizing their budget and satisfying their preferences.

What are the limitations of using Lagrange Multipliers with Two Constraints?

While Lagrange Multipliers with Two Constraints can be a powerful tool for solving optimization problems, there are some limitations to keep in mind:

  • The method may not yield a unique solution, or it may not yield any solution at all.
  • It may be difficult to interpret the values of the Lagrange multipliers, especially in problems with more than two constraints.
  • The method can become computationally intensive for complex problems with many variables and constraints.

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