Can Lagrange Multipliers solve optimization problems with multiple constraints?

In summary: f(x,y,z,t,w,u,v,w) = (x-u)*(y-v)*(z-t)*(w-u+v)subject to the constraints:x/2+y+3z+3t+2.5w<=301800<=130x+150y+200z+70t+110a<=3000
  • #1
lynxman72
16
0
Hi all, I was wondering how to go about solving an optimization problem for a function f(x,y,z) where the two constraint equations are given by:

a is less than or equal to g(x,y,z) is less than or equal to b
(a and b are two distinct numbers)
h(x,y,z) is less than or equal to c
(c is another number distinct from a and b)

Can Lagrange Multipliers solve this problem? In other words, is there some trick to put the constraint equation in the standard form for which Lagrange Multipliers works? Any help is appreciated. Thanks
 
Physics news on Phys.org
  • #2
lynxman72 said:
Hi all, I was wondering how to go about solving an optimization problem for a function f(x,y,z) where the two constraint equations are given by:
a is less than or equal to g(x,y,z) is less than or equal to b
(a and b are two distinct numbers)
h(x,y,z) is less than or equal to c
(c is another number distinct from a and b)
Can Lagrange Multipliers solve this problem? In other words, is there some trick to put the constraint equation in the standard form for which Lagrange Multipliers works? Any help is appreciated. Thanks


When you have one constraint, you write it as (let p be the lambda thing or whatever you use)...

Note: f_x is the partial derivative with respect to x.

f_x (x,y,z) = p * g_x (x,y,z)
f_y (x,y,z) = p * g_y (x,y,z)
f_z (x,y,z) = p * g_z (x,y,z)
g(x,y,z) = k

... where k is the constraint.

For multiple constraints, simply write as (where w is your lambda_2)...

f_x (x,y,z) = p * g_x (x,y,z) + w * h_x (x,y,z)
f_y (x,y,z) = p * g_y (x,y,z) + w * h_y (x,y,z)
f_z (x,y,z) = p * g_z (x,y,z) + w * h_z (x,y,z)
g(x,y,z) = k
h(x,y,z) = j

...where k and j are the constraints.

See the pattern here? You can add as many as you want, but I do know that it's not fun solving, for most of them anyways. :tongue:

If you have a <= or >= as a constraint, you simply find the critical points within the boundaries, evaluate them, and evaluate all the points on the boundary, and then take the largest value. This is basically the same thing as finding a maximum/minimum.
 
  • #3
Jason, thanks for the response. I understand how to use the Lagrange multipliers for an equality constraint. I think I didn't describe my problem clearly, the inequality constraints are on the variables, not on the function itself...here is my specific problem: the function to be maximized is

f(x,y,z,t,w)=ln((y^2-x^2)(z^2-t^2)w^3))+.8x-1.2y-20z/17+14t/17-w^3/pi^3

subject to the constraints:
x/2+y+3z+3t+2.5w<=30
1800<=130x+150y+200z+70t+110a<=3000

I computed all of the partials (for any values of the variables) and set them equal to zero and found (2,3,7/3,10/3/pi) to be the only critical point but this gives a value of -3 which doesn't make much sense for the problem I'm working with, so I don't think I went about it right...
 
  • #4
lynxman72 said:
Jason, thanks for the response. I understand how to use the Lagrange multipliers for an equality constraint. I think I didn't describe my problem clearly, the inequality constraints are on the variables, not on the function itself...here is my specific problem: the function to be maximized is
f(x,y,z,t,w)=ln((y^2-x^2)(z^2-t^2)w^3))+.8x-1.2y-20z/17+14t/17-w^3/pi^3
subject to the constraints:
x/2+y+3z+3t+2.5w<=30
1800<=130x+150y+200z+70t+110a<=3000
I computed all of the partials (for any values of the variables) and set them equal to zero and found (2,3,7/3,10/3/pi) to be the only critical point but this gives a value of -3 which doesn't make much sense for the problem I'm working with, so I don't think I went about it right...

Did you check the values around the boundaries as well?

You must check those at the critical points AND those on the boundary, and then take the maximum value.
 
  • #5
That's my problem, I don't understand how to do that; I have two constraint equations for five variables and so I don't know how to figure out the boundaries of each variable...
 
  • #6
I tihnk the thing you need are called slack variables. google for details about them cos it's been too long since i did anything them with them to know i'll get it right.
 
  • #7
http://www.mpri.lsu.edu/textbook/Chapter2-b.htm#Inequality

Here's one using Lagrange Multiplies with inequality constraints like you are talking about.
 
  • #8
Thanks for the link. Where I'm confused is what to do about having both an upper and lower bound as my constraints, if it was just upper bound I see what to do but I don't know how to handle both.
 

What is the concept of Lagrange multipliers?

Lagrange multipliers are a mathematical tool used to optimize a function subject to one or more constraints. It allows for the finding of extreme values, such as maximum or minimum, of a function while taking into account any constraints that may apply.

How do Lagrange multipliers work?

To use Lagrange multipliers, you first set up the function that you want to optimize, called the objective function, and the constraints that must be satisfied. The Lagrange multiplier is then introduced to the equation, and the partial derivatives of the objective function and the constraints are set equal to each other. This results in a system of equations that can be solved to find the optimal solution.

What are some real-world applications of Lagrange multipliers?

Lagrange multipliers have a wide range of applications in various fields, including physics, economics, and engineering. They can be used to optimize production processes, determine the most efficient use of resources, and solve problems with multiple variables and constraints.

What are the limitations of using Lagrange multipliers?

One limitation of Lagrange multipliers is that they only work for differentiable functions. Additionally, they may not always produce a unique solution, and the process can be time-consuming for complex problems. In some cases, alternative methods may be more efficient.

Can Lagrange multipliers be extended to handle problems with more than one constraint?

Yes, Lagrange multipliers can be extended to handle multiple constraints. This is done by introducing a Lagrange multiplier for each constraint and setting up a system of equations with all of the partial derivatives of the objective function and constraints. The resulting solutions will be the optimal values that satisfy all of the constraints.

Similar threads

Replies
1
Views
926
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
467
Replies
9
Views
1K
Replies
2
Views
938
Replies
9
Views
2K
  • Classical Physics
Replies
7
Views
876
Replies
9
Views
2K
Replies
1
Views
1K
Back
Top