Lagrange Multiplier -> Find the maximum.

In summary, to find the maximum value M of the given function with the given constraints, the Lagrange multiplier method is used. However, due to the presence of sign restrictions, the conditions for a maximum are stricter and additional steps need to be taken to find the correct solution. These include analyzing the unconstrained problem, locating boundary points of the feasible set, and checking for the satisfaction of Karush-Kuhn-Tucker conditions.
  • #1
s3a
818
8
Lagrange Multiplier --> Find the maximum.

Homework Statement


Find the maximum value, M, of the function f(x,y) = x^4 y^9 (7 - x - y)^4 on the region x >= 0, y >= 0, x + y <= 7.

Homework Equations


Lagrange multiplier method and the associated equations.


The Attempt at a Solution


Firstly, my handwriting is ugly because I wrote this for myself before thinking that I would need to post it online but I think it should be legible nonetheless. If it isn't, tell me and I will rewrite it from scratch.

Secondly, the question doesn't force me to use the Lagrange Multiplier method but I chose it thinking it's the best way since it seems like a nice method so if I am wrong in choosing it, tell me. Using the Lagrange Multiplier method, I get M = 0. That's an extremum alright but it's a minimum and not a maximum given the set of constraints.

Any input in helping me figure out what I did wrong would be greatly appreciated!
Thanks in advance!
 

Attachments

  • MyWork.jpg
    MyWork.jpg
    53.5 KB · Views: 571
Physics news on Phys.org
  • #2


s3a said:

Homework Statement


Find the maximum value, M, of the function f(x,y) = x^4 y^9 (7 - x - y)^4 on the region x >= 0, y >= 0, x + y <= 7.

Homework Equations


Lagrange multiplier method and the associated equations.

The Attempt at a Solution


Firstly, my handwriting is ugly because I wrote this for myself before thinking that I would need to post it online but I think it should be legible nonetheless. If it isn't, tell me and I will rewrite it from scratch.

Secondly, the question doesn't force me to use the Lagrange Multiplier method but I chose it thinking it's the best way since it seems like a nice method so if I am wrong in choosing it, tell me. Using the Lagrange Multiplier method, I get M = 0. That's an extremum alright but it's a minimum and not a maximum given the set of constraints.

Any input in helping me figure out what I did wrong would be greatly appreciated!
Thanks in advance!

In problems of this type you cannot always set dL/dx = 0 and dL/dy = 0, due to the presence of sign restrictions x >= 0 and y >= 0. In fact, the conditions (in a MAX problem) are that (i) dL/dx <= 0; (ii) x >= 0; (iii) either dL/dx = 0 or x = 0. Similar conditions hold for y. Also, for constraint g(x,y) <= 0, either g = 0 or the Lagrange multiplier = 0. Because of these either/or restrictions, problems with inequality constraints are harder to deal with than equality-constrained problems. Typically, in small problems done by hand, we first try to gain some insight into where the optimum might lie; for example, if we look first at the unconstrained problem and the max is feasible, we are done (no extra work needed). However, if the unconstrained max lies outside the feasible set, we can try to locate boundary points of the feasible set where the constrained max must lie. Alternatively, we might try to impose the conditions dL/dx = 0, dL/dy = 0 and g = 0 to see what they give us: if we get a feasible solution and the Lagrange multiplier has the correct sign, we are done; otherwise, we must change one or more of our assumptions and try again. Sometimes we make several incorrect guesses before we finally get the right combination, so we might waste a lot of paper! You know you have a candidate for the constrained optimum when you satisfy the so-called Karush-Kuhn-Tucker conditions; see, eg.,
https://netfiles.uiuc.edu/angelia/www/ge330fall09_nlpkkt_l26.pdf or
https://engineering.purdue.edu/ME697Y/KKT.pdf .

RGV
 
Last edited by a moderator:

Related to Lagrange Multiplier -> Find the maximum.

1. What is a Lagrange multiplier?

A Lagrange multiplier is a mathematical tool used in multivariate calculus to optimize a function subject to constraints.

2. How do you find the maximum using Lagrange multipliers?

To find the maximum using Lagrange multipliers, you first set up the Lagrangian function by adding the constraint equations multiplied by a Lagrange multiplier to the original function. Then, you take the partial derivatives of the Lagrangian with respect to all variables and set them equal to zero. Solving this system of equations will give you the values for the variables that will maximize the function.

3. What are the assumptions for using Lagrange multipliers?

The assumptions for using Lagrange multipliers are that the function being optimized is continuous, the constraint equations are differentiable, and the constraint equations are independent.

4. Can Lagrange multipliers be used to find the minimum?

Yes, Lagrange multipliers can be used to find both the maximum and minimum of a function subject to constraints. This is done by setting up the Lagrangian function and solving for the minimum using the same process as finding the maximum.

5. Are there any limitations to using Lagrange multipliers?

One limitation of using Lagrange multipliers is that it can only be used for constrained optimization problems. It cannot be used for unconstrained problems. Additionally, it may not always give the global maximum or minimum, but rather a local maximum or minimum.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
645
  • Calculus and Beyond Homework Help
Replies
16
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
740
Replies
1
Views
891
  • Calculus and Beyond Homework Help
Replies
18
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
971
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
678
  • Calculus and Beyond Homework Help
Replies
10
Views
826
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Back
Top