Lagrange Multiplier -> Find the maximum.

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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!
 

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Ray Vickson
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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 [Broken] or
https://engineering.purdue.edu/ME697Y/KKT.pdf .

RGV
 
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