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Lagrange multi. 
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#1
Nov903, 12:56 PM

P: n/a

We have started to do Lagrange Multi. in my class and my book has a very short section on how to solve these. I was wondering if someone couls help.
The problem is f(x,y)=x^2y^2 with the constraint x^2+y^2=1. I have found the partial derv. but I am not sure on what else to do. Any help would be sweet, later. 


#2
Nov903, 01:21 PM

P: 837

If you have a function f(x,y), and a constraint g(x,y) = 0, then to find the constrained extrema you set the partial derivatives of f + λg to zero, and solve for x and y. Remember to take partial derivatives not only with respect to x and y, but also with respect to λ; otherwise, you won't impose the constraint.



#3
Nov903, 01:37 PM

P: n/a

In my problem I have the constraint equal to 1, should I adjust it and have it equal to 0?



#4
Nov903, 01:38 PM

P: 837

Lagrange multi.



#5
Nov903, 01:48 PM

P: n/a

So then for my problem I would get something like
the partial of X: 2X+(lag. symbol)2X=0 thus getting (Lag. Symbol)= 1 


#6
Nov903, 01:52 PM

P: 837




#7
Nov903, 01:58 PM

P: n/a

Then take that 1 and 0 and plug them into the orig. equation of F(X)
and get (+1,0) or (0,+1) or (0,0). Then I have the conditions to make this either a max or min. Value. Is this correct? 


#8
Nov903, 02:02 PM

P: 837




#9
Nov903, 02:08 PM

P: n/a

Alright, then for my next problem I have
F(X);x^2y and G(X);x^2+2y^2=6 The partials of x I get 2xy+(Lang.)2x=0, giving me (Lang.)=y, can this be true? 


#10
Nov903, 02:15 PM

P: 837

That's one solution (again, another is x=0), but you'll have to impose the other derivative constraints too, and you'll find that, in that case, they restrict what y can be.



#11
Nov903, 02:23 PM

P: n/a

I have thought about the three equations
Partial x 2xy=(Lang.)2x Partial y x^2=(Lang.)4y and x^2+2y^2=6 and am looking for numbers that satisfy all equations. So I would get something like... 


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