Lagrange multipliers with two constraintsby aclotm81 Tags: constraints, lagrange, lagrange multiplier 

#1
Dec1009, 02:59 PM

P: 2

1. The problem statement, all variables and given/known data
By using the Lagrange multipliers find the extrema of the following function: f(x,y)=x+y subject to the constraints: x^{2}+y^{2}+z^{2}=1 y+z=1 2. The attempt at a solution Using lambda = 1/(2x) I got x=yz and y=1z plugging that into the first constraint, I got: 6y^26y+1=0 which makes y=0.5+(3^{1/2}/6) I got the same thing when solving for z, which means x=0 and lambda = infinity, which doesn't make sense. 



#2
Dec1009, 04:07 PM

Sci Advisor
HW Helper
Thanks
P: 25,165

You've also got z=1y. So if you choose the root y=(3+sqrt(3))/2 you have to choose z=(3sqrt(3))/2 not the other root for z. You can't mix and match any two roots with each other.




#3
Dec1009, 04:31 PM

P: 2

Ah, I forgot to distribute the negative! I hate when that happens...it's all worked out now, thanks a lot!



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