Lagrange multipliers with two constraints


by aclotm81
Tags: constraints, lagrange, lagrange multiplier
aclotm81
aclotm81 is offline
#1
Dec10-09, 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:

x2+y2+z2=1
y+z=1


2. The attempt at a solution
Using lambda = 1/(2x) I got x=y-z and y=1-z
plugging that into the first constraint, I got:
6y^2-6y+1=0 which makes y=0.5+-(31/2/6)

I got the same thing when solving for z, which means x=0 and lambda = infinity, which doesn't make sense.
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Dick
Dick is offline
#2
Dec10-09, 04:07 PM
Sci Advisor
HW Helper
Thanks
P: 25,165
You've also got z=1-y. So if you choose the root y=(3+sqrt(3))/2 you have to choose z=(3-sqrt(3))/2 not the other root for z. You can't mix and match any two roots with each other.
aclotm81
aclotm81 is offline
#3
Dec10-09, 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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