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.
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
Dick
Dick is offline
#2
Dec10-09, 04:07 PM
Sci Advisor
HW Helper
Thanks
P: 25,173
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!


Register to reply

Related Discussions
Lagrange multipliers Calculus 1
LaGrange Multipliers Calculus & Beyond Homework 3
Lagrange Multipliers. Calculus 0
Lagrange multipliers Calculus & Beyond Homework 1
Lagrange multipliers Calculus 11