Lagrange multipliers with two constraints

[aclotm81]

Homework Statement

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=12. 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+-(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.

 

[Dick]

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]

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