# Lagrange Multipliers HELP!

reklaws89
We're suppose to minimize f(x,y,z)=x^2+y^2+z^2 subject to 2x+y+2z=9.

I only ever remember learning how to do f(x,y) would it be the same equation? Thus, f(x,y,$$\lambda$$) = f(x,y) + $$\lambda$$ g(x,y)? Meaning f(x,y,z,$$\lambda$$) = x^2+y^2+z^2 + $$\lambda$$ (2x+y+2z-9) and then continue solving for each variable from there?

Any help?

Homework Helper
"Lagrange Multipliers" doesn't apply to a single equation.

"Lagrange Multipliers" says that at a min or max of f(x,y,z), subject to the additional condition that g(x,y,z)= constant, the two gradients must be parallel: one is a multiple of the other. $\nabla f= \lambda \nabla g$. That is the same as saying $\nabla \left(f(x,y,z)+ \lambda g(x,y,z)\right)= 0$ for all $\lambda$ s what you suggest will work.

Please do not post the same thing more than once. I have merged your two posts.

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reklaws89
so you're saying that if i use f(x,y,z,$$\lambda$$) = x^2+y^2+z^2 + $$\lambda$$ (2x+y+2z-9) that i will be able to find the value of each after taking partial derivatives of each variable. So would I just have three points?

p.s. - Sorry!!

Homework Helper
so you're saying that if i use f(x,y,z,$$\lambda$$) = x^2+y^2+z^2 + $$\lambda$$ (2x+y+2z-9) that i will be able to find the value of each after taking partial derivatives of each variable. So would I just have three points?

p.s. - Sorry!!
Yes, take the gradient of f, set it equal to 0 and the three partial derivatives give you three equations to solve for x, y, z (eliminating $\lambda$, reduces that to 2 equations but you also know 2x+ 2y+ 2z= 9.)

I'm not sure what you mean by "just have three points". There is one point on that plane that minimizes f.

reklaws89
Yea, I misspoke. I meant one point with three numbers meaning a x-value, y-value, and z-value.

reklaws89
Yes, take the gradient of f, set it equal to 0 and the three partial derivatives give you three equations to solve for x, y, z (eliminating $\lambda$, reduces that to 2 equations but you also know 2x+ 2y+ 2z= 9.)

what do you mean eliminating $\lambda$ and reducing to two equations.

f_x = 2x+2$\lambda$
f_y = 2y+$\lambda$
f_z = 2z+2$\lambda$
2_$\lambda$ = 2x+y+2z-9

and from there find the value of each and put them back into f(x,y,z)=x^2+y^+z^2

YOu haven't set them equal to 0 yet! If $2x+ 2\lambda= 0$ and $2z+ 2\lambda= 0$, what do you get when you eliminate $\lambda$?
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