Optimizing Multivariable Functions with Lagrange Multipliers

[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?

 

[HallsofIvy]

"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.

 

[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!

 

[HallsofIvy]

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

 

[HallsofIvy]

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?