Optimizing Multivariable Functions with Lagrange Multipliers

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Discussion Overview

The discussion revolves around the optimization of a multivariable function using Lagrange multipliers, specifically focusing on minimizing the function f(x,y,z) = x^2 + y^2 + z^2 subject to the constraint 2x + y + 2z = 9. Participants explore the application of Lagrange multipliers in this context, including the formulation of the function with the constraint and the process of finding critical points.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant questions whether the method for f(x,y,z) is similar to that for f(x,y) and proposes using f(x,y,z,λ) = x^2 + y^2 + z^2 + λ(2x + y + 2z - 9).
  • Another participant clarifies that Lagrange multipliers require the gradients of f and g to be parallel, indicating that the gradients must satisfy ∇f = λ∇g.
  • There is a discussion about taking partial derivatives of f with respect to x, y, and z, and whether this will yield three equations to solve for the variables.
  • One participant expresses confusion about the elimination of λ and how it leads to a reduction in the number of equations.
  • Another participant emphasizes the need to set the partial derivatives equal to zero to find critical points.

Areas of Agreement / Disagreement

Participants generally agree on the use of Lagrange multipliers and the need to set partial derivatives to zero, but there is some confusion regarding the elimination of λ and the interpretation of the resulting equations. The discussion remains somewhat unresolved as participants clarify their understanding of the method.

Contextual Notes

There are limitations in the clarity of the steps involved in applying Lagrange multipliers, particularly regarding the elimination of λ and the relationship between the equations derived from the partial derivatives.

reklaws89
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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,[tex]\lambda[/tex]) = f(x,y) + [tex]\lambda[/tex] g(x,y)? Meaning f(x,y,z,[tex]\lambda[/tex]) = x^2+y^2+z^2 + [tex]\lambda[/tex] (2x+y+2z-9) and then continue solving for each variable from there?

Any help?
 
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"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. [itex]\nabla f= \lambda \nabla g[/itex]. That is the same as saying [itex]\nabla \left(f(x,y,z)+ \lambda g(x,y,z)\right)= 0[/itex] for all [itex]\lambda[/itex] 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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so you're saying that if i use f(x,y,z,[tex]\lambda[/tex]) = x^2+y^2+z^2 + [tex]\lambda[/tex] (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!
 
reklaws89 said:
so you're saying that if i use f(x,y,z,[tex]\lambda[/tex]) = x^2+y^2+z^2 + [tex]\lambda[/tex] (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 [itex]\lambda[/itex], 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.
 
Yea, I misspoke. I meant one point with three numbers meaning a x-value, y-value, and z-value.
 
HallsofIvy said:
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 [itex]\lambda[/itex], reduces that to 2 equations but you also know 2x+ 2y+ 2z= 9.)

what do you mean eliminating [itex]\lambda[/itex] and reducing to two equations.

f_x = 2x+2[itex]\lambda[/itex]
f_y = 2y+[itex]\lambda[/itex]
f_z = 2z+2[itex]\lambda[/itex]
2_[itex]\lambda[/itex] = 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 [itex]2x+ 2\lambda= 0[/itex] and [itex]2z+ 2\lambda= 0[/itex], what do you get when you eliminate [itex]\lambda[/itex]?
 

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