Lagrange Multipliers HELP!

  • Thread starter reklaws89
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
HallsofIvy
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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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  • #3
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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!!
 
  • #4
HallsofIvy
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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!!
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.
 
  • #5
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Yea, I misspoke. I meant one point with three numbers meaning a x-value, y-value, and z-value.
 
  • #6
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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
 
  • #7
HallsofIvy
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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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