Maximizing and Minimizing Functions with Lagrange Multipliers

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Discussion Overview

The discussion revolves around finding the global maximum and minimum values of the function z = x² + 2y² constrained to the circle defined by x² + y² = 1. Participants explore the application of Lagrange multipliers and alternative methods to solve the problem, including derivative analysis and substitution.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant outlines their initial approach using Lagrange multipliers but encounters a contradiction in their calculations.
  • Another participant questions the assumption that both x and y must be non-zero, suggesting that this may lead to incorrect conclusions.
  • A participant proposes specific values for x and y (x=0, y=±1 and y=0, x=±1) as potential solutions, indicating that one set may yield the minimum value.
  • There is a suggestion to substitute the constraint into the function and take derivatives to find critical points, although this diverges from the Lagrange multipliers method.
  • A later reply emphasizes the importance of not dividing by variables that could be zero, reinforcing the need to consider cases where either x or y is zero.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions made in the problem, particularly regarding the values of x and y. There is no consensus on the best approach to apply Lagrange multipliers or the validity of the initial calculations.

Contextual Notes

Participants highlight potential limitations in their reasoning, particularly concerning the assumption that both variables can be non-zero simultaneously, which leads to contradictions in the equations derived from the Lagrange multipliers method.

Pepsi24chevy
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Hey guys, i need some help with this problem. It goes as follow: Find the global max and min valves of the fuction z=x^2+2y^2 on the circle x^2+y^2=1. Ok and here is what i have done. I found the derivites and have done (K is the lagrange constant)
2x=2xK K=1
4y=2yK K=2

Then i set them equal and get 1=2? I know this is not right, but my teacher said do it by lagrange multipliers, so any help is appreciated. Thanks.
 
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Why do you assume that both x and y must be different from zero?
 
[tex]\{x=0;y=\pm 1\} \mbox{and} \{y=0;x=\pm 1\}[/tex] should be the solution...I think the second delivers the minimum value...

Daniel.
 
dextercioby said:
[tex]\{x=0;y=\pm 1\} \mbox{and} \{y=0;x=\pm 1\}[/tex] should be the solution...I think the second delivers the minimum value...

Daniel.
From looking at it I can see how you get those values but how would i do this mathamatically? WOuld i just take 1=2=0? Then go back and plug is or what. I am just trying to understand this, thanks for helping me out.
 
Just substitute [tex]x^2+y^2 =1[/tex] into [tex]z= x^2+ y^2+y^2= 1+ y^2[/tex]. Then just take the derivative of this in terms of y. Once you know the y value where the maximum happens just solve for x such that [tex]x^2= 1- y^2[/tex]

Oh and since you are on the unit circle you also need to check the bound [tex]y=\pm 1[/tex]
 
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I got to do this by lagrange multipliers though, that is what is messing me up because I don't see how it can be done by using this method.
 
Pepschevy:
You've got 3 equations in 3 unknowns:

[tex]2x=2xK, 4y=2yK, x^{2}+y^{2}=1[/tex]
Agreed?
Your flaw has been to divide the first equation with 2x, and the second equation with 2y.
This is only possible to do at the same time if neither x or y is 0!

But, furthermore:
You have proven that if we assume this we get 1=K=2, that is, 1=2 a contradiction.

Thus, you must conclude:
Either x or y MUST be zero!
 
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