Max/Min of f(x,y,z) with Lagrange Multipliers

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Homework Help Overview

The problem involves finding the maximum and minimum values of the function f(x, y, z) = x² - 2y + 2z² using Lagrange multipliers, subject to the constraint x² + y² + z². Participants are exploring the implications of the equations derived from the method of Lagrange multipliers.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the equations resulting from the gradient of the function and the constraint. There is a focus on the interpretation of the Lagrange multiplier λ and whether certain values can be assumed or if cases need to be split. Concerns are raised about the validity of canceling terms in equations.

Discussion Status

The discussion is active, with participants offering insights on the handling of the equations. There is an emphasis on careful treatment of terms in the equations to avoid losing potential solutions. No consensus has been reached on the best approach yet, but guidance has been provided regarding the manipulation of the equations.

Contextual Notes

Participants are navigating the constraints of the problem and the assumptions inherent in the use of Lagrange multipliers. There is an acknowledgment of the potential pitfalls in simplifying the equations without considering all cases.

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Homework Statement


Use Lagrange multipliers to ¯nd the maximum and mini-
mum value(s), if they exist, of
f(x; y; z) = x^2 -2y + 2z^2
subject to the constraint x^2+y^2+z^2

Homework Equations


The Attempt at a Solution



Basically after I find the gradient of the functions I get this.

2x=2x lamda
-2=2y lamda
4z=2z lamda.

One case lamda equals 1 while the other it equals 2. Does this mean that the Lagrange Mult can't be used?
 
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2x*lambda=2x doesn't necessarily mean lambda=1. It might mean x=0. You have to start splitting it into cases.
 


I got it thanks a lot Dick.
 


When you have equations of the form [itex]2x = \lambda \cdot 2x[/itex] , it is dangerous to simply "cancel out" the "x" common to both sides of the equation. By removing the factor, you may be removing a solution to the equation along with it. Generally speaking, one should not divide out a factor unless one is sure it cannot be zero.

It is safer to re-write these equations in the form [itex]2x - \lambda \cdot 2x = 0 \Rightarrow 2x ( 1 - \lambda ) = 0[/itex] and solve the two implied equations from there.
 

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