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

In summary, the conversation discusses the use of Lagrange multipliers to find the maximum and minimum values of a function subject to a constraint. The equations for the gradient are provided and it is noted that one should be careful when dividing out a common factor.
  • #1
xdrgnh
417
0

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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  • #2


2x*lambda=2x doesn't necessarily mean lambda=1. It might mean x=0. You have to start splitting it into cases.
 
  • #3


I got it thanks a lot Dick.
 
  • #4


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.
 

What is the purpose of using Lagrange Multipliers in finding the max/min of a function?

Lagrange Multipliers are used to find the maximum or minimum value of a function subject to constraints. This method allows us to solve optimization problems where the constraints are not explicitly given in the form of equations, but rather as a set of conditions.

How does the Lagrange Multiplier method work?

The Lagrange Multiplier method involves finding the critical points of a function f(x,y,z) when the constraint function g(x,y,z) is satisfied. This is done by setting up a system of equations using the Lagrange Multiplier and solving for the critical points.

What are the conditions for using the Lagrange Multiplier method?

The conditions for using the Lagrange Multiplier method are that the function f(x,y,z) and the constraint function g(x,y,z) must be continuous and differentiable, and the constraint function should not be singular at any point.

Can Lagrange Multipliers be used for functions of more than three variables?

Yes, the Lagrange Multiplier method can be extended to functions of more than three variables. The same principles apply, but the equations become more complex.

Are there any limitations to using the Lagrange Multiplier method?

One limitation of the Lagrange Multiplier method is that it may not always yield a global maximum or minimum. It can only find the critical points, which may or may not be the desired extreme values. Additionally, the method may become more computationally intensive for higher dimensional functions.

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