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Lagrange Multipliers


Date: 01/08/98 at 08:14:16
From: aleja
Subject: Lagrange Multipliers

Dear Dr. Math,

I have a problem with Lagrange Multipliers. I'd like to know if you
could answer me about this. If yes, I'll send you the problem (it's 
long).

Thank you!
                                                                            
Alejandro Pelorosso
Buenos Aires, Argentina


Date: 01/08/98 at 17:10:48
From: Doctor Anthony
Subject: Re: Lagrange Multipliers

These are notes I sent to another enquirer about Lagrange Multipliers; 
they may answer the questions you have about the topic.

This method is best illustrated by an example. I shall look for the 
stationary values of f(x,y) = 2x^2 + 3y^2  subject to the constraint 
2x + y = 1.

The general problem is to find stationary points of f(x,y) subject to 
constraint g(x,y) = 0  [Note that the constraint must be written in 
this form]. So for our problem g(x.y) = 2x + y - 1.

At stationary points of f(x,y) we have

 df = part(df/dx)*dx + part(df/dy)*dy = 0

This implies that the vector [part(df/dx), part(df/dy)] is
perpendicular to the vector [dx, dy]

Since g(x,y) = 0 we can write

 dg = part(dg/dx)*dx + part(dg/dy)*dy = 0

Thus the vector [part(dg/dx), part(dg/dy)] is also perpendicular to 
the vector [dx, dy]. This implies that the vector [part(df/dx), 
part(df/dy)] is parallel to the vector [part(dg/dx), part(dg/dy)] and 
that we can find a number 'k' such that

 [part(df/dx), part(df/dy)] - k[part(dg/dx), part(dg/dy)]
                                     = [0, 0]

This can be summarized by writing

      phi(x,y) = f(x,y) - kg(x,y)

Then f(x,y) will have a stationary point subject to constraint 
g(x,y) = 0 when part(d(phi)/dx) = 0, part(d(phi)/dy) = 0 and 
g(x,y) = 0

This gives three equations to find x, y and k.

k is called the Lagrange multiplier and phi is called the auxiliary 
function.

Applying these ideas to our problem, we have

  f(x,y) = 2x^2 + 3y^2  and g(x,y) = 2x + y - 1

The auxiliary function is

  phi(x,y) = f(x,y) - kg(x,y)
           = 2x^2 + 3y^2 - k(2x + y -1)

Then:

  part(d(phi)/dx) = 4x - 2k = 0       (1)

  part(d(phi)/dy) = 6y - k = 0        (2)

           g(x,y) = 2x + y -1 = 0     (3)

Solving (1), (2) and (3) for k, x and y we get

  k = 6/7   x = 3/7    y = 1/7

So at the stationary point x = 3/7  y = 1/7  we have

f(x,y) = 2*(3/7)^2 + 3*(1/7)^2
       = 18/49 + 3/49
       = 21/49
       = 3/7

To determine the nature of the stationary point you look at 
neighbouring points to see if f(x,y) has increased or decreased or not 
changed (saddle point).

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 01/08/98 at 21:52:00
From: aleja
Subject: Re: Lagrange Multipliers

Hello again, Dr. Math!

I asked today about Lagrange Multipliers, and was surprised at your 
quick answer giving me theory. Thanks!

However, I can't solve this problem. It was given on an exam :( Here 
it is:

Find the points of the function g(x,y) = 1/xy nearest to the (0,0,0) 
point.

Using Lagrange Multipliers, and the distance formula is x^2+y^2+z^2

Thank you very much! 

Alejandro Pelorosso
Buenos Aires, Argentina.


Date: 01/09/98 at 08:36:34
From: Doctor Anthony
Subject: Re: Lagrange Multipliers

Geometrically I will consider a surface  z = g(x,y), and we require 
the coordinates of points on this surface that are closest to (0,0,0).

We can write the equation of the surface in the form  1/(xy) - z = 0  
(this is the constraint).

and we require to minimize  f(x,y,z) = x^2 + y^2 + z^2

The auxiliary equation is:

 phi(x,y,z) = f(x,y,z) - L(1/(xy) - z)  where L = Lagrange multiplier.

            = x^2 + y^2 + z^2 - L(1/(xy) - z)

We then find the coordinates (x,y,z) of the minimum point and L, the 
Lagrange multiplier from the equations:

  part(d(phi)/dx) = 0     part(d(phi)/dy) = 0   part(d(phi)/dz) = 0

and the equation  1/(xy) - z = 0

  (1) 2x + L/(y.x^2) = 0

  (2) 2y + L/(x.y^2) = 0

  (3) 2z + L = 0

  (4) 1/(xy) - z = 0

Fairly trivially, these equations are satisfied by x=y=z = 1  and 
L = -2

So a point of nearest approach to (0,0,0) of the surface z = 1/(xy) 
has coordinates (1,1,1)

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 01/10/98 at 09:55:26
From: aleja
Subject: Re: Lagrange Multipliers

Please forgive my bothering you again, but my real problem on this
point is to solve the nonlinear system. I can't understand your answer 
about the (1,1,1) point. Could you please explain how you did it?

I have solved several systems on other problems but sometimes I can't
get the x,y,z, and L values, so I would appreciate it if you could
explain all of that to me.

Thanks again, and greetings!
Alejandro


Date: 01/10/98 at 11:47:28
From: Doctor Anthony
Subject: Re: Lagrange Multipliers

I will go through the method for finding x, y, z and L from the four 
equations shown below.

   (1)  2x + L/(y.x^2) = 0
  
   (2)  2y + L/(x.y^2) = 0
  
   (3)  2z + L = 0
  
   (4)  1/(xy) - z = 0


  From (3) z = -L/2  and then from (4)  

        1/(xy) + L/2 = 0

                 L/2 = - 1/(xy)

                   L = -2/(xy)

Now using this value of L, equations (1) and (2) can be written

  2x - 2/(y^2.x^3) = 0  ->   x = 1/(y^2.x^3)     x^4 = 1/y^2 ... (5)

  2y - 2/(x^2.y^3) = 0  ->     y = 1/(x^2.y^3)     y^4 = 1/x^2 ... (6)

Taking square roots of (5) and (6)

               x^2 =  1/y   from this  y = 1/x^2  .... (7) 

 and from (6)  y^2 =  1/x  ..... (8)

 but from (5)  y^2 =  1/x^4  ..... (9)

and from (8) and (9)  x^4 = x

                      x^3 = 1   

so x = 1, then from(8) y=1, and from (4), z=1  and from (3) L = -2.
                       
-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
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