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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
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;

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

-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!

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/
```
Associated Topics:
College Calculus
College Definitions
High School Calculus
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