Lagrange Multipliers

Lagrange Multipliers

Suppose that we have a function f(x,y) that we want to maximize in the restricted domain g(x,y) = c for some constant c.  Then we can look at the level curves of f and seek the largest level curve that intersects the curve g(x,y) = c.  It is not hard to see that these curves will be tangent.  Hence the gradient vectors will be parallel.

 Theorem Let f(x,y) be differentiable and g(x,y) = c define a smooth curve.  Then the maximum and minimum of f subject to the constraint g occur at the critical points of           F(x,y)  =  f(x,y) - l g(x,y) for some constant l.

Example:

Find the extrema of

f(x,y) = x2 - y2

subject to the constraint

y - x2 = 0

Solution:

We have

F(x,y)  =   (x2 - y2) - l (y - x2

Now take partial derivatives and set them equal to zero:

Fx(x,y)  =  2x - l2x  =  0

Fy(x,y)  =  -2y - l  =  0

This gives us the three equations:

2x = -l2x,    -2y = l(1),     and    y - x2 = 0

the first equation gives us (for x nonzero)

l = -1

Hence the second equation becomes

-2y = -1

so that

y = 1/2

the third equation gives us

1/2 - x2 = 0

Hence

x = /2

For x = 0, we see that y = 0.  Hence the two possible local extrema are

(/2,1/2)    and    (0,0)

Plugging into f(x,y), we see that

f(/2,1/2) = 1/4

and

f(0,0) = 0

Hence 1/4 is the local maximum and 0 is the local minimum.

Example 2

Find the distance from the origin to the surface

xyz = 8

Solution

We minimize

D = x2 + y2 + z2

subject to the constraint

xyz = 8

We have

F(x,y)  =   ( x2 + y2 + z2) - l (xyz)

Now take partial derivatives and set them equal to zero:

Fx(x,y)  =  2x - lyz  =  0

Fy(x,y)  =  2y - lxz  =  0

Fz(x,y)  =  2z - lxy  =  0

Which give the four equations

2x = l yz,        2y = lxz,        2z = lxy,    xyz  =  8

or

l = 2x/yz = 2y/xz = 2z/xy        xyz  =  8

or

2x2 = 2y2 = 2z2        xyz  =  8

Hence

x = ± y = ± z

Plugging into the last equations gives

±x3 = 8    or    x = ±2

We get the points

(2,2,2), (2,-2,-2), (-2,-2,2), (-2,2,-2)

these all have distance from the origin.

Two Constraints

Example:

Maximize

x2 + y2 + z2

on the intersection of the two surfaces:

xyz = 1    and     x2 + y2 + 2z2 = 4

Solution:

When there are two constraints we use

F(x,y,z)  =  f(x,y,z) - ag(x,y,z) + bh(x,y,z)

and find the critical points of F.  We get

F(x,y,z)  =  (x2 + y2 + z2) - a(xyz) - b(x2 + y2 + 2z2)

and take partial derivatives

Fx  =  2x - ayz - b2x
Fy  =  2y - axz - b2y
Fz  =  2z - axy - b4z

which gives the five equations

2x = ayz + 2bx,     2y = axz + 2by,     2z = axy + 4bz,

xyz = 1,   and     x2 + z = 1

Multiply the first equation by x, the second by y and the third by z to get

2x2 = axyz + 2bx2,     2y2 = axyz + 2by2,     2z2 = axyz + 4bz2,

Solving each for axyz gives

axyz = 2x2  - 2bx2 = 2y2  - 2by2 = 2z2  - 4bz2

This gives that

2x2(1 - b) = 2y2(1 - b) = 2z2 (1 - 2b)

This first equality gives

x = ±y

Using the last of the original equations to solve for x2 gives

x2  = 1 - z

Substituting and dividing by 2 gives

(1 - z)(1 - b) = z2 (1 - 2b)

Using xyz  =  1 and substituting gives

±(1 - z)z  =  1

so

z2 - z + 1  =  0    or    z2 + z - 1  =  0

The first has no solution and the second has solutions

z  =  -1/2 ± 1/2

Now

and

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