Chap14_Sec8 - Lagrange Multiplier.ppt

PARTIAL DERIVATIVES
In Example 6 in Section 14.7, we maximized
a volume function V = xyz subject to
the constraint 2xz + 2yz + xy = 12—which
expressed the side condition that the surface
area was 12 m2.

In this section, we present Lagrange’s method
for maximizing or minimizing a general
function f(x, y, z) subject to a constraint
(or side condition) of the form g(x, y, z) = k.
PARTIAL DERIVATIVES

14.8
Lagrange Multipliers
In this section, we will learn about:
Lagrange multipliers for two and three variables,
and given one and two constraints.
PARTIAL DERIVATIVES

LAGRANGE MULTIPLIERS
It’s easier to explain the geometric basis
of Lagrange’s method for functions of two
variables.

So, we start by trying to find the extreme
values of f(x, y) subject to a constraint of
the form g(x, y) = k.
 In other words, we seek the extreme values
of f(x, y) when the point (x, y) is restricted to lie
on the level curve g(x, y) = k.
LAGRANGE MULTIPLIERS—TWO VARIABLES

The figure shows this curve together with
several level curves of f.
 These have the
equations f(x, y) = c,
where c = 7, 8, 9,
10, 11

To maximize f(x, y) subject to g(x, y) = k
is to find:
 The largest value of c
such that the level
curve f(x, y) = c
intersects g(x, y) = k.

It appears that this happens when these
curves just touch each other—that is, when
they have a common tangent line.
 Otherwise, the value
of c could be
increased further.

This means that the normal lines at
the point (x0 , y0) where they touch are
identical.
 So the gradient vectors are parallel.
 That is,
for some scalar λ.
0 0
0 0
( , ) ( , )
f x y g x y

  

This kind of argument also applies to the
problem of finding the extreme values of
f(x, y, z) subject to the constraint g(x, y, z) = k.
 Thus, the point (x, y, z) is restricted to lie on
the level surface S with equation g(x, y, z) = k.
LAGRANGE MULTIPLIERS—THREE VARIABLES

Instead of the level curves in the previous
figure, we consider the level surfaces
f(x, y, z) = c.
 We argue that, if the maximum value of f is
f(x0, y0, z0) = c, then the level surface f(x, y, z) = c
is tangent to the level surface g(x, y, z) = k.
 So, the corresponding gradient vectors are parallel.

This intuitive argument can be
made precise as follows.

Suppose that a function f has an extreme
value at a point P(x0, y0, z0) on the surface S.
 Then, let C be a curve with vector equation
r(t) = <x(t), y(t), z(t)> that lies on S and passes
through P.

If t0 is the parameter value corresponding to
the point P, then
r(t0) = <x0, y0, z0>
The composite function
h(t) = f(x(t), y(t), z(t))
represents the values that f
takes on the curve C.

f has an extreme value at (x0, y0, z0).
So, it follows that h has an extreme value at t0.
 Thus, h’(t0) = 0.

However, if f is differentiable, we can use
the Chain Rule to write:
 
       
   
   
0
0 0 0 0 0 0 0 0
0 0 0 0
0 0 0 0
0 '
, , ' , , '
, , '
, , '
x y
z
h t
f x y z x t f x y z y t
f x y z z t
f x y z t

 

  r

This shows that the gradient vector
is orthogonal to the tangent
vector r’(t0) to every such curve C.
 
0 0 0
, ,
f x y z


However, we already know from Section 14.6
that the gradient vector of g, ,
is also orthogonal to r’(t0) for every such
curve.
 See Equation 18 from Section 6.
 This means that the gradient vectors
and must be parallel.
 
0 0 0
, ,
g x y z

 
0 0 0
, ,
f x y z

 
0 0 0
, ,
g x y z


Therefore, if , there is
a number λ such that:
 The number λ in the equation is called
a Lagrange multiplier.
 The procedure based on Equation 1 is as follows.
 
0 0 0
, , 0
g x y z
 
   
0 0 0 0 0 0
, , , ,
f x y z g x y z

  
LAGRANGE MULTIPLIER Equation 1

LAGRANGE MULTIPLIERS—METHOD
To find the maximum and minimum values
of f(x, y, z) subject to the constraint
g(x, y, z) = k [assuming that these extreme
values exist and on the surface
g(x, y, z) = k], we proceed as follows.
g
  0

a. Find all values of x, y, z, and λ such that
and
b. Evaluate f at all the points (x, y, z) that
result from step a.
 The largest of these values is the maximum value of f.
 The smallest is the minimum value of f.
LAGRANGE MULTIPLIERS—METHOD
   
 
, , , ,
, ,
f x y z g x y z
g x y z k

  


In deriving Lagrange’s method, we
assumed that .
 In each of our examples, you can check that
at all points where g(x, y, z) = k.
0
g
 
0
g
 
LAGRANGE’S METHOD

If we write the vector equation
in terms of its components, then the equations
in step a become:
fx = λgx fy = λgy fz = λgz g(x, y, z) = k
 This is a system of four equations in the four unknowns
x, y, z, and λ.
 However, it is not necessary to find explicit values for λ.
LAGRANGE’S METHOD
f g

  

For functions of two variables,
the method of Lagrange multipliers is
similar to the method just described.
LAGRANGE’S METHOD

To find the extreme values of f(x, y) subject
to the constraint g(x, y) = k, we look for values
of x, y, and λ such that:
 This amounts to solving three equations
in three unknowns:
fx = λgx fy = λgy g(x, y) = k
     
, , and ,
f x y g x y g x y k

   
LAGRANGE’S METHOD

Our first illustration of Lagrange’s method
is to reconsider the problem given in
Example 6 in Section 14.7
LAGRANGE’S METHOD

A rectangular box without a lid is to be
made from 12 m2 of cardboard.
 Find the maximum volume of such a box.
LAGRANGE’S METHOD Example 1

As in Example 6 in Section 14.7, we let
x, y, and z be the length, width, and height,
respectively, of the box in meters.
 Then, we wish to maximize V = xyz
subject to the constraint
g(x, y, z) = 2xz + 2yz +xy = 12

Using the method of Lagrange multipliers,
we look for values of x, y, z, and λ
such that:
and ( , , ) 12
V g g x y z

   

This gives the equations
Vx = λgx
Vy = λgy
Vz = λgz
2xz + 2yz + xy = 12

The equations become:
yz = λ(2z + y)
xz = λ(2z + x)
xy = λ(2x + 2y)
2xz + 2yz + xy = 12
LAGRANGE’S METHOD E. g. 1—Eqns. 2-5

There are no general rules for solving
systems of equations.
 Sometimes, some ingenuity is required.

In this example, you might notice that
if we multiply Equation 2 by x, Equation 3 by y,
and Equation 4 by z, then left sides of
the equations will be identical.

Doing so, we have:
xyz = λ(2xz + xy)
xyz = λ(2yz + xy)
xyz = λ(2xz + 2yz)

We observe that λ ≠ 0 because λ = 0
would imply yz = xz = xy = 0 from Equations
2, 3, and 4.
This would contradict Equation 5.

Therefore, from Equations 6 and 7,
we have
2xz + xy = 2yz + xy
which gives xz = yz.
 However, z ≠ 0 (since z = 0 would give V = 0).
 Thus, x = y.

From Equations 7 and 8,
we have
2yz + xy = 2xz + 2yz
which gives 2xz = xy.
 Thus, since x ≠ 0, y = 2z.

If we now put x = y = 2z in Equation 5,
we get:
4z2 + 4z2 + 4z2 = 12
 Since x, y, and z are all positive, we therefore
have z = 1, and so x = 2 and y = 2.
 This agrees with our answer in Section 14.7

Another method for solving the system
of equations 2–5 is to solve each of Equations
2, 3, and 4 for λ and then to equate
the resulting expressions.
LAGRANGE’S METHOD

Find the extreme values of the function
f(x, y) = x2 + 2y2 on the circle x2 + y2 = 1.
 We are asked for the extreme values of f
subject to the constraint
g(x, y) = x2 + y2 = 1

Using Lagrange multipliers, we solve
the equations and g(x, y) = 1.
These can be written as:
fx = λgx
fy = λgy
g(x, y) = 1
f g

  

They can also be written as:
2x = 2xλ
4y = 2yλ
x2 + y2 = 1

From Equation 9, we have
x = 0 or λ = 1
 If x = 0, then Equation 11 gives y = ±1.
 If λ = 1, then y = 0 from Equation 10;
so, then Equation 11 gives x = ±1.

Therefore, f has possible extreme values
at the points
(0, 1), (0, –1), (1, 0), (–1, 0)
 Evaluating f at these four points,
we find that:
f(0, 1) = 2 f(0, –1) = 2 f(1, 0) = 1 f(–1, 0) = 1

Therefore, the maximum value of f
on the circle x2 + y2 = 1 is:
f(0, ±1) = 2
The minimum value is:
f(±1, 0) = 1

Checking with the figure, we see that
these values look reasonable.

LAGRANGE’S METHOD
The geometry behind the use of Lagrange
multipliers in Example 2 is shown here.
 The extreme values
of f(x, y) = x2 + 2y2
correspond to the
level curves that
touch the circle
x2 + y2 = 1

Find the extreme values of
f(x, y) = x2 + 2y2 on the disk x2 + y2 ≤ 1
 According to the procedure in Equation 9
in Section 14.7, we compare the values of f
at the critical points with values at the points
on the boundary.

Since fx = 2x and fy = 4y, the only critical
point is (0, 0).
 We compare the value of f at that point
with the extreme values on the boundary
from Example 2:
f(0, 0) = 0 f(±1, 0) =1 f(0, ±1) = 2

Therefore, the maximum value of f
on the disk x2 + y2 ≤ 1 is:
f(0, ±1) = 2
The minimum value is:
f(0, 0) = 0

Find the points on the sphere
x2 + y2 + z2 = 4 that are closest to
and farthest from the point (3, 1, –1).

The distance from a point (x, y, z) to the point
(3, 1, –1) is:
 However, the algebra is simpler if we
instead maximize and minimize the square
of the distance:
     
2 2 2
3 1 1
d x y z
     
     
2
2 2 2
( , , )
3 1 1
d f x y z
x y z

     

The constraint is that the point (x, y, z)
lies on the sphere, that is,
g(x, y, z) = x2 + y2 + z2
= 4

According to the method of Lagrange
multipliers, we solve:
, 4
f g g

   

LAGRANGE’S METHOD
That gives:
2(x – 3) = 2xλ
2(y – 1) = 2yλ
2(z + 1) = 2zλ
x2 + y2 + z2 = 4
E. g. 4—Eqns. 12-15

The simplest way to solve these equations
is to solve for x, y, and z in terms of λ from
Equations 12, 13, and 14, and then
substitute these values into Equation 15.

From Equation 12, we have:
x – 3 = xλ or x(1 – λ) = 3 or
 Note that 1 – λ ≠ 0 because λ = 1 is impossible
from Equation 12.
3
1
x




Similarly, Equations 13 and 14
give:
1 1
1 1
y z
 
  
 

So, from Equation 15, we have:
 This gives (1 – λ)2 = 11/4, 1 – λ = ± .
 Thus,
   
 
 
2
2 2
2 2 2
1
3 1
4
1 1 1
  

  
  
11/ 2
11
1
2
  

These values of λ then give the corresponding
points (x, y, z):
 It’s easy to see that f has a smaller value
at the first of these points.
6 2 2 6 2 2
, , and , ,
11 11 11 11 11 11
   
  
   
   

Thus, the closest point is:
The farthest is:
 
6/ 11,2/ 11, 2/ 11

 
6/ 11, 2/ 11,2/ 11
 

LAGRANGE’S METHOD
The figure shows the sphere and
the nearest point in Example 4.
 Can you see how to
find the coordinates
of P without using
calculus?

TWO CONSTRAINTS
Suppose now that we want to find
the maximum and minimum values of
a function f(x, y, z) subject to two constraints
(side conditions) of the form g(x, y, z) = k
and h(x, y, z) = c.

TWO CONSTRAINTS
Geometrically, this means:
 We are looking for the extreme values of f
when (x, y, z) is restricted to lie on the curve
of intersection C of the level surfaces
g(x, y, z) = k and
h(x, y, z) = c.

Suppose f has such an extreme value at
a point P(x0, y0, z0).
We know from the beginning of this section
that is orthogonal to C at P.
f

TWO CONSTRAINTS

However, we also know that is orthogonal
to g(x, y, z) = k and is orthogonal to
h(x, y, z) = c.
So, and are both orthogonal to C.
g

h

TWO CONSTRAINTS
g
 h


This means that the gradient vector
is in the plane determined
by and
 We assume that these gradient vectors
are not zero and not parallel.
 
0 0 0
, ,
f x y z

TWO CONSTRAINTS
 
0 0 0
, ,
g x y z
  
0 0 0
, ,
h x y z


So, there are numbers λ and μ
(called Lagrange multipliers)
such that:
 
   
0 0 0
0 0 0 0 0 0
, ,
, , , ,
f x y z
g x y z h x y z
 

   
TWO CONSTRAINTS Equation 16

In this case, Lagrange’s method is to look
for extreme values by solving five equations
in the five unknowns
x, y, z, λ, μ
TWO CONSTRAINTS

These equations are obtained by writing
Equation 16 in terms of its components and
using the constraint equations:
fx = λgx + μhx fy = λgy + μhy fz = λgz + μhz
g(x, y, z) = k h(x, y, z) = c
TWO CONSTRAINTS

Find the maximum value of the function
f(x, y, z) = x + 2y + 3z on the curve of
intersection of the plane x – y + z = 1
and the cylinder x2 + y2 = 1
TWO CONSTRAINTS Example 5

We maximize the given function subject
to the constraints
g(x, y, z) = x – y + z = 1
h(x, y, z) = x2 + y2 = 1

The Lagrange condition is
So, we solve the equations
1 = λ + 2xμ
2 = –λ + 2yμ
3 = λ
x – y + z = 1
x2 + y2 = 1
f g h
 
    
TWO CONSTRAINTS E. g. 5—Eqns. 17-21

Putting λ = 3 (from Equation 19)
in Equation 17, we get 2xμ = –2.
Thus, x = –1/μ.
 Similarly, Equation 18 gives y = 5/(2μ).

Substitution in Equation 21 then
gives:
 Thus,
2 2
1 25
1
4
 
 
2 29
4 , 29 / 2
 
  

Then,
and, from Equation 20,
2/ 29
x 
1
1 7 / 29
z x y
  
 
5/ 29
y  

The corresponding values of f are:
 Hence, the maximum value of f
on the given curve is:
2 5 7
2 3 1 3 29
29 29 29
   
     
   
   
3 29


TWO CONSTRAINTS
The cylinder x2 + y2 = 1
intersects the plane
x – y + z = 1 in an
ellipse.
 Example 5 asks for
the maximum value of f
when (x, y, z) is restricted
to lie on the ellipse.

Chap14_Sec8 - Lagrange Multiplier.ppt

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