Lagrange Multipliers
To find the maximum or minimum of f (x , y , z )
under the condition g (x , y , z )  0,
we solve the system of equations:
f (x ,, y , z )   . g (x , y , z )

 g (x , y , z )  0
To find the maximum or minimum of f (x , y , z )
under the conditions g (x , y , z )  0 and h (x , y , z )  0,
we solve the system of equations:
f (x ,, y , z )   .g (x , y , z )   . h (x , y , z )

 g (x , y , z )  0
h (x , y , z )  0

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Exercise. Find the closesest points to the origion on
the surface y  x  1.
2
2
Solution. We should find the minimum of x 2  y 2  z 2 ,
when y - x  1.So it is enough to find the mininum of
2
2
f (x , y , z )  x  y  z when g (x , y , z )  y - x  1  0.
2
2
2
g (x , y , z )  y 2  x 2  1
f (x , y , z )  x  y  z
2
2
2
2
2
 g ( p )  (2x )i  (2 y ) j
 f ( p )  (2x )i  (2 y ) j  (2z )k
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f ( p )  g ( p )

g ( p )  0

 


(2x )i  (2 y ) j  (2z )k    ( 2x )i  (2 y ) j
 2
2
y

x
1  0


(1)
 2x  2 x  x  0    1
 2 y  2 y  y  0    1
(2)


z  0
2
2

y

x
1  0
(3)

(2)
(3)
  1 
 y  0 
 x 2  1 impossible.
(2) 

1
(2)
(1)
(3)

   1 
 x  0 
 y 2  1  y  1
 the points are (0,1, 0) or
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(0, 1, 0)
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


f (0,1, 0)  f (0, 1, 0)  1.
To see that whether the points (0,1,0) and (0,-1,0)
are minimum of maximum, we choose another point
on the surface y 2  x 2  1. The point ( 3 , 2,1) is on this surface.
Now f ( 3, 2,1)  8  1  f (0,1, 0)  f (0, 1, 0).
So the points (0,1, 0) and (0, 1, 0) are the mimimum points.
Thereis no maximum for x  y  z when y - x  1  0,
becuae there is no restrction for the value of z under the
2
2
2
2
2
condition y - x  1  0.
2
2
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Exercise. Find the maximum of xyz , on the line
of the intersection of the two planes
x  y  z  8 , x  y  z  4.
Solution. f (x , y , z )  xyz , g (x , y , z )  x  y  z  8, h (x , y , z )  x  y  z  4.
We should solve the system of equations:
f (x ,, y , z )   .g (x , y , z )   . h (x , y , z )

 g (x , y , z )  0
h (x , y , z )  0

f (x , y , z )  xyz  f (x , y , z )  ( yz )i  (xz ) j  (xy )k .
g (x , y , z )  x  y  z  4  0  g (x , y , z )  i  j  k .
h (x , y , z )  x  y  z  8  0  h (x , y , z )  i  j  k .
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( yz )i  (xz ) j  (xy )k   (i  j  k )   (i  j  k )

x  y  z  4  0
x  y  z  8  0



(1)
 yz    
xz    
(2) 

(6).
  xz  xy
xy




(3)


x  y  z  4  0
(4) 

  2x  12  x  6
(5) 
x  y  z  8  0
(6)
x  6  6z  6 y  z  y . (7)
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z y
(4)
z  y , x  6  6  y  y  4  0  y  1  z  1.
The point is (6, 1, 1). Note that the point (6,0,-2) is also
on the intersetion of the two planes.
f (6, 1, 1)  6  ( 1)  ( 1)  6  0  6  0  (-2)=f(6,0,-2),
so (6, 1, 1) is the maximum point.
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Exercise.
Solution. Suppose 2x and 2 y and 2z are the length, the width and
the height of the cube. So the vertices of the cube are ( x ,  y ,  z ).
Thus the volume of the cube is
V  2z  2 y  2x  8xyz .
z
y
( x, y)
y
x
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x
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so it is enough to find the maximum of xyz ,
when ( x ) 2  9( y ) 2  ( z ) 2  9
f (x , y , z )  xyz , g (x , y , z )  x  9 y  z  9  0
(1)
 yz  2 x
 xz  18 y
(2)
f ( x, y, z )  g ( x, y, z ) 


(3)
 g ( x, y , z )  0
 xy  2 z
 x 2  9 y 2  z 2  9 (4)
yz
xz
y x
(1), (2) 

 
 9 y 2  x 2 (5)
2 x 18 y
x 9y
xz
xy
z
y
2
2
(2), (3) 


  9y  z
(6)
18 y 2 z
9y z
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2
2
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(5), (6)  z  x
2
2
2
x  x  x 9  0  x   3
2
2
2
 z   3,
1
1
9y  3  y  
8 3
 3 8 3
3
3
2
Maximum Volume
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