Lagrange Multipliers
Say we want to find the point on the line      which is
closest to the origin.
  È  
We can minimize      .
Look at the contours (level curves) for      which
are circles of the form      
y = 2x + 5
5
r=3
-2.5
r=1
What we want is the contour which is tangent to the line.
Note that the equation      is just the level curve
for  a b       where   
 a b gives a vector normal to the contour through the point
a b  is a vector normal to the contours     
Therefore the vectors  and  are parallel at this point where
the minimum occurs! This means
   (or    where  is called the
Lagrange multiplier .
  a  b   a b
    a   b  a b 
   and             
    
        a  b          
       and    
Ans: The point closest to the origin is a   b
Check
    
          a  b

   a  b  

                
Theorem
Let  and  have continuous 1st partial derivatives such that  has an
extremum at the point a b on the smooth curve a b  
If  a b   then there exists a constant  such that
 a b  a b
 is called the Lagrange multiplier .
Example:
Find 2 positive numbers  and  whose sum is 60 such that the
product   is as large as possible.
The side condition is      and the function
to be maximized is  a b   
  a  b and   a b
   
   and    
         
a  b       or    
   is obviously not the max    
gives the maximum          
and    The maximum is a bab.
We can even solve more complicated problems.
Example (Optimization with 2 constraints)
The plane        intersects the paraboloid  =  
in an ellipse. Find the point on this ellipse which is closest to the origin.
Find the point on the ellipse which is the farthest from the origin.
ellipse
The function to be minimized is         
There are 3 surfaces to consider.
         
 a   b        
a   b      
At the point where the minimum occurs,  is normal to the
curve of intersection.
Also,  and  are normal to the curve.
If the 3 vectors are normal to the curve at the same point,
then z must lie on the plane determined by  and g
assuming  and  are not parallel .
This means that      
  a   b   a  b   a   b
    
a   b  a  b  a   b
What we are going to arrive at is 5 eq's in 5 unknowns.
    
    
    
      
     
Using 1st eq's
     and      
              
 a  b  a  b     or   
                 

   which is impossible. So   1.

This means that    
       
                    
 a  ba  b        or .
If           ab   and pt. = a     b
If         ab   and pt. = a  b
 a     b        
 a  b        
Ans: The closest point is a  b and the point a     b
is the farthest.