The example, continued 

One way to say the relation is that the contour of  and the constraint
curve are tangent to each other at each of those points.   (That is, the two
curves have the same tangent line!)

This observation holds in general for the following reason.  Suppose that we
can parametrize a portion of the constraint curve near a point  where
the restriction of has a critical point as a curve  .  (This is always
possible when  by a more advanced result called the  Implicit
Function Theorem.)   Say  If the restricted function has a
critical point at  , then  

     at   However, by the multivariable Chain Rule we discussed earlier, this implies 

   In other words, the gradient vector of  f  at  is orthogonal to the tangent 

line to the constraint curve at that point.  Since the gradient vector of  g  at  

that point is also orthogonal to the tangent line to the constraint curve,  

we have that  the two vectors  and  
 

(called a Lagrange multiplier )  such that  

Hence we get the ``method of Lagrange multipliers" for solving the constrained 

optimization problem: 

   1.  Set up the Lagrange equations :

        

         

   2.  Solve them simultaneously for  x, y,  (and sometimes λ)  

   3.  By comparing the values of  f  at all of the points 

        found in step 2,  we can determine the
        maximum and minimum values of  f  on the constraint curve.

        (Note: it is often the case that the values of  λ  in the  

        solutions of the Lagrange equations are not relevant,  

        and we will not use them for anything.  Still, there are 

        times when it might be convenient to solve for  λ  
        and use those values to get  x  and  y.

Here is how this ``program'' works in our example above:

   1.  The Lagrange equations are

          The hardest step is almost always actually solving 

        the Lagrange equations.  A good general strategy is 

        to try to use the equations to eliminate variables and 

        reduce down to an equation in one variable (if possible).
        For some equations, though, this may require a good deal 

        of cleverness to carry through.  This is an example where 

        that is the case.  

        Here note that the first two equations allow us to  

        eliminate  λ:  

             

        So  Then we can eliminate the  x  in the constraint  

        equation by substituting for the   

        So

              

>

 

(1.1)

 

>

 

(1.2)

 

So this is a ``hard'' example where there are no rational roots and 

where using a numerical root-finding method is probably needed.
The Maple command for this is the  fsolve  command.  To set up  

for the fact that there are several different  y - values.  We collect 

them in a list: 

>

 

(1.3)

 

    For each of these  y - values, there is exactly one corresponding  x - value, 

    given by       We can have Maple do all the calculations for us like this, using
    the ``for loop'' structure to build up a corresponding list of  x-values.
 

>

 

    Next, we print out the 4 points and the values of  f  at each one:
 

>

 

 

 

 

 

 

 

 

	(1.4)

 

So we see that

   the maximum value of  f  on the constraint curve is  about 3.66,  achieved at the
   third critical point in this list:   [] , and

   the minimum value is about  - 8.14,  achieved at the last one:  .  

Note how this agrees with our graph of the restriction of  f  to the constraint set  

before(!)   The two other points give a local minimum and a local maximum that are not 

global extrema on the constraint curve.