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Lagrange Multipliers

We are going to take a look at another way of optimizing a function subject to given constraint(s). The constraint(s) may be the equation(s) that describe the boundary of a region although in this section we won’t concentrate on those types of problems since this method just requires a general constraint and doesn’t really care where the constraint came from. We want to optimize ( i.e.  find the minimum and maximum value of) a function,  f ( x , y , z ) f ( x , y , z ) , subject to the constraint  g ( x , y , z ) = k g ( x , y , z ) = k . Again, the constraint may be the equation that describes the boundary of a region or it may not be. The process is actually fairly simple, although the work can still be a little overwhelming at times. Notice that the system of equations from the method actually has four equations, we just wrote the system in a simpler form. To see this let’s take the first equation and put in the definition of the gradient vector to see what we get. ⟨ f...
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Computing Extreme Values over a region

If  f ( x , y ) f ( x , y )  is continuous in some closed, bounded set  D D  in  R 2 R 2  then there are points in  D D ,  ( x 1 , y 1 ) ( x 1 , y 1 )  and  ( x 2 , y 2 ) ( x 2 , y 2 )  so that  f ( x 1 , y 1 ) f ( x 1 , y 1 )  is the absolute maximum and  f ( x 2 , y 2 ) f ( x 2 , y 2 )  is the absolute minimum of the function in  D D . 3 Step Way to Find Extreme Values Find all the critical points of the function that lie in the region  D D  and determine the function value at each of these points. Find all extrema of the function on the boundary. This usually involves the Calculus I approach for this work. The largest and smallest values found in the first two steps are the absolute minimum and the absolute maximum of the function. Example Find the absolute minimum and absolute maximum of  f ( x , y ) = 2 x 2 − y 2 + 6 y f ( x , y ) = 2 x 2 − y 2 + 6 y  on the disk of radius 4,  x...
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