local_extrema

local_extrema - Local Extrema Previously we have taken the...

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L ocal E xtrema Previously we have taken the partial derivative of a function f ( x , y ). But those partial derivatives were themselves functions and so we can take their partial derivatives. Local Maximums and Local Minimums of Functions f ( x , y ) has a local maximum at the point ( x 0 , y 0 ) if f ( x 0 , y 0 ) ¥ f ( x , y ) for all points ( x , y ) near ( x 0 , y 0 ) f ( x , y ) has a local minimum at the point ( x 0 , y 0 ) if f ( x 0 , y 0 ) § f ( x , y ) for all points ( x , y ) near ( x 0 , y 0 ) Recall that the gradient vector points in a direction where the function increases. Suppose that the point ( x 0 , y 0 ) is a local maximum of the function f ( x , y ) which is not on the boundary of the domain. If the vector f ( x 0 , y 0 ) is defined and non-zero, then we can increase f ( x , y ) by moving in the direction of f ( x 0 , y 0 ). But we just said that the point ( x 0 , y 0 ) is a local maximum, thus there is no direction that we can travel in to make the function larger. Hence, we must have that f ( x 0 , y 0 ) = 0 . Similarly, if ( x 0 , y 0 ) were a local minimum, it would follow that f ( x 0 , y 0 ) = 0 as well. Using our results from above, we also have a useful relation between local maxima and local minima to first-order partial derivatives. Local Extrema Theorem If f ( x , y ) has a local maximum or a local minimum at ( x 0 , y 0 ) and the first-order partial derivatives of f ( x , y ) exist there, then f x ( x 0 , y 0 ) = 0 and f y ( x 0 , y 0 ) = 0. This leads us to the definition of a critical point for a function. As is the case with one- variable calculus, we allow for the possibility that a point is neither a local maximum nor a local minimum. Critical Points of a Function A point ( x 0 , y 0 ) is said to be a critical point of the function f ( x , y ) if the gradient, f ( x 0 , y 0 ), is either 0 or undefined. 1
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Example 1: Find and analyze the critical point(s) of f ( x , y ) = x 2 + xy + y 2 . Solution: f x ( x , y ) = 2 x + y and f y ( x , y ) = x + 2 y . Setting these equations equal to 0, we have two equations: 2 x + y = 0 and x + 2 y = 0 We can solve for x and y by multiplying the second equation by -2 and adding the equations. Doing this, we have that -3 y = 0, and so y = 0. Plugging that into the first equation, we see that x = 0. Thus, the only critical point is (0, 0). Notice that f (0, 0) = 0. Thus, if f ( x , y ) is always positive or zero near (0, 0), then we would have that (0, 0) is a local minimum. If f ( x , y ) is always negative or zero near (0, 0), then we would have a local maximum. Looking at a sketch of the graph (Figure 1 below), it would appear that (0, 0, 0) is a local minimum. Figure 1: Sketch of f ( x , y ) = x 2 + xy + y 2 We can confirm our suspicion by completing the square on the function. That is, we have that 2 22 13 (, ) 24 2 f xy x x y y x y y     . Notice that this is a sum of two squares, so it is always greater than or equal to zero. Thus, the critical point is a local (and in fact, global) minimum.
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local_extrema - Local Extrema Previously we have taken the...

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