topics_15_7_Minimum_and_Maximum_Values_Part_I

# topics_15_7_Minimum_and_Maximum_Values_Part_I - 15.7...

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Unformatted text preview: 15.7. Maximum and Minimum Values. Part I. Critical points. Classifying of a critical point as a point of loc max, a point of loc min, or a saddle point. Problem 7/ page 967. Find the loc maximum and loc minimum values; and the saddle points of the function f ( x,y ) = x 4 + y 4- 4 xy + 2 . Plot in an appropriate way in order to verify your result. Solution. (1) We compute the critical points. Obviously, for the given function, the first order partial derivatives exist at any point on the xy- plane. Hence, each critical point ( x,y ) of f must satisfy: f x ( x,y ) = 4 x 3- 4 y = 0 , f y ( x,y ) = 4 y 3- 4 x = 0 or x 3 = y and y 3 = x . If we subtract both equalities we obtain x 3- y 3 = y- x ⇒ ( x- y )( x 2 + xy + y 2 ) = y- x ⇒ ( x- y )( x 2 + xy + y 2 + 1) = 0 . Now, by using that x 2 + y 2 ≥ 2 | xy | we obtain that x 2 + y 2 + xy +1 ≥ 2 | xy | + xy +1 > 0 we conclude that x 2 + y 2 + xy + 1 6 = 0 and from here each solution of f x ( x,y ) = 0 ,f y ( x,y ) = 0 must satisfy x = y . Then x 3- x = 0 ⇔ x ( x- 1)( x + 1) = 0 and from here all critical points are ( , ) , ( 1 , 1 ) , (- 1 ,- 1 ) . (2) In order to classify the critical points we compute: f xx ( x,y ) = 12 x 2 , f xy ( x,y ) =- 4 , f yy ( x,y ) = 12 y 2 . (a) Take the first critical point ( x,y ) = (0 , 0). We compute f xx (0 , 0) = 0 , f xx (0 , 0) f xy (0 , 0) f yx (0 , 0) f yy (0 , 0) =- 4- 4 =- 16 < and from here ( , ) is a saddle point . Note that we conclude a saddle point only from the negative value of the determinant. Here is the plot of the surface around the saddle point (0 , 0) and the corresponding tangent plane. As you see the surface graph is crossing the tangent plane:and the corresponding tangent plane....
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topics_15_7_Minimum_and_Maximum_Values_Part_I - 15.7...

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