Explanation Given information: f ( x , y ) = x y e − x 2 − y 2 Formula used: Second derivative test: If P = ( a , b ) is a critical point of f ( x , y ) , then D ( a , b ) > 0 , f x x ( a , b ) > 0 ⇒ f ( a , b ) is a local minimum point D ( a , b ) > 0 , f x x ( a , b ) < 0 ⇒ f ( a , b ) is a local maximum point D ( a , b ) < 0 ⇒ f ( a , b ) is a saddle point D ( a , b ) < 0 ⇒ test fails Where, D = f x x ( a , b ) f y y ( a , b ) − ( f x y ( a , b ) ) 2 Calculation: To find the critical points of f ( x , y ) = x y e − x 2 − y 2 the formula ∇ f = ( f x , f y ) = ( 0 , 0 ) Thus, ∇ f = ( d ( x y e − x 2 − y 2 ) d x , d ( x y e − x 2 − y 2 ) d y ) = ( 0 , 0 ) ∇ f = ( e − x 2 − y 2 ( 1 − 2 x 2 ) , e − x 2 − y 2 ( 1 − 2 y 2 ) ) = ( 0 , 0 ) ⇒ e − x 2 − y 2 ( 1 − 2 x 2 ) = 0 & e − x 2 − y 2 ( 1 − 2 y 2 ) = 0 ⇒ ( 1 − 2 x 2 ) = 0 & ( 1 − 2 y 2 ) = 0 ⇒ x = ± 1 2 & y = ± 1 2 Thus critical points are ( 1 2 , 1 2 ) , ( − 1 2 , 1 2 ) , ( 1 2 , − 1 2 ) , ( − 1 2 , − 1 2 ) Now to find if they are local minimum, local maximum or saddle points by the use of second derivative test First to find D, find f x x , f y y & f x y D = f x x ( a , b ) f y y ( a , b ) − ( f x y ( a , b ) ) 2 Thus, f x x = d d x f x = d d x ( ( 1 − 2 x 2 ) e − x 2 − y 2 ) = e − x 2 − y 2 ( 4 x 3 − 6 x ) f y y = d d y f y = d d y ( ( 1 − 2 y 2 ) e − x 2 − y 2 ) = e − x 2 − y 2 ( 4 y 3 − 6 y ) , and f x y = d d x f y = d d x ( ( 1 − 2 y 2 ) e − x 2 − y 2 ) = e − x 2 − y 2 ( − 2 x + 2 x y 2 ) Now, check for ( a , b ) = ( 1 2 , 1 2 ) , f x x ( 1 2 , 1 2 ) = e − ( 1 2 ) 2 − ( 1 2 ) 2 ( 4 ( 1 2 ) 3 − 6 ( 1 2 ) ) = e − 1 ( − 4 2 ) ≺ 0 f y y ( 1 2 , 1 2 ) = e − ( 1 2 ) 2 − ( 1 2 ) 2 ( 4 ( 1 2 ) 3 − 6 ( 1 2 ) ) = e − 1 ( − 4 2 ) ≺ 0 f x y ( 1 2 , 1 2 ) = e − ( 1 2 ) 2 − ( 1 2 ) 2 ( − 2 ( 1 2 ) + 2 ( 1 2 ) ( 1 2 ) 2 ) = e − 1 ( − 1 2 ) ≺ e − 1 ( − 4 2 ) ≺ 0 Thus, D = f x x ( 1 2 , 1 2 ) f y y ( 1 2 , 1 2 ) − ( f x y ( 1 2 , 1 2 ) ) 2 D = ( e − 1 ( − 4 2 ) ) 2 − ( e − 1 ( − 1 2 ) ) 2 ≻ 0 As, D > 0 , f x x ( 1 2 , 1 2 ) < 0 thus, function has local maxima at ( 1 2 , 1 2 ) Now for ( a , b ) = ( − 1 2 , 1 2 ) , f x x ( − 1 2 , 1 2 )

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 14.7, Problem 15E

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Find critical points of f(x,y)=xye−x2−y2, then find whether they are local minimum, local maximum or saddle points.

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Chapter 14.7, Problem 14E

Chapter 14.7, Problem 16E

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Chapter 14 Solutions

Calculus: Early Transcendentals

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E

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Find critical points of f ( x , y ) = x y e − x 2 − y 2 , then find whether they are local minimum , local maximum or saddle points.

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Find critical points of f(x,y)=xye−x2−y2, then find whether they are local minimum, local maximum or saddle points.

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Chapter 14 Solutions