Explanation Given: The following function is given: f ( x , y ) = ( x + 3 y ) e y − x 2 Let P = ( a , b ) be a critical point of f ( x , y ) . Let the second order derivatives f x x , f y y , f x y are continuous near P . Let D be the discriminant of the function f ( x , y ) and is defined as D ( x , y ) = f x x f y y − f x y 2 (a) If D > 0 , then f ( a , b ) is a local minimum or local maximum. (b) If D < 0 , then ( a , b ) is a saddle point. (c) If D = 0 , then the test is inconclusive. (d) If D > 0 and f x x ( a , b ) > 0 , then f ( a , b ) is a local minimum. (e) If D > 0 and f x x ( a , b ) < 0 , then f ( a , b ) is a local maximum. Calculation: Consider the following function: f ( x , y ) = ( x + 3 y ) e y − x 2 Differentiating, f x = ( 1 − 6 x y − 2 x 2 ) e y − x 2 f y = ( x + 3 y + 3 ) e y − x 2 For critical points, putting f x = 0 , f y = 0 ( 1 − 6 x y − 2 x 2 ) e y − x 2 = 0 and ( x + 3 y + 3 ) e y − x 2 = 0 1 − 6 x y − 2 x 2 = 0 , x + 3 y + 3 = 0 ( e y − x 2 ≠ 0 ) 1 − 6 x

3rd Edition

ISBN: 9781319030728

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 14.7, Problem 23E

To determine

The critical points of the function.

Whether they are local minimum, local maximum or saddle points using the Second Derivative Test.

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Chapter 14.7, Problem 22E

Chapter 14.7, Problem 24E

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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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The critical points of the function. Whether they are local minimum , local maximum or saddle points using the Second Derivative Test.

The critical points of the function. Whether they are local minimum, local maximum or saddle points using the Second Derivative Test.

Want to see the full answer?

Chapter 14 Solutions

The critical points of the function.

Whether they are local minimum, local maximum or saddle points using the Second Derivative Test.