Explanation Given: The following function is given: f ( x , y ) = ( x + y ) ln ( x 2 + y 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 + y ) ln ( x 2 + y 2 ) Differentiating, f x = 2 x ( x + y ) + ( x 2 + y 2 ) ln ( x 2 + y 2 ) x 2 + y 2 f y = 2 y ( x + y ) + ( x 2 + y 2 ) ln ( x 2 + y 2 ) x 2 + y 2 For critical points, putting f x = 0 , f y = 0 2 x ( x + y ) + ( x 2 + y 2 ) ln ( x 2 + y 2 ) x 2 + y 2 = 0 and 2 y ( x + y ) + ( x 2 + y 2 ) ln ( x 2 + y 2 ) x 2 + y 2 = 0 ⇒ x = y = a So, the critical point is ( a , a )

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Chapter 14.7, Problem 20E

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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.

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Chapter 14.7, Problem 19E

Chapter 14.7, Problem 21E

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Chapter 14 Solutions

CALCULUS 4E (HC) W/ ACHIEVE ACCESS

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.

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Chapter 14 Solutions

The critical points of the function.

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