Explanation Given: The function is f ( x , y ) = ln ( 1 + x 2 + y 2 ) (1) Result used: Second derivative test “Let D ( x , y ) = f x x ( x , y ) f y y ( x , y ) − f 2 x y ( x , y ) . Then, (a) If D ( a , b ) > 0 and f x x < 0 then the function f has a relative maximum at the critical point ( a , b ) . (b) If D ( a , b ) > 0 and f x x > 0 then the function f has a relative minimum at the critical point ( a , b ) . (c) If D ( a , b ) < 0 then the function f has a no relative extrema at the critical point ( a , b ) . The point ( a , b , f ( a , b ) ) is said to be a saddle point. (d) If D ( a , b ) = 0 then there will be no result.” Calculation: Obtain the critical point of the function . The partial derivatives of the given function is computed as follows, Partially differentiate the equation (1) with respect to x . ∂ ∂ x f ( x , y ) = ∂ ∂ x ( ln ( 1 + x 2 + y 2 ) ) = 1 1 + x 2 + y 2 ( 2 x ) ∂ ∂ x f ( x , y ) = 2 x 1 + x 2 + y 2 (2) Partially differentiate the equation (1) with respect to y . ∂ ∂ y f ( x , y ) = ∂ ∂ y ( ln ( 1 + x 2 + y 2 ) ) = 1 1 + x 2 + y 2 ( 2 y ) ∂ ∂ y f ( x , y ) = 2 y 1 + x 2 + y 2 (3) Thus, the partial derivatives are, f x = 2 x 1 + x 2 + y 2 and f y = 2 y 1 + x 2 + y 2 . Find the critical points by solving the equation f x ( x , y ) = 0 and f y ( x , y ) = 0 . The equation (2) written as ( 2 x ) 1 + x 2 + y 2 = 0 . 2 x = 0 ( 1 + x 2 + y 2 ) x = 0 Thus, the value of x is 0. The equation (3) can be written as ( 2 y ) 1 + x 2 + y 2 = 0 . 2 y = 0 ( 1 + x 2 + y 2 ) y = 0 Thus, the value of y is 0. Therefore, the critical point of the given function is ( 0 , 0 ) . Partially differentiate the equation (2) with respect to x . ∂ 2 ∂ x 2 f ( x , y ) = ∂ ∂ x ( ( 2 x ) 1 + x 2 + y 2 ) (4) Use the mention formula ∂ u ∂ v = v ∂ ∂ u − u ∂ ∂ v v 2 , to obtain the value of ∂ 2 ∂ x 2 f ( x , y ) , Here v = 1 + x 2 + y 2 and u = 2 x . The equation (4) can be written as follows, ∂ 2 ∂ x 2 f ( x , y ) = ( 1 + x 2 + y 2 ) ( 2 ) − ( 2 x ) ( 2 x ) ( 1 + x 2 + y 2 ) 2 f x x = 2 + 2 x 2 + 2 y 2 − 4 x 2 ( 1 + x 2 + y 2 ) 2 = 2 ( 1 − x 2 + y 2 ) ( 1 + x 2 + y 2 ) 2 Partially differentiate the equation (3) with respect to y . ∂ 2 ∂ y 2 f ( x , y ) = ∂ ∂ y ( ( 2 y ) 1 + x 2 + y 2 ) (5) Use the mentioned formula ∂ u ∂ v = v ∂ ∂ u − u ∂ ∂ v v 2 and take the following value of u and v as taken below. Here v = 1 + x 2 + y 2 and u = 2 y . Then the equation (5) becomes as follows, ∂ 2 ∂ x 2 f ( x , y ) = ( 1 + x 2 + y 2 ) ( 2 ) − ( 2 y ) ( 2 y ) ( 1 + x 2 + y 2 ) 2 = 2 + 2 x 2 + 2 y 2 − 4 y 2 ( 1 + x 2 + y 2 ) 2 f y y = 2 ( 1 + x 2 − y 2 ) ( 1 + x 2 + y 2 ) 2 Similarly, partially differentiate the equation (2) with respect to y

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

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ISBN: 9780357465400

Author: Tan

Publisher: CENGAGE L

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Chapter 8.3, Problem 19E

To determine

The critical points and relative extrema of the function; Classify the nature of each point if possible.

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Chapter 8.3, Problem 18E

Chapter 8.3, Problem 20E

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Chapter 8 Solutions

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

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The critical points and relative extrema of the function; Classify the nature of each point if possible.

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The critical points and relative extrema of the function; Classify the nature of each point if possible.

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Chapter 8 Solutions