Explanation Given: The function is f ( x , y ) = e x y (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 : Compute the partial derivative of the equation (1) with respect to x . ∂ ∂ x f ( x , y ) = ∂ ∂ x ( e x y ) = e x y ( y ) = y e x y Thus, ∂ ∂ x f ( x , y ) = y e x y (2) Compute the partial derivative of the equation (1) with respect to y . ∂ ∂ y f ( x , y ) = ∂ ∂ y ( e x y ) = e x y ( x ) = x e x y Hence, f ( x , y ) = x e x y (3) Substitute f x ( x , y ) = 0 and f y ( x , y ) = 0 in the equation (2) and equation (3), Then equation (2) becomes, e x y ( y ) = 0 y = 0 Thus, the value of y is y = 0 . And the equation (3) becomes, e x y ( x ) = 0 x = 0 Thus, the value of x = 0 . Therefore, the critical point of the given function is ( 0 , 0 ) . Apply the second derivative test : Compute the partial derivative of the equation (2) with respect to x . ∂ 2 ∂ x 2 f ( x , y ) = ∂ ∂ x ( e x y ( y ) ) = ( y ) e x y ( y ) + e x y ( 0 ) = e x y ( y 2 ) Compute the partial derivative of the equation (2) with respect to y . ∂ ∂ x y f ( x , y ) = ∂ ∂ y ( e x y ( y ) ) = ( y ) e x y ( x ) + e x y ( 1 ) = e x y ( x y ) + e x y Compute the partial derivative of the equation (3) with respect to y . ∂ 2 ∂ y 2 f ( x , y ) = ∂ ∂ y ( e x y ( x ) ) = ( x ) e x y ( x ) + e x y ( 0 ) = e x y ( x ) 2 Therefore, the second derivatives of the function are f x x = e x y ( y ) 2 , f x y = e x y ( x y ) + e x y and f y y = e x y ( x ) 2

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Author: Tan

Publisher: INTER CENG

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Chapter 8.3, Problem 18E

To determine

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

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Chapter 8.3, Problem 17E

Chapter 8.3, Problem 19E

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Take the derivative

Find the critical point of the function. Then use the Second Derivative Test to classify the nature of this point, if possible. (If an answer does not exist, enter DNE.) f(x, у) %3D х? +2ху + 2у2 - 6х + 10у + 8 critical point (х, у) classification ---Select--- Finally, determine the relative extrema of the function. (If an answer does not exist, enter DNE.) relative minimum value relative maximum value

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

Solution Summary: The author explains how to find the critical points and the relative extrema of the function.

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

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Chapter 8 Solutions