Explanation Given: The function is f ( x , y ) = 2 y 3 − 3 y 2 − 12 y + 2 x 2 − 6 x + 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, check the values of D ( a , b ) as below. (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 ( 2 y 3 − 3 y 2 − 12 y + 2 x 2 − 6 x + 2 ) = ∂ ∂ x ( 2 y 3 ) − ∂ ∂ x ( 3 y 2 ) − ∂ ∂ x ( 12 y ) + ∂ ∂ x ( 2 x 2 ) − ∂ ∂ x ( 6 x ) + ∂ ∂ x ( 2 ) = 4 x − 6 Thus, ∂ ∂ x f ( x , y ) = 4 x − 6 (2) Compute the partial derivative of the equation (1) with respect to y . ∂ ∂ y f ( x , y ) = ∂ ∂ y ( 2 y 3 − 3 y 2 − 12 y + 2 x 2 − 6 x + 2 ) = 6 y 2 − 6 y − 12 Hence, ∂ ∂ y f ( x , y ) = 6 ( y 2 − y − 2 ) (3) Obtain the critical points by equating f x ( x , y ) = 0 and f y ( x , y ) = 0 . Then the equation (2) becomes, 4 x − 6 = 0 (4) And the equation (3) becomes, 6 ( y 2 − y − 2 ) = 0 y 2 − y − 2 = 0 (5) Obtain the value of x from the equation (4). 4 x = 6 x = 3 2 Obtain the value of y from equation (5). y 2 − y − 2 = 0 ( y + 1 ) ( y − 2 ) = 0 Thus, the values of y are y = − 1 and y = 2 . Therefore, the critical points of the given function are ( 3 2 , − 1 ) and ( 3 2 , 2 ) . Apply the second derivative test : Compute the partial derivative of the equation (2) with respect to x . ∂ 2 ∂ x 2 f ( x , y ) = ∂ ∂ x ( 4 x − 6 ) = 4 Compute the partial derivative of the equation (2) with respect to y. ∂ 2 ∂ x y f ( x , y ) = ∂ ∂ y ( 4 x − 6 ) = 0 Compute the partial derivative of the equation (3) with respect to y . ∂ 2 ∂ y 2 f ( x , y ) = ∂ ∂ y ( 6 y 2 − 6 y − 12 ) = 12 y − 6 = 6 ( 2 y − 1 ) Thus, the second derivatives of the function are f x x = 4 , f y y = 6 ( 2 y − 1 ) and f x y = 0

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Chapter 8.3, Problem 10E

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

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

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