Explanation Result used: Second Derivative Test: “Suppose the second partial derivatives of f are continuous on a disk with center ( a , b ) , and suppose that f x ( a , b ) = 0 and f y ( a , b ) = 0 (that is ( a , b ) is a critical point of f ). Let D = D ( a , b ) = f x x ( a , b ) f y y ( a , b ) − [ f x y ( a , b ) ] 2 ( a ) If D > 0 and f x x ( a , b ) > 0 , then f ( a , b ) is a local minimum . ( b ) If D > 0 and f x x ( a , b ) < 0 , then f ( a , b ) is a local maximum. ( c ) If D < 0 , then f ( a , b ) is not a local maximum or minimum and it is called a saddle point”. Given: The function is, f ( x , y ) = − ( x 2 − 1 ) 2 − ( x 2 y − x − 1 ) 2 . Calculation: Take the partial derivative in the given function with respect to x and obtain f x . ∂ f ∂ x = ∂ ∂ x [ − ( x 2 − 1 ) 2 − ( x 2 y − x − 1 ) 2 ] = − ∂ ∂ x [ ( x 2 − 1 ) 2 ] − ∂ ∂ x [ ( x 2 y − x − 1 ) 2 ] = − 2 ( x 2 − 1 ) ( 2 x ) − 2 ( x 2 y − x − 1 ) ( 2 x y − 1 ) Thus, ∂ f ∂ x = − 2 ( x 2 − 1 ) ( 2 x ) − 2 ( x 2 y − x − 1 ) ( 2 x y − 1 ) . (1) Take the partial derivative in the given function with respect to y and obtain f y . ∂ f ∂ y = ∂ ∂ y [ − ( x 2 − 1 ) 2 − ( x 2 y − x − 1 ) 2 ] = − ∂ ∂ y [ ( x 2 − 1 ) 2 ] − ∂ ∂ y [ ( x 2 y − x − 1 ) 2 ] = 0 − 2 ( x 2 y − x − 1 ) ( x 2 ) = − 4 x 2 ( x 2 y − x − 1 ) Thus, ∂ f ∂ y = − 2 x 2 ( x 2 y − x − 1 ) . (2) Solve the equations (1) and (2) and find the values of x and y . From the equation (2), − 2 x 2 ( x 2 y − x − 1 ) = 0 ⇒ 2 x 2 = 0 or ( x 2 y − x − 1 ) = 0 ⇒ x = 0 or y = x + 1 x 2 From the result there are two possibilities such as either x = 0 or y = x + 1 x 2 . Suppose x = 0 then the equation (1), f x ( x , y ) = − 2 ( x 2 − 1 ) ( 2 x ) − 2 ( x 2 y − x − 1 ) ( 2 x y − 1 ) f x ( 0 , y ) = − 2 ( 0 − 1 ) ( 2 ( 0 ) ) − 2 ( 0 − 0 − 1 ) ( 2 ( 0 ) y − 1 ) = 0 − 2 ( − 1 ) ( − 1 ) = − 2 Thus, there are no critical points for x = 0 . So, substitute y = x + 1 x 2 in the equation (1), f x ( x , y ) = − 2 ( x 2 − 1 ) ( 2 x ) − 2 ( x 2 y − x − 1 ) ( 2 x y − 1 ) f x ( x , x + 1 x 2 ) = − 4 ( x ) ( x 2 − 1 ) − 2 ( x 2 ( x + 1 x 2 ) − x − 1 ) ( 2 x ( x + 1 x 2 ) − 1 ) = − 4 x ( x 2 − 1 ) − ( 2 x − 2 + 2 x + 2 ) ( 2 x + 2 x − 1 ) = − 4 x ( x 2 − 1 ) Thus, the value of x = ± 1 . Therefore, the critical points of the function f ( x , y ) = − ( x 2 − 1 ) 2 − ( x 2 y − x − 1 ) 2 are ( 1 , 2 ) and ( − 1 , 0 )

2nd Edition

ISBN: 9781337772198

Author: Stewart

Publisher: VST

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Chapter 11.7, Problem 29E

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To show: The function f(x,y)=−(x2−1)2−(x2y−x−1)2 has only two critical points and it has two local maxima on the same points.

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Chapter 11 Solutions

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The function f ( x , y ) = − ( x 2 − 1 ) 2 − ( x 2 y − x − 1 ) 2 has only two critical points and it has two local maxima on the same points.

Solution Summary: The author explains that the function f(x,y)=- (x2-1 ) 2 has only two critical points and it has two local maxima

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To show: The function f(x,y)=−(x2−1)2−(x2y−x−1)2 has only two critical points and it has two local maxima on the same points.

Chapter 11 Solutions