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 ( a , b ) = f x x ( a , b ) f y y ( a , b ) − [ f x y ( a , b ) ] 2 ( a ) If D ( a , b ) > 0 and f x x ( a , b ) > 0 , then f ( a , b ) is a local minimum. ( b ) If D ( a , b ) > 0 and f x x ( a , b ) < 0 , then f ( a , b ) is a local maximum. ( c ) If D ( a , b ) < 0 , then f ( a , b ) is a saddle point. ( d ) If D ( a , b ) = 0 , then the test is inconclusive”. Calculation: The given function is, f ( x , y ) = x 4 + y 4 − 16 x y . Take the partial derivative in the given function with respect to x and obtain f x . ∂ f ∂ x = ∂ ∂ x ( x 4 + y 4 − 16 x y ) = ∂ ∂ x ( x 4 ) + ∂ ∂ x ( y 4 ) − ∂ ∂ x ( 16 x y ) = 4 x 3 + 0 − 16 y ( 1 ) = 4 x 3 − 16 y Thus, ∂ f ∂ x = 4 x 3 − 16 y (1) Take the partial derivative in the given function with respect to y and obtain f y . ∂ f ∂ y = ∂ ∂ y ( x 4 + y 4 − 16 x y ) = ∂ ∂ y ( x 4 ) + ∂ ∂ y ( y 4 ) − ∂ ∂ y ( 16 x y ) = 0 + 4 y 3 − 16 x ( 1 ) = 4 y 3 − 16 x Thus, ∂ f ∂ y = 4 y 3 − 16 x (2) Set the above partial derivatives to 0 and find the values of x and y . From the equation (1), 4 x 3 − 16 y = 0 4 x 3 = 16 y y = 4 16 x 3 y = x 3 4 Substitute y = x 3 4 in the equation (2), 4 ( x 3 4 ) 3 − 16 x = 0 4 x 9 64 − 16 x = 0 x ( x 8 − 256 ) = 0 x = 0 , x = ± 2 Substitute x = 0 in y = x 3 4 , y = ( 0 ) 3 4 = 0 Substitute x = 2 in y = x 3 4 , y = ( 2 ) 3 4 = 8 4 = 2 Substitute x = − 2 in y = x 3 4 , y = ( − 2 ) 3 4 = − 8 4 = − 2 Thus, the critical points are, ( 0 , 0 ) , ( 2 , 2 ) and ( − 2 , − 2 ) . Take the partial derivative of the equation (1) with respect to x and obtain f x x . ∂ 2 f ∂ x 2 = ∂ ∂ x ( 4 x 3 − 16 y ) = ∂ ∂ x ( 4 x 3 ) − ∂ ∂ x ( 16 y ) = 4 ( 3 x 2 ) − 0 = 12 x 2 Hence, ∂ 2 f ∂ x 2 = 12 x 2 . Take the partial derivative of the equation (2) with respect to y and obtain f y y . ∂ 2 f ∂ y 2 = ∂ ∂ y ( 4 y 3 − 16 x ) = ∂ ∂ y ( 4 y 3 ) − ∂ ∂ y ( 16 x ) = 4 ( 3 y 2 ) − 0 = 12 y 2 Hence, ∂ 2 f ∂ y 2 = 12 y 2

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 12, Problem 84RE

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To find: The critical points for the function f(x,y)=x4+y4−16xy and the local maximum, local minimum and saddle point.

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Chapter 12, Problem 83RE

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Calculus: Early Transcendentals (2nd Edition)

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