Explanation Given Information: The expression is f ( x , y ) = x 3 + y 3 − 3 x − 12 y . Formula used: The D can be calculated by: D = f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 Where f x x , f y y and f x y are the second partial derivatives. Condition for f to has a maximum at ( a , b ) , if D > 0 and f x x < 0 . Condition for f to has a minimum at ( a , b ) , if D > 0 and f x x > 0 . And f has neither a maximum nor a minimum at ( a , b ) , if D < 0 . Calculation: Consider the expression, f ( x , y ) = x 3 + y 3 − 3 x − 12 y Find f x f x = ∂ ∂ x ( x 3 + y 3 − 3 x − 12 y ) = 3 x 2 + 0 − 3 − 0 = 3 x 2 − 3 Find f y f y = ∂ ∂ y ( x 3 + y 3 − 3 x − 12 y ) = 0 + 3 y 2 − 0 − 12 = 3 y 2 − 12 Find f x x f x x = ∂ ∂ x [ ∂ ∂ x ( x 3 + y 3 − 3 x − 12 y ) ] = ∂ ∂ x ( 3 x 2 + 0 − 3 − 0 ) = ∂ ∂ x ( 3 x 2 − 3 ) = 6 x Find f y y f y y = ∂ ∂ y ( 3 y 2 − 12 ) = 6 y Find f x y f x y = ∂ ∂ y [ ∂ ∂ x ( x 3 + y 3 − 3 x − 12 y ) ] = ∂ ∂ y ( 3 x 2 + 0 − 3 − 0 ) = ∂ ∂ y ( 3 x 2 − 3 ) = 0 Solve the system of equations f x = 0 and f y = 0 3 x 2 − 3 = 0 … … ( 1 ) 3 y 2 − 12 = 0 … … ( 2 ) Solve equation ( 1 ) for x , we get 3 x 2 − 3 = 0 3 x 2 = 3 x 2 = 1 x = ± 1 Solve equation ( 2 ) for y , we get 3 y 2 − 12 = 0 3 y 2 = 12 y 2 = 4 y = ± 2 Thus, ( − 1 , − 2 ) and ( 1 , 2 ) are possiblepoints for maximum or minimum

Calculus and Its Applications (11th Edition)

11th Edition

ISBN: 9780321979391

Author: Marvin L. Bittinger, David J. Ellenbogen, Scott J. Surgent

Publisher: PEARSON

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Chapter 6, Problem 9T

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The relative maximum and minimum values for the equation, f ( x , y ) = x 3 + y 3 − 3 x − 12 y .

Solution Summary: The author calculates the relative maximum and minimum values for the equation f(x,y)=x3+y