Explanation Given: The function is, f ( x , y ) = x 2 + 4 y 3 − 6 x y − 1 . Approach: “For a function z = f ( x , y ) , let f x x , f y y , and f x y all exist in a circular region contained in the xy -plane with center ( a , b ) . Further, let f x ( a . b ) = 0 and f y ( a . b ) = 0 . Define the number D , known as the discriminant, by D = f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 Then, a. f ( a , b ) is a relative maximum if D > 0 and f x x ( a , b ) < 0 . b. f ( a , b ) is a relative minimum if D > 0 and f x x ( a , b ) > 0 . c. f ( a , b ) is a saddle point (neither a maximum nor a minimum) if D < 0 . d. if D = 0 , the test gives no information.” Calculation: Consider the given function, f ( x , y ) = x 2 + 4 y 3 − 6 x y − 1 . Obtain the values f x ( x , y ) , and f y ( x , y ) as follows. f x ( x , y ) = ∂ ∂ x ( x 2 + 4 y 3 − 6 x y − 1 ) = 2 x − 6 y and, f y ( x , y ) = ∂ ∂ y ( x 2 + 4 y 3 − 6 x y − 1 ) = 12 y 2 − 6 x A function has relative extrema at a point, if f x ( x , y ) = 0 2 x − 6 y = 0 And if f y ( x , y ) = 0 , f y ( x , y ) = 0 12 y 2 − 6 x = 0 Solve the equation 2 x − 6 y = 0 and 12 y 2 − 6 x = 0 using substitution, 12 y 2 − 6 x = 0 12 y 2 − 6 ( 3 y ) = 0 ( ∵ 2 x − 6 y = 0 ) 12 y 2 − 18 y = 0 y = 0 or y = 3 2 Thus, the solutions are, x = 0 , y = 0 and x = 9 2 , y = 3 2

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ISBN: 9780321964458

Author: Lial

Publisher: PEARSON EDUCATION (COLLEGE)

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Chapter 9.3, Problem 15E

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The points for relative extrema and identify any saddle point for the given function.

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Chapter 9.3, Problem 14E

Chapter 9.3, Problem 16E

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Chapter 9 Solutions

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The points for relative extrema and identify any saddle point for the given function.

Solution Summary: The author explains how to find the points for relative extrema and identify any saddle point for the given function.

To find: The points for relative extrema and identify any saddle point for the given function.

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Chapter 9 Solutions

To find:

The points for relative extrema and identify any saddle point for the given function.