Explanation Given information: The function f ( x , y ) = x 2 + k x y + y 2 . Consider the value of k = 0 and k ≠ 0 . Critical point: An interior point of the domain of a function f ( x , y ) where both f x and f y are zero or where one or both or where one or both of f x and f y do not exist is a critical point of f . Theorem:11 Suppose that f ( x , y ) and its first and second partial derivative is continuous throughout a disk centered at ( a , b ) and that f x ( a , b ) = f y ( a , b ) = 0 . i. f has a local maximum at ( a , b ) if f x x < 0 and f x x f y y − f x y 2 > 0 at ( a , b ) . ii. f has a local minimum at ( a , b ) if f x x > 0 and f x x f y y − f x y 2 > 0 at ( a , b ) . iii. f has a saddle point at ( a , b ) if f x x f y y − f x y 2 < 0 at ( a , b ) . Calculation: Consider k = 0 . Provide the function as shown below: f ( x , y ) = x 2 + k x y + y 2 (1) Differentiate Equation (1) with respect to x . ∂ f ∂ x = ∂ ∂ x ( x 2 + k x y + y 2 ) = 2 x + k y (2) Substitute 0 for k in Equation (2). ∂ f ∂ x = 2 x + ( 0 ) y = 2 x Differentiate Equation (1) with respect to y

14th Edition

ISBN: 9780134654874

Author: WEIR

Publisher: VST

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Chapter 14.7, Problem 45E

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EBK THOMAS' CALCULUS

Chapter 14.1, Problem 1E Chapter 14.1, Problem 2E Chapter 14.1, Problem 3E Chapter 14.1, Problem 4E Chapter 14.1, Problem 5E Chapter 14.1, Problem 6E Chapter 14.1, Problem 7E Chapter 14.1, Problem 8E Chapter 14.1, Problem 9E Chapter 14.1, Problem 10E

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