To determine Calculate the local maxima, minima , and saddle point for the function. Explanation Given information: The function is f ( x , y ) = x 3 + 3 x y 2 − 15 x + y 3 − 15 y 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: Provide the function as shown below: f ( x , y ) = x 3 + 3 x y 2 − 15 x + y 3 − 15 y (1) Differentiate Equation (1) with respect to x . ∂ f ∂ x = ∂ ∂ x ( x 3 + 3 x y 2 − 15 x + y 3 − 15 y ) = 3 x 2 + 3 y 2 − 15 (2) Differentiate Equation (1) with respect to y . ∂ f ∂ y = ∂ ∂ y ( x 3 + 3 x y 2 − 15 x + y 3 − 15 y ) = 6 x y + 3 y 2 − 15 (3) Find the critical point as shown below: Solve Equations (2) and (3) to get values, The critical points are ( 2 , 1 ) , ( − 2 , − 1 ) , ( 0 , 5 ) , ( 0 , − 5 ) . Differentiate Equation (2) with respect to x at ( 2 , 1 ) . ∂ 2 f ∂ x 2 = ∂ ∂ x ( 3 x 2 + 3 y 2 − 15 ) = 6 x (4) Substitute 2 for x and 1 for y . ∂ 2 f ∂ x 2 = 6 ( 2 ) = 12 Differentiate Equation (3) with respect to y at ( 2 , 1 ) . ∂ 2 f ∂ y 2 = ∂ ∂ y ( 6 x y + 3 y 2 − 15 ) = 6 x + 6 y (5) Substitute 2 for x and 1 for y . ∂ f ∂ y ( 2 , 1 ) = 6 ( 2 ) + 6 ( 1 ) = 12 + 6 = 18 Differentiate Equation (2) with respect to y . ∂ 2 f ∂ y ∂ x = ∂ ∂ y ( 3 x 2 + 3 y 2 − 15 ) = 6 y (6) Substitute 2 for x and 1 for y . ∂ 2 f ∂ y ∂ x ( 2 , 1 ) = 6 ( 1 ) = 6 By applying condition 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 ) . Substitute 12 for f x x , 18 for f x y , and 6 for f x y 2 . f x x f y y − f x y 2 > 0 12 ( 18 ) − 6 2 > 0 180 > 0 Thus, the function has a local minimum. Find the local minimum for the function at point ( 2 , 1 ) . f ( x , y ) ( 2 , 1 ) = ( 2 ) 3 + 3 ( 2 ) ( 1 ) 2 − 15 ( 2 ) + ( 1 ) 3 − 15 ( 1 ) = 8 + 6 − 30 + 1 − 15 = − 30 Thus, the function has a local minimum is − 30 _

MyLab Math plus Pearson eText -- Standalone Access Card -- for Thomas' Calculus: Early Transcendentals (14th Edition)

14th Edition

ISBN: 9780134764528

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Chapter 14.7, Problem 17E

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