Explanation Definition used: “The Lagrange multipliers defined as ∇ f ( x , y , z ) = λ ∇ g ( x , y , z ) . This equation can be expressed as f x = λ g x , f y = λ g y , f z = λ g z and g ( x , y , z ) = k ”. Calculation: The given plane equation is, x − y + z = 2 . The distance d between ( 1 , 1 , 1 ) and any point ( x , y , z ) is, d = ( x − 1 ) 2 + ( y − 1 ) 2 + ( z − 1 ) 2 d 2 = ( x − 1 ) 2 + ( y − 1 ) 2 + ( z − 1 ) 2 Thus, f ( x , y , z ) = ( x − 1 ) 2 + ( y − 1 ) 2 + ( z − 1 ) 2 . Thus, the minimize function f ( x , y , z ) = ( x − 1 ) 2 + ( y − 1 ) 2 + ( z − 1 ) 2 subject to the constraint g ( x , y , z ) = x − y + z = 2 . The Lagrange multipliers ∇ f ( x , y , z ) = λ ∇ g ( x , y , z ) is computed as follows. ∇ f ( x , y , z ) = λ ∇ g ( x , y , z ) 〈 f x , f y , f z 〉 = λ 〈 g x , g y , g z 〉 〈 f x [ ( x − 1 ) 2 + ( y − 1 ) 2 + ( z − 1 ) 2 ] , f y [ ( x − 1 ) 2 + ( y − 1 ) 2 + ( z − 1 ) 2 ] , f z [ ( x − 1 ) 2 + ( y − 1 ) 2 + ( z − 1 ) 2 ] 〉 = λ <

Calculus: Early Transcendentals, 2nd Edition

2nd Edition

ISBN: 9780321965165

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

Publisher: PEARSON

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Chapter 12.9, Problem 45E

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To find: The point on the plane x−y+z=2. which is closest to the point (1,1,1) by using the method of Lagrange multipliers.

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Chapter 12.9, Problem 44E

Chapter 12.9, Problem 46E

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Chapter 12 Solutions

Calculus: Early Transcendentals, 2nd Edition

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The point on the plane x − y + z = 2 . which is closest to the point ( 1 , 1 , 1 ) by using the method of Lagrange multipliers.

Solution Summary: The author calculates the plane equation x-y+z=2 by using the method of Lagrange multipliers.

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To find: The point on the plane x−y+z=2. which is closest to the point (1,1,1) by using the method of Lagrange multipliers.

Want to see the full answer?

Chapter 12 Solutions