Explanation Given information: The plane x + 2 y + z = 8 . Concept Used: Let x , y , z be point lying on the plane 2 x + y + 3 z = 6 which is closest to the origin. Distance of point x , y , z from the origin is x 2 + y 2 + z 2 . To find the points closest to the origin, minimize the distance x 2 + y 2 + z 2 , therefore objective function is x 2 + y 2 + z 2 . Values of x , y and z at which x 2 + y 2 + z 2 is minimized is the same as the values at which x 2 + y 2 + z 2 2 = x 2 + y 2 + z 2 is minimized. So, the objective function is f x , y , z = x 2 + y 2 + z 2 . Calculation: As point x , y , z lies on the plane x + 2 y + z = 8 , therefore, x + 2 y + z = 8 becomes a constraint. Write the constraint equation x + 2 y + z = 8 as x + 2 y + z − 8 = 0 . Let g x , y , z = x + 2 y + z − 8 . Now, form the Lagrange function as, L x , y , z = f x , y , z − λ g x , y , z = x 2 + y 2 + z 2 − λ x + 2 y + z − 8 = x 2 + y 2 + z 2 − λ x − 2 λ y − λ z + 8 λ Compute the first partial derivatives of L x . L x = ∂ L ∂ x = ∂ x 2 + y 2 + z 2 − λ x − 2 λ y − λ z + 8 λ ∂ x = 2 x − λ Thus, L x = 2 x − λ . Compute the first partial derivatives of L y . L y = ∂ L ∂ y = ∂ x 2 + y 2 + z 2 − λ x − 2 λ y − λ z + 8 λ ∂ y = 2 y − 2 λ = 2 y − λ Thus, L y = 2 y − λ . Compute the first partial derivatives of L z . L z = ∂ L ∂ z = ∂ x 2 + y 2 + z 2 − λ x − 2 λ y − λ z + 8 λ ∂ z = 2 z − λ Thus, L z = 2 z − λ . Compute the first partial derivatives of L λ . L λ = ∂ L ∂ λ = ∂ x 2 + y 2 + z 2 − λ x − 2 λ y − λ z + 8 λ ∂ λ = − x − 2 y − z + 8 Thus, L λ = − x − 2 y − z + 8

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Author: BITTINGER

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Chapter 6.5, Problem 21E

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To find: Point lying on the plane x+2y+z=8 which is closest to the origin.

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Chapter 6.5, Problem 20E

Chapter 6.5, Problem 22E

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Point lying on the plane x + 2 y + z = 8 which is closest to the origin.

Solution Summary: The author explains that the plane x+2y+z=8 is closest to the origin and minimizes the distance.

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To find: Point lying on the plane x+2y+z=8 which is closest to the origin.

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