Explanation Result used: Method of Lagrange Multipliers: 1. Rewrite the problem in the form of maximize (or minimize) the function z = f ( x , y ) subject to the constraint g ( x , y ) = 0 . 2. Form the function F ( x , y , λ ) = f ( x , y ) + λ g ( x , y ) . 3. Compute the critical points of F ; That is, the solution of the system of the equation F x ( x , y , z ) = 0 F y ( x , y , z ) = 0 F z ( x , y , z ) = 0 4. if ( x 0 , y 0 , λ 0 ) is the only critical point of the function F , then the point ( x 0 , y 0 ) is the solution of the given hypothesis. If F has more than one critical point, compute the value of f ( x , y ) at ( x 0 , y 0 ) for each critical point ( x 0 , y 0 , λ 0 ) of F . then assume that the largest value of these values is the maximum value of f ( x , y ) , subject to constraint g ( x , y ) = 0 , and the smallest is the minimum of f ( x , y ) , subject to the constraint g ( x , y ) = 0 . Calculation: The multivariable function is f ( x , y ) = x 2 + y 2 + z 2 . The constraint is 2 x − y + 3 z = − 28 . Rewrite the constraint of the form g ( x , y ) = 0 as follows. 2 x − y + 3 z = − 28 2 x − y + 3 z + 28 = 0 The function f ( x , y ) = x 2 + y 2 + z 2 subject to constraint 2 x − y + 3 z + 28 = 0 . The form of the function should be F ( x , y , λ ) = f ( x , y ) + λ g ( x , y ) . Substitute f ( x , y ) = x 2 + y 2 + z 2 and g ( x , y ) = 2 x − y + 3 z + 28 in F ( x , y , λ ) = f ( x , y ) + λ g ( x , y ) , F ( x , y , λ ) = x 2 + y 2 + z 2 + λ ( 2 x − y + 3 z + 28 ) = x 2 + y 2 + z 2 + 2 x λ − y λ + 3 λ z + 28 λ Thus, the function F ( x , y , λ ) = x 2 + y 2 + z 2 + 2 x λ − y λ + 3 λ z + 28 λ . The partial derivative of the function F ( x , y , z , λ ) with respect to x is, F x ( x , y , z , λ ) = ∂ ∂ x ( x 2 + y 2 + z 2 + 2 x λ − y λ + 3 λ z + 28 λ ) = ∂ ∂ x ( x 2 ) + ∂ ∂ x ( y 2 ) + ∂ ∂ x ( z 2 ) + ∂ ∂ x ( 2 x λ ) − ∂ ∂ x ( y λ ) + ∂ ∂ x ( 3 λ z ) + ∂ ∂ x ( 28 λ ) = 2 x + 0 + 0 + ( 2 λ ) ∂ ∂ x ( x ) − 0 + 0 + 0 = 2 x + 2 λ Thus, the partial derivative of the function F x ( x , y , z , λ ) = 2 x + 2 λ . The partial derivative of the function F ( x , y , z , λ ) with respect to y is, F y ( x , y , z , λ ) = ∂ ∂ y ( x 2 + y 2 + z 2 + 2 x λ − y λ + 3 λ z + 28 λ ) = ∂ ∂ y ( x 2 ) + ∂ ∂ y ( y 2 ) + ∂ ∂ y ( z 2 ) + ∂ ∂ y ( 2 x λ ) − ∂ ∂ y ( y λ ) + ∂ ∂ y ( 3 λ z ) + ∂ ∂ y ( 28 λ ) = 0 + 2 y + 0 + 0 + 0 − ( λ ) ∂ ∂ y ( y ) 0 + 0 + 0 = 2 y − λ Thus, the partial derivative of the function F y ( x , y , λ ) = 2 y − λ

Calculus for Business Economics Life Sciences and Social Sciences Plus NEW

13th Edition

ISBN: 9780321925138

Author: Raymond Barnett

Publisher: PEARSON

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Chapter 7.4, Problem 17E

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Minimum of the function f(x,y)=x2+y2+z2 subject to the 2x−y+3z=−28 by using Lagrange multipliers.

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Minimum of the function f ( x , y ) = x 2 + y 2 + z 2 subject to the 2 x − y + 3 z = − 28 by using Lagrange multipliers.

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Minimum of the function f(x,y)=x2+y2+z2 subject to the 2x−y+3z=−28 by using Lagrange multipliers.

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