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 , z ) = x 2 + y 2 + z 2 . The constraint is x + y + 3 z = 55 . Rewrite the constraint of the form g ( x , y , z ) = 0 as follows. x + y + 3 z = 55 x + y + 3 z − 55 = 0 The function f ( x , y , z ) = x 2 + y 2 + z 2 subject to constraint x + y + 3 z − 55 = 0 . The form of the function should be F ( x , y , z , λ ) = f ( x , y , z ) + λ g ( x , y , z ) . Substitute f ( x , y , z ) = x 2 + y 2 + z 2 and g ( x , y , z ) = x + y + 3 z − 55 in F ( x , y , z , λ ) = f ( x , y , z ) + λ g ( x , y , z ) , F ( x , y , z , λ ) = x 2 + y 2 + z 2 + λ ( x + y + 3 z − 55 ) = x 2 + y 2 + z 2 + x λ + y λ + 3 z λ − 55 λ Thus, the function F ( x , y , z , λ ) = x 2 + y 2 + z 2 + x λ + y λ + 3 z λ − 55 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 + x λ + y λ + 3 z λ − 55 λ ) = ∂ ∂ x ( x 2 ) + ∂ ∂ x ( y 2 ) + ∂ ∂ x ( z 2 ) + ∂ ∂ x ( x λ ) + ∂ ∂ x ( y λ ) + ∂ ∂ x ( 3 z λ ) − ∂ ∂ x ( 55 λ ) = 2 x + 0 + 0 + ( λ ) ∂ ∂ x ( x ) + 0 + 0 − 0 = 2 x + λ Thus, the partial derivative of the function F x ( x , y , z , λ ) = 2 x + 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 + x λ + y λ + 3 z λ − 55 λ ) = ∂ ∂ y ( x 2 ) + ∂ ∂ y ( y 2 ) + ∂ ∂ y ( z 2 ) + ∂ ∂ y ( x λ ) + ∂ ∂ y ( y λ ) + ∂ ∂ y ( 3 z λ ) − ∂ ∂ y ( 55 λ ) = 0 + 2 y + 0 + 0 + ( λ ) ∂ ∂ y ( y ) + 0 − 0 = 2 y + λ Thus, the partial derivative of the function F y ( x , y , λ ) = 2 y + λ

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Author: Barnett

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Chapter 7.4, Problem 17E

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To find: Minimum of the function f(x,y,z)=x2+y2+z2 subject to the x+y+3z=55.

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Chapter 7.4, Problem 16E

Chapter 7.4, Problem 18E

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Minimum of the function f ( x , y , z ) = x 2 + y 2 + z 2 subject to the x + y + 3 z = 55 .

Solution Summary: The author explains the method of Lagrange Multipliers: Rewrite the problem in the form of maximize (or minimize) the function z=f(x,y,lambda

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To find: Minimum of the function f(x,y,z)=x2+y2+z2 subject to the x+y+3z=55.

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Chapter 7 Solutions