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 + 10 y + 2 z . The constraint is x 2 + y 2 + z 2 = 105 . Rewrite the constraint of the form g ( x , y , z ) = 0 as follows. x 2 + y 2 + z 2 = 105 x 2 + y 2 + z 2 − 105 = 0 The function f ( x , y , z ) = x + 10 y + 2 z subject to constraint x 2 + y 2 + z 2 − 105 = 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 + 10 y + 2 z and g ( x , y , z ) = x 2 + y 2 + z 2 − 105 in F ( x , y , z , λ ) = f ( x , y , z ) + λ g ( x , y , z ) , F ( x , y , z , λ ) = x + 10 y + 2 z + λ ( x 2 + y 2 + z 2 − 105 ) = x + 10 y + 2 z + λ x 2 + λ y 2 + λ z 2 − 105 λ Thus, the function F ( x , y , z , λ ) = 300 − x 2 − 2 y 2 − z 2 + 4 x λ + y λ + z λ − 70 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 + 10 y + 2 z + λ x 2 + λ y 2 + λ z 2 − 105 λ ) = ∂ ∂ x ( x ) + ∂ ∂ x ( 10 y ) + ∂ ∂ x ( 2 z ) + ∂ ∂ x ( λ x 2 ) + ∂ ∂ x ( λ y 2 ) + ∂ ∂ x ( λ z 2 ) − ∂ ∂ x ( 105 λ ) = 1 + 0 + 0 + ( λ ) ∂ ∂ x ( x 2 ) + 0 + 0 − 0 = 1 + 2 x λ Thus, the partial derivative of the function F x ( x , y , z , λ ) = 1 + 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 + 10 y + 2 z + λ x 2 + λ y 2 + λ z 2 − 105 λ ) = ∂ ∂ y ( x ) + ∂ ∂ y ( 10 y ) + ∂ ∂ y ( 2 z ) + ∂ ∂ y ( λ x 2 ) + ∂ ∂ y ( λ y 2 ) + ∂ ∂ y ( λ z 2 ) − ∂ ∂ y ( 105 λ ) = 0 + ( 10 ) ∂ ∂ y ( y ) + 0 + 0 + ( λ ) ∂ ∂ y ( y 2 ) + 0 − 0 = 10 + 2 λ y Thus, the partial derivative of the function F y ( x , y , λ ) = 10 + 2 λ y . The partial derivative of the function F ( x , y , z , λ ) with respect to z is, F z ( x , y , z , λ ) = ∂ ∂ z ( x + 10 y + 2 z + λ x 2 + λ y 2 + λ z 2 − 105 λ ) = ∂ ∂ z ( x ) + ∂ ∂ z ( 10 y ) + ∂ ∂ z ( 2 z ) + ∂ ∂ z ( λ x 2 ) + ∂ ∂ z ( λ y 2 ) + ∂ ∂ z ( λ z 2 ) − ∂ ∂ z ( 105 λ ) = 0 + 0 + ( 2 ) ∂ ∂ z ( z ) + 0 + 0 + ( λ ) ∂ ∂ z ( z 2 ) − 0 = 2 + 2 z λ Thus, the partial derivative of the function F z ( x , y , z , λ ) = 2 + 2 z λ

Pearson eText for College Mathematics for Business, Economics, Life Sciences, and Social Sciences -- Instant Access (Pearson+)

14th Edition

ISBN: 9780137553341

Author: Raymond Barnett, Michael Ziegler

Publisher: PEARSON+

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Chapter 14.4, Problem 21E

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To find: Maximum and minimum value of the function f(x,y,z)=x+10y+2z subject to x2+y2+z2=105.

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Chapter 14.4, Problem 20E

Chapter 14.4, Problem 22E

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Pearson eText for College Mathematics for Business, Economics, Life Sciences, and Social Sciences -- Instant Access (Pearson+)

Ch. 14.1 - Matched Problem 1 Find C(20, 10) for the cost...Ch. 14.1 - Prob. 2MP Ch. 14.1 - Prob. 3MP Ch. 14.1 - Prob. 4MP Ch. 14.1 - Prob. 5MP Ch. 14.1 - Prob. 6MP Ch. 14.1 - Prob. 7MP Ch. 14.1 - Prob. 1ED Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E

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Maximum and minimum value of the function f ( x , y , z ) = x + 10 y + 2 z subject to x 2 + y 2 + z 2 = 105 .

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To find: Maximum and minimum value of the function f(x,y,z)=x+10y+2z subject to x2+y2+z2=105.

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Chapter 14 Solutions