Explanation Formula used: Lagrange Multipliers: To obtain the relative extremum of the function f ( x , y ) subject to the constraint g ( x , y ) = 0 (assuming that these extreme values exist), (1) Form an auxiliary function F ( x , y , λ ) = f ( x , y ) + λ g ( x , y ) called the Lagrangian function where the variable λ is called the Lagrange multiplier. (2) Solve the system that consists of the equations F x = 0 , F y = 0 and F λ = 0 for all values of x , y , and λ . (3) The solutions found in step (2) are candidates for the extremum of f . Calculation: Consider the function g ( x , y ) = x y − 1 . It is given that x y = 1 . Hence, obtain the maximum of f ( x , y ) = x 2 + 4 y 2 constrained to the function g ( x , y ) = x y − 1 . Frame an auxiliary function F for the given function f ( x , y ) = x 2 + 4 y 2 subject to g ( x , y ) = x y − 1 . F ( x , y , λ ) = f ( x , y ) + λ g ( x , y ) = x 2 + 4 y 2 + λ ( x y − 1 ) Differentiate F ( x , y , λ ) = x 2 + 4 y 2 + λ ( x y − 1 ) with respect to x , F x = ∂ ∂ x ( x 2 + 4 y 2 + λ ( x y − 1 ) ) = ∂ ∂ x ( x 2 + 4 y 2 ) + ∂ ∂ x ( λ ( x y − 1 ) ) = 2 x + λ ( y ) = 2 x + λ y By the step (2) of the above formula it is observed that 2 x + λ y = 0 . Compute the term F y from the Lagrangian function. Differentiate F ( x , y , λ ) = x 2 + 4 y 2 + λ ( x y − 1 ) with respect to y , F y = ∂ ∂ y ( x 2 + 4 y 2 + λ ( x y − 1 ) ) = ∂ ∂ y ( x 2 + 4 y 2 ) + ∂ ∂ y ( λ ( x y − 1 ) ) = 4 ( 2 y ) + λ ( x ) = 8 y + λ x By the step (2) of the above formula it is observed that 8 y + λ x = 0 . Compute the term F λ from the Lagrangian function. Differentiate F ( x , y , λ ) = x 2 + 4 y 2 + λ ( x y − 1 ) with respect to λ , F λ = ∂ ∂ λ ( x 2 + 4 y 2 + λ ( x y − 1 ) ) = ∂ ∂ λ ( x 2 + 4 y 2 ) + ∂ ∂ λ ( λ ( x y − 1 ) ) = x y − 1 By the step (2) of the above formula it is observed that x y − 1 = 0

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 8.5, Problem 5E

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To minimize: The function f(x,y)=x2+4y2 subjects to the constraint xy=1 .

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Chapter 8.5, Problem 4E

Chapter 8.5, Problem 6E

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Use Lagrange multipliers to find the maximum and minimum values of the function f(x,y,z) = x² + y² +2² subject to the constraint x²+y²+2+xy=12. You may assume ftakes both a maximum and minimum value subject to the constraint.

Chapter 8 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

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To minimize: The function f ( x , y ) = x 2 + 4 y 2 subjects to the constraint x y = 1 .

Solution Summary: The author explains how to minimize the function f(x,y,lambda ) with the Lagrange multiplier.

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To minimize: The function f(x,y)=x2+4y2 subjects to the constraint xy=1 .

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Chapter 8 Solutions