Explanation Given information: To maximize the function f ( x , y ) = e x y subject to x 2 + y 2 = 3 using method of Lagrange Multipliers. Formula used: For any two function f ( x , y ) and g ( x , y ) that satisfy hypotheses of Lagrange multipliers, such that the function f ( x , y ) has an extremum when subjected to a constraint g ( x , y ) = c . We perform the following steps to find the maximum and minimum value of the function. Step 1: Equating the partial derivatives to solve equation for x , y and λ , f x ( x , y ) = λ g x ( x , y ) , f y ( x , y ) = λ g y ( x , y ) and g ( x , y ) = c Step 2: We substitute all the points found in above step and the boundary points from the constraint in the function f ( x , y ) to evaluate the extrema. The derivative with respect to a variable for the function of two or more than two variables is partially differentiating with respect to that variable and keeping all other variables as constants. Calculation: Consider the function, f ( x , y ) = e x y subject to x 2 + y 2 = 3 . Partial derivative of f ( x , y ) = e x y with respect to x , ∂ ∂ x f ( x , y ) = ∂ ∂ x e x y = y e x y Partial derivative of f ( x , y ) = e x y with respect to y , ∂ ∂ y f ( x , y ) = ∂ ∂ y e x y = x e x y Partial derivative of g ( x , y ) = x 2 + y 2 with respect to x , ∂ ∂ x g ( x , y ) = ∂ ∂ x ( x 2 + y 2 ) = 2 x Partial derivative of f ( x , y ) = e x y with respect to x , ∂ ∂ y g ( x , y ) = ∂ ∂ y ( x 2 + y 2 ) = 2 y For any two function f ( x , y ) and g ( x , y ) that satisfy hypotheses of Lagrange multipliers, such that the function f ( x , y ) has an extremum when subjected to a constraint g ( x , y ) = c . We perform the following steps to find the maximum and minimum value of the function. Step 1: Equating the partial derivatives to solve equation for x , y and λ , f x ( x , y ) = λ g x ( x , y ) , f y ( x , y ) = λ g y ( x , y ) and g ( x , y ) = c The equation f x ( x , y ) = λ g x ( x , y ) gives, y e x y = λ 2 x λ = y e x y 2 x The equation f y ( x , y ) = λ g y ( x , y ) gives, x e x y = λ 2 y λ = x e x y 2 y Equating the values of λ we get, y e x y 2 x = x e x y 2 y y 2 = x 2 Using the equation, g ( x , y ) = c that is x 2 + y 2 = 3

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 11.8, Problem 14PS

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To calculate:The constrained extrema to maximize the function f(x,y)=exy subject to x2+y2=3 using method of Lagrange Multipliers.

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Chapter 11.8, Problem 13PS

Chapter 11.8, Problem 15PS

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The constrained extrema to maximize the function f ( x , y ) = e x y subject to x 2 + y 2 = 3 using method of Lagrange Multipliers.

Solution Summary: The author calculates the constrained extrema to maximize the function f(x,y)=exy.

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To calculate:The constrained extrema to maximize the function f(x,y)=exy subject to x2+y2=3 using method of Lagrange Multipliers.

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Chapter 11 Solutions