Explanation Result used: Method of Lagrange Multipliers “The below four steps are the procedure for 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 : 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: Step 1: Maximize and minimize f ( x , y ) = x 2 − y 2 Subject to: g ( x , y ) = x 2 + y 2 − 25 = 0 Step 2: Form the function F by substituting f ( x , y ) = x 2 − y 2 and g ( x , y ) = x 2 + y 2 − 25 in F ( x , y , λ ) = f ( x , y ) + λ g ( x , y ) , F ( x , y , λ ) = ( x 2 − y 2 ) + λ ( x 2 + y 2 − 25 ) = x 2 − y 2 + λ x 2 + λ y 2 − 25 λ Thus, the function F ( x , y , λ ) = x 2 − y 2 + λ x 2 + λ y 2 − 25 λ . Step 3: In order to find the critical points, first find the partial derivatives. Compute the partial derivative F x ( x , y , λ ) . That is, differentiate the function F ( x , y , λ ) with respect to x by taking y and λ are constants. F x ( x , y , λ ) = ∂ ∂ x ( x 2 − y 2 + λ x 2 + λ y 2 − 25 λ ) = ∂ ∂ x ( x 2 ) − ∂ ∂ x ( y 2 ) + ∂ ∂ x ( λ x 2 ) + ∂ ∂ x ( λ y 2 ) − ∂ ∂ x ( 25 λ ) = ∂ ∂ x ( x 2 ) − 0 + ( λ ) ∂ ∂ x ( x 2 ) + 0 − 0 = 2 x + 2 x λ Thus, the partial derivative of the function with respect to x is F x ( x , y , λ ) = 2 x + 2 x λ . Compute the partial derivative F y ( x , y , λ ) . That is, differentiate the function F ( x , y , λ ) with respect to y by taking x and λ are constants. F y ( x , y , λ ) = ∂ ∂ y ( x 2 − y 2 + λ x 2 + λ y 2 − 25 λ ) = ∂ ∂ y ( x 2 ) − ∂ ∂ y ( y 2 ) + ∂ ∂ y ( λ x 2 ) + ∂ ∂ y ( λ y 2 ) − ∂ ∂ y ( 25 λ ) = 0 − ∂ ∂ y ( y 2 ) + 0 + ( λ ) ∂ ∂ y ( y 2 ) − 0 = − 2 y + 2 y λ Thus, the partial derivative of the function with respect to y is F y ( x , y , λ ) = − 2 y + 2 λ y . Compute the partial derivative F λ ( x , y , λ )

College Mathematics for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

14th Edition

ISBN: 9780134674148

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen, Christopher J. Stocker

Publisher: PEARSON

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Chapter 14.4, Problem 14E

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To find: The maximum and minimum value of the function f(x,y)=x2−y2 subject to the constraint x2+y2=25 using Lagrange multipliers.

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Chapter 14 Solutions

College Mathematics for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

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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The maximum and minimum value of the function f ( x , y ) = x 2 − y 2 subject to the constraint x 2 + y 2 = 25 using Lagrange multipliers.

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To find: The maximum and minimum value of the function f(x,y)=x2−y2 subject to the constraint x2+y2=25 using Lagrange multipliers.

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Chapter 14 Solutions