Explanation Given: The rectangular area enclosed with 400 feet of fencing if no fencing is needed along one side. Definition used: Lagrange Multipliers 1. The form of the constraint is g ( x , y ) = 0 . 2. The form of the Lagrange function F ( x , y , λ ) = f ( x , y ) − λ ⋅ g ( x , y ) . 3 Form the system of equations F x ( x , y , λ ) = 0 , F y ( x , y , λ ) = 0 and F λ ( x , y , λ ) = 0 , compute the solution of these system and also the relative extrema among these solution. Calculation: The area of the rectangle is given by A = x y , where x is a width of rectangle and y is the length of rectangle. Therefore, the multivariable function f ( x , y ) = x y . Since the area enclosed within 400 feet of fencing where no fencing is needed along one side, the constraint becomes 2 x + y = 400 . The constraint is in the form g ( x , y ) = 0 . Thus, the given constraint can be written as g ( x , y ) = 2 x + y − 400 The form of the Lagrange function is, F ( x , y , λ ) = f ( x , y ) − λ ⋅ g ( x , y ) . Substitute f ( x , y ) = x y and g ( x , y ) = 2 x + y − 400 in F ( x , y , λ ) = f ( x , y ) − λ ⋅ g ( x , y ) , F ( x , y , λ ) = x y − λ ( 2 x + y − 400 ) = x y − 2 x λ − λ y + 400 λ Thus, the Lagrange function is F ( x , y , λ ) = x y − 2 x λ − λ y + 400 λ . Find the partial derivative of the function F ( x , y , λ ) with respect to x , where y and λ are constants. F x ( x , y , λ ) = ∂ ∂ x ( x y − 2 x λ − λ y + 400 λ ) = ∂ ∂ x ( x y ) − ∂ ∂ x ( λ 2 x ) − ∂ ∂ x ( λ y ) + ∂ ∂ x ( 400 λ ) = ( y ) ∂ ∂ x ( x ) − ( 2 λ ) ∂ ∂ x ( x ) − ∂ ∂ x ( λ y ) + ∂ ∂ x ( 400 λ ) Simplify the terms and obtain the value as follows, F x ( x , y , λ ) = ( y ) ( 1 ) − ( 2 λ ) ( 1 ) − 0 + 0 = y − 2 λ Therefore, the partial derivative with respect to x is F x ( x , y , λ ) = y − 2 λ . Find the partial derivative of the function F ( x , y , λ ) with respect to y , where x and λ are constants. F y ( x , y , λ ) = ∂ ∂ y ( x y − 2 x λ − λ y + 400 λ ) = ∂ ∂ y ( x y ) − ∂ ∂ y ( 2 λ x ) − ( λ ) ∂ ∂ y ( y ) + ∂ ∂ y ( 400 λ ) = ( x ) ∂ ∂ y ( y ) − 0 − ( λ ) ∂ ∂ y ( y ) + 0 = x − λ Therefore, the partial derivative with respect to y is F y ( x , y , λ ) = x − λ

Finite Mathematics and Calculus with Applications Books a la Carte Plus MyLab Math Package (10th Edition)

10th Edition

ISBN: 9780133935592

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 17, Problem 106RE

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Chapter 17 Solutions

Finite Mathematics and Calculus with Applications Books a la Carte Plus MyLab Math Package (10th Edition)

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W2.23x+6dx Ch. 17.6 - Prob. 3WE Ch. 17.6 - Prob. 4WE Ch. 17.6 - Prob. 5WE Ch. 17.6 - Prob. 6WE Ch. 17.6 - Prob. 1E Ch. 17.6 - Prob. 2E Ch. 17.6 - Prob. 3E Ch. 17.6 - Prob. 4E Ch. 17.6 - Prob. 5E Ch. 17.6 - Prob. 6E Ch. 17.6 - Prob. 7E Ch. 17.6 - Prob. 8E Ch. 17.6 - Prob. 9E Ch. 17.6 - Prob. 10E Ch. 17.6 - Prob. 11E Ch. 17.6 - Prob. 12E Ch. 17.6 - Prob. 13E Ch. 17.6 - Prob. 14E Ch. 17.6 - Prob. 15E Ch. 17.6 - Prob. 16E Ch. 17.6 - Prob. 17E Ch. 17.6 - Prob. 18E Ch. 17.6 - Prob. 19E Ch. 17.6 - Prob. 20E Ch. 17.6 - Prob. 21E Ch. 17.6 - Prob. 22E Ch. 17.6 - Prob. 23E Ch. 17.6 - Prob. 24E Ch. 17.6 - Prob. 25E Ch. 17.6 - Prob. 26E Ch. 17.6 - Prob. 27E Ch. 17.6 - Prob. 28E Ch. 17.6 - Prob. 29E Ch. 17.6 - Prob. 30E Ch. 17.6 - Prob. 31E Ch. 17.6 - Prob. 32E Ch. 17.6 - Prob. 33E Ch. 17.6 - Prob. 34E Ch. 17.6 - Prob. 35E Ch. 17.6 - Prob. 36E Ch. 17.6 - Prob. 37E Ch. 17.6 - Prob. 38E Ch. 17.6 - Prob. 39E Ch. 17.6 - Prob. 40E Ch. 17.6 - Prob. 41E Ch. 17.6 - Prob. 42E Ch. 17.6 - Prob. 43E Ch. 17.6 - Use the region R with the indicated boundaries to...Ch. 17.6 - Prob. 45E Ch. 17.6 - Prob. 46E Ch. 17.6 - Prob. 47E Ch. 17.6 - Prob. 48E Ch. 17.6 - Prob. 49E Ch. 17.6 - Prob. 50E Ch. 17.6 - Prob. 51E Ch. 17.6 - Prob. 52E Ch. 17.6 - Prob. 53E Ch. 17.6 - Prob. 54E Ch. 17.6 - Prob. 55E Ch. 17.6 - Prob. 56E Ch. 17.6 - Prob. 57E Ch. 17.6 - Prob. 58E Ch. 17.6 - Prob. 59E Ch. 17.6 - Prob. 60E Ch. 17.6 - Prob. 61E Ch. 17.6 - Prob. 62E Ch. 17.6 - Prob. 63E Ch. 17.6 - Prob. 64E Ch. 17.6 - Prob. 65E Ch. 17.6 - Prob. 66E Ch. 17.6 - Prob. 67E Ch. 17.6 - Prob. 68E Ch. 17.6 - Prob. 69E Ch. 17.6 - Time In Exercise 39 of Section 17.3, we saw that...Ch. 17.6 - Profit In Exercise 38 of Section 17.3, we saw that...Ch. 17 - Determine whether each of the following statements...Ch. 17 - Prob. 2RE Ch. 17 - Prob. 3RE Ch. 17 - Prob. 4RE Ch. 17 - Prob. 5RE Ch. 17 - Prob. 6RE Ch. 17 - Prob. 7RE Ch. 17 - Prob. 8RE Ch. 17 - Prob. 9RE Ch. 17 - Prob. 10RE Ch. 17 - Prob. 11RE Ch. 17 - Prob. 12RE Ch. 17 - Prob. 13RE Ch. 17 - Prob. 14RE Ch. 17 - Prob. 15RE Ch. 17 - Prob. 16RE Ch. 17 - Prob. 17RE Ch. 17 - Prob. 18RE Ch. 17 - Prob. 19RE Ch. 17 - Prob. 20RE Ch. 17 - Prob. 21RE Ch. 17 - Prob. 22RE Ch. 17 - Prob. 23RE Ch. 17 - Prob. 24RE Ch. 17 - Prob. 25RE Ch. 17 - Prob. 26RE Ch. 17 - Prob. 27RE Ch. 17 - Prob. 28RE Ch. 17 - Prob. 29RE Ch. 17 - Prob. 30RE Ch. 17 - Prob. 31RE Ch. 17 - Prob. 32RE Ch. 17 - Prob. 33RE Ch. 17 - Prob. 34RE Ch. 17 - Prob. 35RE Ch. 17 - Prob. 36RE Ch. 17 - Prob. 37RE Ch. 17 - Prob. 38RE Ch. 17 - Prob. 39RE Ch. 17 - Prob. 40RE Ch. 17 - Prob. 41RE Ch. 17 - Prob. 42RE Ch. 17 - Prob. 43RE Ch. 17 - Prob. 44RE Ch. 17 - Prob. 45RE Ch. 17 - Prob. 46RE Ch. 17 - Prob. 47RE Ch. 17 - Prob. 48RE Ch. 17 - Prob. 49RE Ch. 17 - Prob. 50RE Ch. 17 - Prob. 51RE Ch. 17 - Prob. 52RE Ch. 17 - Prob. 53RE Ch. 17 - Prob. 54RE Ch. 17 - Prob. 55RE Ch. 17 - Prob. 56RE Ch. 17 - Prob. 57RE Ch. 17 - Prob. 58RE Ch. 17 - Evaluate dz using the given information. 59.z =...Ch. 17 - Prob. 60RE Ch. 17 - Prob. 61RE Ch. 17 - Prob. 62RE Ch. 17 - Prob. 63RE Ch. 17 - Prob. 64RE Ch. 17 - Prob. 65RE Ch. 17 - Prob. 66RE Ch. 17 - Prob. 67RE Ch. 17 - Prob. 68RE Ch. 17 - Prob. 69RE Ch. 17 - Prob. 70RE Ch. 17 - Prob. 71RE Ch. 17 - Prob. 72RE Ch. 17 - Prob. 73RE Ch. 17 - Prob. 74RE Ch. 17 - Prob. 75RE Ch. 17 - Prob. 76RE Ch. 17 - Prob. 77RE Ch. 17 - Prob. 78RE Ch. 17 - Prob. 79RE Ch. 17 - Prob. 80RE Ch. 17 - Prob. 81RE Ch. 17 - Prob. 82RE Ch. 17 - Prob. 83RE Ch. 17 - Evaluate each double integral. 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The maximum rectangular area by using Lagrange multipliers.

Solution Summary: The author calculates the maximum rectangular area by using Lagrange multipliers. The constraint is g(x,y,lambda )=0.

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Chapter 17 Solutions