12.10 Notes - The largest value yields the maximum of f...

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Calculus III Section 12.10 – Lagrange Multipliers Lagrange’s Theorem Let f and g have continuous first partial derivatives such that f has an extremum at a point ) , ( 0 0 y x on the smooth constraint curve c y x g = ) , ( . If 0 ) , ( 0 0 y x g , then there is a real number λ such that ) , ( ) , ( 0 0 0 0 y x g y x f = . We can use the Method of Lagrange Multipliers to find the extreme values of a function f subject to a constraint. Method of Lagrange Multipliers Let f and g satisfy the hypothesis of Lagrange’s Theorem, and let f have a minimum or maximum subject to the constraint c y x g = ) , ( . To find the minimum or maximum of f use the following steps. 1. Simultaneously solve the equations ) , ( ) , ( 0 0 0 0 y x g y x f = and c y x g = ) , ( by solving the following system of equations. c y x g y x g y x f y x g y x f y y x x = = = ) , ( ) , ( ) , ( ) , ( ) , ( 2. Evaluate f at each solution point obtained in the first step.
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Unformatted text preview: The largest value yields the maximum of f subject to the constraint c y x g = ) , ( , and the smallest value yields the minimum of f subject to the constraint c y x g = ) , ( . Example: Minimize 2 2 ) , ( y x y x f + = subject to the constraint 15 4 2 =-+ y x . Maximize xyz z y x f = ) , , ( subject to the constraint 6 =-+ + z y x . The Method of Lagrange Multipliers can be extended to optimize a problem involving two constraints g and h : h g f ∇ + ∇ = ∇ μ λ Maximize xyz z y x f = ) , , ( subject to the constraints 5 2 2 = + z x and 2 =-y x ....
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This note was uploaded on 01/13/2012 for the course MATH 2574 taught by Professor Pamelasatterfield during the Spring '12 term at NorthWest Arkansas Community College.

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12.10 Notes - The largest value yields the maximum of f...

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