notes-lagrange

# notes-lagrange - OPMT 5701 Optimization with Constraints...

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OPMT 5701 Optimization with Constraints The Lagrange Multiplier Method Sometimes we need to to maximize (minimize) a function that is subject to some sort of constraint. For example Maximize z = f ( x, y ) subject to the constraint x + y 100 Forth isk indofprob lemthereisatechn ique ,or trick, developed for this kind of problem known as the Lagrange Multiplier method . This method involves adding an extra variable to the problem called the lagrange multiplier, or λ . We then set up the problem as follows: 1. Create a new equation form the original information L = f ( x, y )+ λ (100 x y ) or L = f ( x, y )+ λ [ Zero ] 2. Then follow the same steps as used in a regular maximization problem L x = f x λ =0 L y = f y λ =0 L ∂λ =100 x y =0 3. In most cases the λ will drop out with substitution. Solving these 3 equations will give you the constrained maximum solution Example 1: Suppose z = f ( x, y )= xy. and the constraint is the one from above. The problem then becomes L = xy + λ (100 x y ) Now take partial derivatives, one for each unknown, including λ L x = y λ =0 L y = x λ =0 L ∂λ =100 x y =0 Starting with the f rst two equations, we see that x = y and λ drops out. From the third equation we can easily f nd that x = y =50 and the constrained maximum value for z is z

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Example 2: Maximize u =4 x 2 +3 xy +6 y 2 subject to x + y =56 Set up the Lagrangian Equation: L =4 x 2 +3 xy +6 y 2 + λ (56 x y ) Take the f rst-order partials and set them to zero L x =8 x +3 y λ =0 L y =3 x +12 y λ =0 L λ =5 6 x y =0 From the f rst two equations we get 8 x +3 y =3 x +12 y x =1 . 8 y Substitute this result into the third equation 56 1 . 8 y y =0 y =2 0 therefore x =36 λ =348 Example 3: Cost minimization A
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## This note was uploaded on 03/03/2012 for the course ECON 345 taught by Professor Wendywu during the Fall '11 term at Wilfred Laurier University .

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notes-lagrange - OPMT 5701 Optimization with Constraints...

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