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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 For this kind of problem there is a technique, 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 fi rst two equations, we see that x = y and λ drops out. From the third equation we can easily fi nd that x = y = 50 and the constrained maximum value for z is z = xy = 2500 . 1
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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 fi 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 = 56 x y = 0 From the fi 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 = 20 therefore x = 36 λ = 348 Example 3: Cost minimization A fi rm produces two goods, x and y. Due to a government quota, the fi rm must produce subject to the constraint x + y = 42 . The fi rm’s cost functions is c ( x, y ) = 8 x 2 xy + 12 y 2
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