105-11a-ConstrainedOptimization

# 105-11a-ConstrainedOptimization - Week 11 Constrained...

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Unformatted text preview: Week 11 Constrained Optimization Constrained Optimization A constrained optimization problem (in 2 variables) consists of an objective function f ( x,y ) and a constraint g ( x,y ) = 0. The goal is to optimize the objective function subject to the constraint; or in other words, among the points ( a,b ) satisfying g ( a,b ) = 0, find the ones for which f ( a,b ) is the largest or smallest. This kind of situation is quite common. Here are two different kinds of examples. • For instance, suppose you want to know what rectangle with perimeter 4 has the largest area. If you call the sides of the rectangle x and y , then the area is xy and the perimeter is 2 x + 2 y = 4, so you have to solve the constrained optimization problem f ( x,y ) = xy, g ( x,y ) = 2 x + 2 y- 4 = 0 . In this case, you can probably guess that the answer is a square with sides x = y = 1. • An example from economics is optimizing a production function subject to a budget: if x 1 / 2 y 1 / 2 is the number of units produced using x units of labor and y units of capital, and 3 x + 2 y = 100, for what choice of labor and capital will production be highest? Here the constraint can be seen as a budget of 100 dollars, with labor costing 3 dollars per unit and capital 2 dollars per unit. Then you have to solve the constrained optimization problem: f ( x,y ) = x 1 / 2 y 1 / 2 , g ( x,y ) = 3 x + 2 y- 100 = 0 . Turning constrained optimization into just optimization Before looking at the general method for solving constrained optimization problems, first we’ll look at a trick that works for easy examples. • Let’s look at the rectangle example above. From the constraint 2 x +2 y = 4, we can isolate y = 2- x , and plug that into the objective function xy to get a new 1-variable function h ( x ) = 2 x- x 2 . That one is easy to optimize: set h ( x ) = 2- 2 x = 0, so x = 1. Then from the constraint we get y = 1, and that tells us that the square with sides x = y = 1 has the largest area. • For the production function example, doing the same thing gives 3 x + 2 y = 100 ⇒ y = 50- 3 2 x ⇒ h ( x ) = f ( x, 50- 3 2 x ) = r 50 x- 3 2 x 2 ⇒ h ( x ) = 50- 3 x 2 q 50 x- 3 2 x 2 = 0 ⇒ 50- 3 x = 0 ⇒ x = 50 3 , y = 50- 3 2 · 50 3 = 25 ....
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## This note was uploaded on 01/14/2012 for the course MATH 105 taught by Professor Malabikapramanik during the Fall '10 term at The University of British Columbia.

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105-11a-ConstrainedOptimization - Week 11 Constrained...

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