Lectures-ec251-2011-week7-update1

# Lectures-ec251-2011-week7-update1 - Constrained...

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Unformatted text preview: Constrained Optimization: Examples Consumer’s problem : Suppose that a consumer has a utility function U ( x,y ) = x . 5 y . 5 , the price of x is \$2, the price of y is \$3 and the consumer has \$100 in income. How much of the two goods should the consumer purchase to maximize her utility? Firm’s problem : Suppose that a firm’s production function is given by q = K . 5 L . 5 , the price of capital is \$2 and the price of labour is \$5. What is the least cost way for the firm to produce 100 units of output? 1 Relevant Mathematics Both of the above problems have a common mathematical structure: max x 1 ,...,x n f ( x 1 ,x 2 ,...,x n ) subject to g ( x 1 ,x 2 ,...,x n ) = 0 It is possible that instead of maximizing f ( x 1 ,...,x n ) we could be minimizing f ( x 1 ,...,x n ). The above problems can be translated into the above mathematical framework as max ( x,y ) x . 5 y . 5 subject to 2 x + 3 y- 100 = 0 and min ( K,L ) 2 K + 5 L subject to K . 5 L . 5- 100 = 0 2 One approach When the constraint(s) are equalities, we can convert the problem from a constrained optimization to an unconstrained optimization problem by substituting for some of the variables. In the consumer’s problem, we have 2 x +3 y = 100, so x = 50- (3 / 2) y . We can use this relation to substitute for x in the utility function which gives U ( x,y ) = x . 5 y . 5 = (50- (3 / 2) y ) . 5 y . 5 This is now a function of just y and we can now maximize this func- tion with respect to y . It is important to observe that this is an unconstrained optimization problem since we have incorporated the constraint by substituting for x . 3 The first order conditions for the maximization of (50- (3 / 2) y ) . 5 y . 5 gives us 1 2- 3 2 50- 3 2 y- . 5 y . 5 + 1 2 50- 3 2 y . 5 y- . 5 = 0 Solving this gives y = 50 / 3, and since x = 50- (3 / 2) y , we have x = 25. For the firm’s minimization problem, we can proceed similarly: the constraint gives us K 1 / 2 L 1 / 2 = 100 or K = 10000 /L 2 . Therefore, the objective function becomes 20000 /L 2 + 5 L . This can be minimized easily with respect to L , and then the corresponding K found easily. 4 The Lagrangean approach Two reasons for an alternative approach: 1. In some cases, we cannot use substitution easily: for instance, suppose the con straint is x 4 + 5 x 3 y + y 2 x + x 6 + 5 = 0. Here, it is not possible to solve this equation to get x as a function of y or vice versa. 2. In many cases, the economic constraints are written in the form g ( x 1 ,...,x n ) ≤ 0 or g ( x 1 ,...,x n ) ≥ 0. While the Lagrangean tech- nique can be modified to take care of such cases, the substitution technique cannot be modified, or can be modified only with some difficulty....
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Lectures-ec251-2011-week7-update1 - Constrained...

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