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Unformatted text preview: Fall 11, Econ 11 Prof. Mazzocco Problem Set 1 Solution Exercise 1. Let f ( x ) = 4 x + x, where x > 0. Find the x that minimizes f ( x ), be sure to check your answer is indeed a minimum. Solution. Compute the first order condition df dx = 4 x 2 + 1 = 0 , we get x 2 = 4, so x = 2. Check the second order condition d 2 f d 2 x = 8 x 3 , which is positive when x = 2. So the minimizer for f ( x ) is x = 2. Exercise 2. Lagrange Multiplier I Let f ( x 1 ,x 2 ) = log x 1 + 2log x 2 . Find the maximum of f ( x 1 ,x 2 ) subject to x 1 + x 2 = 1. Solution. Set up the Lagrangian L ( x 1 ,x 2 ,λ ) = log x 1 + 2log x 2 + λ (1 x 1 x 2 ) . The FOCs are ∂L ∂x 1 = 1 x 1 λ = 0 (1) ∂L ∂x 2 = 2 x 2 λ = 0 (2) ∂L ∂λ = 1 x 1 x 2 = 0 . (3) Fall 11, Econ 11 Prof. Mazzocco From (1),(2) we get x 1 = 1 /λ , x 2 = 2 /λ . Plug into (3) we get 1 1 λ 2 λ = 0 , hence λ = 3, plug back to x 1 ,x 2 we obtain x 1 = 1 / 3, x 2 = 2 / 3. So the maximizer of f ( x 1 ,x 2 ) subject to the constraint is (...
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This note was uploaded on 10/08/2011 for the course ECONOMICS 11 taught by Professor Mazzocco during the Fall '11 term at UCLA.
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