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HW5Sol

# HW5Sol - Fall 2009 IEOR 160 Industrial Engineering...

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Fall 2009 IEOR 160 Industrial Engineering & Operations Research October 9, 2009 Page 1 of 1 HOMEWORK 5 SOLUTIONS Chapter 12.6 1. We wish to minimize f ( x, y ) = n i =1 ( x i - x ) 2 + ( y i - y ) 2 . ∂f ∂x = - 2 n i =1 ( x i - x ) = 0 for x * = n P i =1 x i n and ∂f ∂y = - 2 n i =1 ( y i - y ) = 0 for y * = n P i =1 y i n . Now H ( x, y ) = 2 n 0 0 2 n . Since H is positive definite f is convex and therefore ( x * , y * ) is the global minimum. 2. We wish to maximize f ( q 1 , q 2 ) = 2( q 1 / 3 1 + q 2 / 3 2 ) - q 1 - 1 . 5 q 2 . ∂f ∂q 1 = 2 3 q - 2 / 3 1 - 1 = 0 for q * 1 = ( 2 3 ) 2 / 3 and ∂f ∂q 2 = 4 3 q - 1 / 3 1 - 1 . 5 = 0 for q * 2 = ( 8 9 ) 3 . The objective function is the sum of concave functions, so we know that the above values for q * 1 and q * 2 will maximize profit. 3. Let f ( q 1 , q 2 ) be the profit when company 1 sells q 1 units and company 2 sells q 2 units. Then f ( q 1 , q 2 ) = (200 - q 1 - q 2 )( q 1 + q 2 ) - q 1 - 0 . 5 q 2 2 . ∂f ∂q 1 = 199 - 2( q 1 + q 2 ) and ∂f ∂q 2 = 200 - 3 q 2 - 2 q 1 . f = 0 for q 1 = 98 . 5 and q 2 = 1. Since H ( q 1 , q 2 ) = - 2 - 2 - 2 - 3 and H is negative definite, we know that (98.5, 1) maximizes profit over all values of q 1 and q 2 .

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HW5Sol - Fall 2009 IEOR 160 Industrial Engineering...

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