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Unformatted text preview: (3) can be written as: B = 100 A (4) plug (4) into (1) and (2) x 1 = 1 : 1 A + 0 : 95(100 A ) = 0 : 15 A + 95 (5) x 2 = 0 : 7 A + 1 : 15(100 A ) = 115 : 45 A (6) 1 maximization problem can be written as max 0 : 75 & ln(0 : 15 & A + 95) + 0 : 25 & ln(115 : 45 & A ) take derivative wrt A and equalize to 0 : 75 & : 15 : 15 & A +95 = : 25 & : 45 115 : 45 & A ) : 15 & A + 95 = 115 : 45 & A ) : 6 A = 20 A = 20 = : 6 = 33 : 333 B = 66 : 666 2...
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This note was uploaded on 10/25/2010 for the course ECO 420K taught by Professor D during the Spring '10 term at University of Texas at Austin.
 Spring '10
 d

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