420KSpring10MT1answer-part2

420KSpring10MT1answer-part2 - (3) can be written as: B =...

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Answers for Midterm 1 1. Transitivity. 2. u represents < if x < y () u ( x ) u ( y ) : MU 1 = 2 x 1 MU 2 = 3 x 2 Then apply the formula MRS = MU 1 =MU 2 MRS = 2 x 2 3 x 1 b) By tangency condition: MRS = p 1 p 2 using what we found in part a" 2 x 2 3 x 1 = p 1 p 2 x 1 = 2 x 2 p 2 3 p 1 (1) we need another equation which is budget equation: p 1 x 1 + p 2 x 2 = m (2) plug (1) into (2) : 2 x 2 p 2 3 + p 2 x 2 = m ) x 2 = 3 m 5 p 2 (3) plug (3) into (1) and get x 1 = 2 m 5 p 1 4. a) PV = 300 + 400 1 : 04 = 684 : 62 b) PV = 300 + 300 1 : 08 + 300 (1 : 08) 2 = 300 ± ( 1 1 1 1 : 08 ) = 4050 : 0 5. MU 1 = 1 x 1 MU 2 = 0 : 9 x 2 MRS = x 2 0 : 9 x 1 MRS = 1 + r ) x 2 0 : 9 x 1 = 1 : 05 ) x 2 = 1 : 05 ± 0 : 9 ± x 1 (1) 1 x 1 + 1 1 : 05 x 2 = 200 + 400 1 : 05 = 580 : 95 (2) Plug (1) into (2) : x 1 + 0 : 9 x 1 = 580 : 95 ) x 1 = 580 : 95 1 : 9 = 305 : 76 so she borrows 305 : 76 ² 200 = 105 : 76 p CE = 0 : 3 ± p 3600 + 0 : 7 ± p 2500 = 53 : 0 CE = 53 2 = 2809 7. max 0 : 75 ln x 1 + 0 : 25 ln x 2 x 1 = 1 : 1 A + 0 : 95 B (1) x 2 = 0 : 7 A + 1 : 15 B (2) A + B = 100 (3)
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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.

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420KSpring10MT1answer-part2 - (3) can be written as: B =...

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