PS2 - Econ 310 Problem Set 2 Lucas Manuelli October 5 2008...

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Econ 310 Problem Set 2 Lucas Manuelli October 5, 2008 Exercise 1 (a) I = 50 H 25000 (b) We simply need to maximize U ( I;H ) = U (50 H 25000 ;H ) = ln(50 H 25000) + 2 ln(5000 H ) taking the derivative and setting equal to zero we see that U 0 = 50 50 H 25000 2 5000 H = 0 Solving for H we get H = 2000 and thus I = 50(2000) 25000 = 75000 : This is truly a max because U 00 = 2500 (50 H 25000) 2 2 (5000 H ) 2 which is always negative. (c) Recomputing for $40 per hour we get that I = 40 H 25000 so U ( I;H ) = U (54 H 25000 ;H ) = ln(40 H 25000) + 2 ln(5000 H ) taking the derivative and setting equal to zero we see that U 0 = 40 40 H 25000 2 5000 H = 0 solving for H we get H = 2083 : 3 and I = 58333 : 3 : This is truly a max because U 00 = 1600 (40 H 25000) 2 2 (5000 H ) 2 which is always negative. 1
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The economic intuition for this result comes from the Slutsky equation. If we let L be the hours of leisure, we have L = 5000 H: So now utility is a function of L and I: Let W be your wage. U 0 ( I;L ) = ln I + 2 ln L subject to the constraint I + WL = 5000 W 25 ; 000 In our example W; the price of leisure, decreased. Using the Slutsky equa- tion the substitution e/ect is positive, since if we were able to stay at the same utility level, a decrease in price would mean an increase in consump- tion. On the other hand, the income e/ect is negative because lowering W also had the e/ect of lowering our purchasing power, thus we have less money to spend on L: In this case the negative income e/ect overpowers the positive substitution e/ect and causes a decrease in consumption. Excercise 2 (a) Our budget constraint is p x x + p y y = I so the Lagrangian is L ( ) = 4 ln( x + 2) + ln( y 1) + ( I p x x p y y ) @ L @x = 4 x + 2 x = 0 @ L @y = 1 y 1 y = 0 @ L = I
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PS2 - Econ 310 Problem Set 2 Lucas Manuelli October 5 2008...

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