# ps3a - Problem Set#3Answer Key 1 Basics Consider the...

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Unformatted text preview: Problem Set #3Answer Key 1 Basics Consider the utility function u = x α y 1- α . Answer: For this utility function, find: 1. Demand functions. Set up the Lagrangian: L = x α y 1- α + λ ( I- p x x- p y y ) This gives the first order conditions: αx α- 1 y 1- α- λp x = 0 (1- α ) x α y- α- λp y = 0 I- p x x- p y y = 0 The first two equations together imply that αy (1- α ) x = p x p y , which also means that y = (1- α ) p x αp y x . Plugging this into the last equation and solving gives the answers: x * = αI p x , y * = (1- α ) I p y 2. Compensated Demand functions. Answer: for this case, we set up the lagrangian “backwards:” L = p x x + p y y + λ ( u- x α y 1- α ) This gives the first order conditions: ∂L ∂x = p x- λαx α- 1 y 1- α = 0 ∂L ∂y = p y- λ (1- α ) x α y- α = 0 ∂L ∂λ = u- x α y 1- α = 0 1 The first two equations together once again imply that αy (1- α ) x = p x p y- they are identical to those from the previous problem!. Thus, we once again have y = (1- α ) p x αp y x . Plugging this into the last equation gives: u = x α parenleftbigg (1- α ) p x αp y x parenrightbigg 1- α Simplifying gives: x c = u parenleftbigg αp y (1- α ) p x parenrightbigg 1- α Repeating the same steps for y gives: y c = u parenleftbigg (1- α ) p x αp y parenrightbigg α 3. The Expenditure function. Answer: to find the expenditure function, we simply use our compensated demand functions from the previous problem in the expression: E = p x x c + p y y c Plugging in gives: E = p x u parenleftbigg αp y (1- α ) p x parenrightbigg 1- α + p y u parenleftbigg (1- α ) p x αp y parenrightbigg α This can be simplified to read: E = up α x p 1- α y bracketleftBigg parenleftbigg α 1- α parenrightbigg 1- α + parenleftbigg 1- α α parenrightbigg α bracketrightBigg 4. The Indirect Utility function. Answer: Here, we plug the regular demand functions back into the utility function, with result: V...
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ps3a - Problem Set#3Answer Key 1 Basics Consider the...

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