u-max-demo

# Finding the mrs from utility functions example find

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Finding the MRS from Utility functions EXAMPLE: Find the total di ff erential for the following utility functions 1. U ( x 1 , x 2 ) = ax 1 + bx 2 where ( a, b > 0) 2. U ( x 1 , x 2 ) = x 2 1 + x 3 2 + x 1 x 2 3. U ( x 1 , x 2 ) = x a 1 x b 2 where ( a, b > 0) 4. U ( x 1 , x 2 ) = α ln c 1 + β ln c 2 where ( α , β > 0) Answers: 1. U x 1 = U 1 = a U x 2 = U 2 = b and dU = U 1 dx 1 + U 2 dx 2 = adx 1 + bdx 2 = 0 If we rearrange to get dx 2 /dx 1 dx 2 dx 1 = U x 1 U x 2 = U 1 U 2 = a b The MRS is the Absolute value of dx 2 dx 1 : MRS = a b 2. U x 1 = U 1 = 2 x 1 + x 2 U x 2 = U 2 = 3 x 2 2 + x 1 and dU = U 1 dx 1 + U 2 dx 2 = (2 x 1 + x 2 ) dx 1 + (3 x 2 2 + x 1 ) dx 2 = 0 2

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Find dx 2 /dx 1 dx 2 dx 1 = U 1 U 2 = (2 x 1 + x 2 ) (3 x 2 2 + x 1 ) The MRS is the Absolute value of dx 2 dx 1 : MRS = (2 x 1 + x 2 ) (3 x 2 2 + x 1 ) iii) U x 1 = U 1 = ax a 1 1 x b 2 U x 2 = U 2 = bx a 1 x b 1 2 and dU = ³ ax a 1 1 x b 2 ´ dx 1 + ³ bx a 1 x b 1 2 ´ dx 2 = 0 Rearrange to get dx 2 dx 1 = U 1 U 2 = ax a 1 1 x b 2 bx a 1 x b 1 2 = ax 2 bx 1 The MRS is the Absolute value of dx 2 dx 1 : MRS = ax 2 bx 1 iv) U c 1 = U 1 = α ³ 1 c 1 ´ dc 1 = ³ α c 1 ´ dc 1 U x 2 = U 2 = β ³ 1 c 2 ´ dc 2 = ³ β c 2 ´ dc 2 and dU = μ α c 1 dc 1 + Ã β c 2 !
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