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The Problem the Individual Solves to Derive the Labor Supply maxc,LU(c,L)=logc1/21ˉL1/3 subject to c+Sˉ=wL\max _ { c , L } U ( c , L ) = \log c - 1 / 2 \frac { 1 } { \bar { \ell } } L ^ { 1 / 3 } \text { subject to } c + \bar { S } = w L

Question 94

Multiple Choice

The problem the individual solves to derive the labor supply is:


A) maxc,LU(c,L) =logc1/21ˉL1/3 subject to c+Sˉ=wL\max _ { c , L } U ( c , L ) = \log c - 1 / 2 \frac { 1 } { \bar { \ell } } L ^ { 1 / 3 } \text { subject to } c + \bar { S } = w L
B) maxctody cfarr U(ctoduy cfuture ) =logctoday +βlogcfuture  subject toc ctoday +cfuture 1+R=Wˉ\max _ { c _ { \text {tody } } c _ { \text {farr } } } U \left( c _ { \text {toduy } } c _ { \text {future } } \right) = \log c _ { \text {today } } + \beta \log c _ { \text {future } } \text { subject toc } c _ { \text {today } } + \frac { c _ { \text {future } } } { 1 + R } = \bar { W }
C) maxcU(c) =logc subject to Pc=wL\max _ { c }U ( c ) = \log c \text { subject to } P c = w L

D) maxc,tU(c,ˉ) =αlogc+(1α) ˉ subject to cwˉ=wL\max _ { c , t } U ( c , \bar { \ell } ) = \alpha \cdot \log c + ( 1 - \alpha ) \cdot \bar { \ell } \text { subject to } c - w \bar { \ell } = w L
E) maxk,LU(K,L) =AˉK13L23 subject to C=rK+wL\max _ { k , L } U ( K , L ) = \bar { A } K ^ { \frac { 1 } { 3 } } L ^ { \frac { 2 } { 3 } } \text { subject to } C = r K + w L

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