So I have been given a utility function = $ 48 R + Ry -R^2$ where $R$ represents leisure hours and $y$ represents labour income. $y=rl$, $r$ is wage rate and $l$ is labour hours. Find hours worked increases with wage. So what I thought I can do was:

I equate MRS = Slope of Budget constraint and I got something like this $(48 + y - 2R)/ R = r $ Then I equated in $R$ and got $R=48+(rl)/r+2$ Can I just now do differential of $R$ with respect to $r$

Can I continue with this method? Have I made some mistake so far? Or is this method completely wrong and is there some other method.

  • $\begingroup$ In my view, this can't be solved as the relationship between $R$ and $l$ is not given, i.e. what's the total amount of hours that are shared between work and leisure? $\endgroup$ – E. Sommer Oct 9 '18 at 8:42
  • $\begingroup$ But I considered $R = L*-l$ where $L*$ is maximum possible hours, some constant $\endgroup$ – Sumukh Sai Oct 9 '18 at 9:04

It's probably easiest to insert everything into Utility $U = U(R)$, such that it depends only on one choice variable:

$U(R) = 48R + Rr(L^{*}-R)-R^2$

Deriving w.r.t. $R$ yields the first order condition:

$48 - Rr + r(L^{*}-R)-2R \overset{!}{=} 0$

Rearranging yields leisure in terms of the wage rate: $R(r) = \frac{48-L^{*}r}{3+r}$

Labor Supply $l(r)$ is hence given by: $l(r) = \frac{3L^{*}+2rL^{*}-48}{3+r}$

Deriving this by $r$ should yield what you're looking for:

$\frac{d l}{d r} = \frac{3L^{*} + 16}{(r+3)^2}$


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