# Hours worked increases with wage

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.

• 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? – E. Sommer Oct 9 '18 at 8:42
• But I considered $R = L*-l$ where $L*$ is maximum possible hours, some constant – 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:

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