<p>It's probably easiest to insert everything into Utility <span class="math-container">$U = U(R)$</span>, such that it depends only on one choice variable:</p>

<p><span class="math-container">$U(R) = 48R + Rr(L^{*}-R)-R^2$</span></p>

<p>Deriving w.r.t. <span class="math-container">$R$</span> yields the first order condition:</p>

<p><span class="math-container">$48 - Rr + r(L^{*}-R)-2R \overset{!}{=} 0$</span></p>

<p>Rearranging yields leisure in terms of the wage rate:
<span class="math-container">$R(r) = \frac{48-L^{*}r}{3+r}$</span></p>

<p>Labor Supply <span class="math-container">$l(r)$</span> is hence given by:
<span class="math-container">$l(r) = \frac{3L^{*}+2rL^{*}-48}{3+r}$</span></p>

<p>Deriving this by <span class="math-container">$r$</span> should yield what you're looking for:</p>

<p><span class="math-container">$\frac{d l}{d r} = \frac{3L^{*} + 16}{(r+3)^2}$</span></p>

CC BY-SA 4.0
2018-10-09T09:21:23.490Z
2018-10-09T09:21:23.490Z
E. Sommer
<p>So I have been given a utility function = <span class="math-container">$ 48 R + Ry -R^2$</span> where <span class="math-container">$R$</span> represents leisure hours and <span class="math-container">$y$</span> represents labour income. <span class="math-container">$y=rl$</span>, <span class="math-container">$r$</span> is wage rate and <span class="math-container">$l$</span> is labour hours. Find hours worked increases with wage.
So what I thought I can do was:</p>

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

<p>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.</p>

CC BY-SA 4.0
2018-10-09T08:42:52.203Z
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
CC BY-SA 4.0
2018-10-09T09:04:07.990Z
But I considered $R = L*-l$ where $L*$ is maximum possible hours, some constant
Sumukh Sai
<labor-economics>
Hours worked increases with wage
Sumukh Sai
# Hours worked increases with wage

(Asked by: username_0 on 2018-10-08)

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.

**Tags:** labor-economics

> **Comments:**
> **username_1**: 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?  
> **username_0**: But I considered $R = L*-l$ where $L*$ is maximum possible hours, some constant  

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## 1 answers

### Answer by username_1 on 2018-10-09. Score: 1 | Accepted answer

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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