Consumption and Leisure
First we'll cover a consumer making a consumption and leisure choice. Recall that they choose a certain amount of hours to work $h$ and use the remaining time for leisure $\ell$ so that
h + \ell = 1
Their income consists of labor income $wh$, income from capital gains $\pi$, and they must pay taxes $T$. Thus their consumption will be
c = w h + \pi - T
They have utility over consumption and leisure given by
u(c,\ell) = \log(c) + \eta \log(\ell)
Here $\eta$ reflects how much the agent cares about leisure relative to consumption.
This problem can be reduced to a choice of hours worked, since from that we can find both consumption and leisure directly using the above equations. The utility gained by working a certain amount $h$ is
v(h) = \log(wh+\pi-T) + \eta \log(1-h)
To find the optimal choice of $h$ we must maximize this function, which we do by finding where its derivative is equal to zero
\align v^{\prime}(h) = \frac{w}{wh+\pi-T} - \frac{\eta}{1-h} = 0
Solving this equation for $h$ we find
\Rightarrow \align \frac{w}{wh+\pi-T} = \frac{\eta}{1-h} \\
\Rightarrow \align w(1-h) = \eta(wh+\pi-T) \\
\Rightarrow \align w + \eta(\pi-T) = \eta w h + w h \\
\Rightarrow \align w + \eta(\pi-T) = (1+\eta) wh \\
\Rightarrow \align h = \frac{w + \eta (\pi-T)}{(1+\eta) w} \\
\Rightarrow \align h = \frac{1}{1+\eta} - \frac{\eta}{1+\eta} \left( \frac{\pi-T}{w} \right)
Knowing the optimal choice for $h$, we can now go through and find $c$ and $\ell$. First for consumption
c \align = w h + \pi - T \\
\align = w \left[ \frac{1}{1+\eta} - \frac{\eta}{1+\eta} \left( \frac{\pi-T}{w} \right) \right] + \pi - T \\
\align = \frac{1}{1+\eta} w - \frac{\eta}{1+\eta} (\pi-T) + (\pi-T) \\
\align = \frac{1}{1+\eta} w + \left[ 1 - \frac{\eta}{1+\eta} \right] (\pi-T) \\
\align = \frac{1}{1+\eta} w + \frac{1}{1+\eta} (\pi-T) \\
\align = \frac{1}{1+\eta} (w+\pi-T)
And for leisure we find
\ell \align = 1 - h \\
\align = 1 - \left[ \frac{1}{1+\eta} - \frac{\eta}{1+\eta} \left( \frac{\pi-T}{w} \right) \right] \\
\align = \left[ 1 - \frac{1}{1+\eta} \right] + \frac{\eta}{1+\eta} \left( \frac{\pi-T}{w} \right) \\
\align = \frac{\eta}{1+\eta} + \frac{\eta}{1+\eta} \left( \frac{\pi-T}{w} \right) \\
\align = \frac{\eta}{1+\eta} \left( 1 + \frac{\pi-T}{w} \right) \\
\align = \frac{\eta}{1+\eta} \left( \frac{w+\pi-T}{w} \right)
See the lecture notes for further interpretation.
Alternative Utility Function
I'm feeling kind of wacky, so let's try another utility function and see what happens. For now let's say that there is no base income so that $\pi-T=0$. So then the budget constraint is simply $c=wh$. Let the new utility function be
u(c,\ell) = \sqrt{c} + \sqrt{\eta\ell}
Again here $\eta$ is some number greater than zero that denotes how much the agent cares about leisure. Proceeding as before
v(h) = \sqrt{wh} + \sqrt{\eta(1-h)}
Taking the derivative, we find
v^{\prime}(h) = \frac{1}{2}\sqrt{\frac{w}{h}} - \frac{1}{2}\sqrt{\frac{\eta}{1-h}} = 0
Solving this for $h$ yields
\Rightarrow \align \frac{1}{2}\sqrt{\frac{w}{h}} = \frac{1}{2}\sqrt{\frac{\eta}{1-h}} \\
\Rightarrow \align \sqrt{\frac{w}{h}} = \sqrt{\frac{\eta}{1-h}} \\
\Rightarrow \align \frac{w}{h} = \frac{\eta}{1-h} \\
\Rightarrow \align w (1-h) = \eta h \\
\Rightarrow \align w = \eta h + wh \\
\Rightarrow \align w = (\eta + w) h \\
\Rightarrow \align h = \frac{w}{w+\eta}
which makes a lot of sense if you think about it. Mapping this into the other variables of interest, we find
c = wh = \frac{w^2}{w+\eta} \qquad \text{and} \qquad \ell = 1 - h = \frac{\eta}{w+\eta}
Contrary to the case with log utility, here we find that hours worked always increases with the wage rate $w$.
Firm Production
Here we are studying the profit maximization problem of a firm that is choosing how much labor $h$ to employ. Given a certain amount of capital $k$ and labor $h$, assume that output $y$ is given by
y = z f(k,h) = z k^{\alpha} h^{1-\alpha}
where $\alpha \in (0,1)$ is a number denoting how important capital is for production (relative to labor). With this, the profit of the firm will be
\pi(h) = z k^{\alpha} h^{1-\alpha} - w h
To maximize this we find the derivative and set it equal to zero
\pi^{\prime}(h) = z k^{\alpha} (1-\alpha) h^{-\alpha} - w = 0
Solving for the optimal $h$ we find
\Rightarrow \align z (1-\alpha) k^{\alpha} h^{-\alpha} = w \\
\Rightarrow \align z (1-\alpha) \left(\frac{k}{h}\right)^{\alpha} = w \\
\Rightarrow \align \left(\frac{k}{h}\right)^{\alpha} = \frac{w}{z(1-\alpha)} \\
\Rightarrow \align \left(\frac{h}{k}\right)^{\alpha} = \frac{z(1-\alpha)}{w} \\
\Rightarrow \align \frac{h}{k} = \left[\frac{z(1-\alpha)}{w}\right]^{\frac{1}{\alpha}} \\
\Rightarrow \align h = k \left[\frac{z(1-\alpha)}{w}\right]^{\frac{1}{\alpha}}
We can then find the output implied by this value
y \align = z k^{\alpha} h^{1-\alpha} \\
\align = z k^{\alpha} \left(k\left[\frac{z(1-\alpha)}{w}\right]^{\frac{1}{\alpha}}\right)^{1-\alpha} \\
\align = z k^{\alpha} k^{1-\alpha} z^{\frac{1-\alpha}{\alpha}} \left(\frac{1-\alpha}{w}\right)^{\frac{1-\alpha}{\alpha}} \\
\align = k z^{\frac{1}{\alpha}} \left(\frac{1-\alpha}{w}\right)^{\frac{1-\alpha}{\alpha}}
So we can see that both labor utilization and output are both linearly proportional to the amount of capital. Finally we can calculate the firm's profit
\pi \align = z k^{\alpha} h^{1-\alpha} - wh = y - wh \\
\align = k z^{\frac{1}{\alpha}} \left(\frac{1-\alpha}{w}\right)^{\frac{1-\alpha}{\alpha}} - w k \left[\frac{z(1-\alpha)}{w}\right]^{\frac{1}{\alpha}} \\
\align = k z^{\frac{1}{\alpha}} \left(\frac{1-\alpha}{w}\right)^{\frac{1-\alpha}{\alpha}} - (1-\alpha) k z^{\frac{1}{\alpha}} \left(\frac{1-\alpha}{w}\right)^{\frac{1-\alpha}{\alpha}} \\
\align = \alpha k z^{\frac{1}{\alpha}} \left(\frac{1-\alpha}{w}\right)^{\frac{1-\alpha}{\alpha}} = \alpha y
Thus firm profits end up being some fraction $\alpha$ of output $y$. The remaining fraction goes to labor income, since $wh = (1-\alpha) y$.
Production Equilibrium
For this section, assume that there is no government taxes or spending, so that $T = 0$. Now we can combine the results from
One way to get around this is to notice that since $\pi = \alpha y$ and $wh = (1-\alpha)y$, combining these yields
\frac{\pi}{wh} = \frac{\alpha}{1-\alpha}
Plugging this into the consumers labor supply function we find
\align h = \frac{1}{1+\eta} - \frac{\eta}{1+\eta} \left(\frac{\pi}{w}\right) \\
\Rightarrow \align h = \frac{1}{1+\eta} - \frac{\eta}{1+\eta} \left(\frac{\alpha}{1-\alpha}\right)h \\
\Rightarrow \align (1-\alpha) (1+\eta) h = (1-\alpha) - \eta \alpha h \\
\Rightarrow \align (1-\alpha+\eta) h = 1 - \alpha \\
\Rightarrow \align h^{\ast} = \frac{1-\alpha}{1-\alpha+\eta}
Finally, we can solve for the equilibrium wage using the firm's labor demand function
\align h^D(w) = h^{\ast} \\
\Rightarrow \align k \left[\frac{z(1-\alpha)}{w}\right]^{\frac{1}{\alpha}} = \frac{1-\alpha}{1-\alpha+\eta} \\
\Rightarrow \align k^{\alpha} \frac{z (1-\alpha)}{w} = \left(\frac{1-\alpha}{1-\alpha+\eta}\right)^{\alpha} \\
\Rightarrow \align w^{\ast} = (1-\alpha+\eta)^{\alpha} (1-\alpha)^{1-\alpha} z k^{\alpha}
Cost Minimization
In problem set 1, I ask about cost minimization. Here the task is to produce a certain amount $y$ at minimal cost. There costs come in the form of both labor $h$ at rate $w$ and capital $k$ at rate $r$. Thus the problem is given by
\min_{k,\ell} \quad \align rk + wh \\
\text{where} \quad \align y = z k^{\alpha} h^{1-\alpha}
As with the consumption and leisure problem, we can either use Lagrangian techniques, or we can substitute the constraint into the objective to turn it into a single variable optimization. So for instance to produce a certain amount $y$ using $k$ capital, we would need
\align y = z k^{\alpha} h^{1-\alpha} \\
\Rightarrow \align h^{1-\alpha} = \frac{y}{z k^{\alpha}} \\
\Rightarrow \align h = \left(\frac{y}{z k^{\alpha}}\right)^{\frac{1}{1-\alpha}} \\
\Rightarrow \align h = \left(\frac{y}{z}\right)^{\frac{1}{1-\alpha}} \left(\frac{1}{k}\right)^{\frac{\alpha}{1-\alpha}}
And thus it would suffice to minimize
\min_k \quad r k + w \left(\frac{y}{z}\right)^{\frac{1}{1-\alpha}} \left(\frac{1}{k}\right)^{\frac{\alpha}{1-\alpha}}
over $k$ and then back out the value of $h$ afterwards.