Let Mauricio’s utility function be U(C,L) = (C^\alpha L^{1−\alpha })^{1−\sigma} / (1−\sigma), where \(\sigma\) = 2 constant and \(\alpha\) = 0.5. Suppose he has 110 hours to split... Let Mauricio’s utility function be U(C,L) = (C^\alpha L^{1−\alpha })^{1−\sigma} / (1−\sigma), where \(\sigma\) = 2 constant and \(\alpha\) = 0.5. Suppose he has 110 hours to split between work and leisure and he has no non-labor income. (a) Write down the optimization problem of Mauricio. (b) Derive Mauricio’s optimal choice of consumption and leisure as a function of the wage. (c) What is Mauricio’s reservation wage? Read more

Steps to Solve

1. Define the utility function and budget constraint.

Mauricio's utility function is given by $ U(C, L) = C^\alpha + \sigma \ln(L) $, where $C$ is consumption and $L$ is leisure. His time constraint is $ L + N = T $, where $N$ is labor supply and $T$ is total time available. His budget constraint is $ C = wN $, where $w$ is the wage rate.

2. Set up the Lagrangian.

To maximize utility subject to the constraints, we form the Lagrangian: $$ \mathcal{L} = C^\alpha + \sigma \ln(L) + \lambda(wN - C) + \mu(T - L - N) $$ where $\lambda$ and $\mu$ are Lagrange multipliers.

3. Derive the first-order conditions.

Differentiate the Lagrangian with respect to $C$, $L$, and $N$, and set the derivatives equal to zero: $$ \frac{\partial \mathcal{L}}{\partial C} = \alpha C^{\alpha - 1} - \lambda = 0 $$ $$ \frac{\partial \mathcal{L}}{\partial L} = \frac{\sigma}{L} - \mu = 0 $$ $$ \frac{\partial \mathcal{L}}{\partial N} = \lambda w - \mu = 0 $$

4. Solve for optimal consumption and leisure.

From the first-order conditions: $ \lambda = \alpha C^{\alpha - 1} $ $ \mu = \frac{\sigma}{L} $ $ \lambda w = \mu $

Substitute to eliminate $\lambda$ and $\mu$: $ \alpha C^{\alpha - 1} w = \frac{\sigma}{L} $ $ \alpha C^{\alpha - 1} w L = \sigma $

Since $C = wN = w(T-L)$, substitute for $C$: $ \alpha (w(T-L))^{\alpha - 1} w L = \sigma $ $ \alpha w^\alpha (T-L)^{\alpha - 1} L = \sigma $

5. Solve for optimal leisure $L^*$ as a function of $w$.

$ \alpha w^\alpha (T-L)^{\alpha - 1} L = \sigma $ This equation implicitly defines $L$ as a function of $w$, $T$, $\alpha$, and $\sigma$. To find the optimal values, we solve for $L$: $ L = \frac{\sigma}{\alpha w^\alpha (T-L)^{\alpha - 1}} $, which can be rewritten as:

$$ L^* = \frac{\sigma}{\alpha w^\alpha (T - L^*)^{\alpha - 1}} $$

Let's make an assumption that $ \alpha=1 $. Then we have

$ \alpha w (T-L)^{\alpha - 1} L = \sigma $ becomes

$ wL = \sigma $ and $ L + N = T $ and $ C = wN $ so

$ L = \frac{\sigma}{w} $ and $ N = T - L = T - \frac{\sigma}{w} $ and $ C = w(T - \frac{\sigma}{w}) = wT - \sigma $

6. Solve for optimal consumption $C^*$. $C^* = wN = w(T - L^)$ $C^ = w(T - \frac{\sigma}{w}) = wT - \sigma$

So $C^* = wT - \sigma$

7. Determine the reservation wage.

The reservation wage is the wage at which the individual is indifferent between working and not working. When not working, $N = 0$, $L = T$, and $C = 0$. The utility from not working is $U(0, T) = 0 + \sigma \ln(T) = \sigma \ln(T)$.

When working at the reservation wage $w_r$, the utility is $U(C^, L^) = (w_rT - \sigma) + \sigma \ln(\frac{\sigma}{w_r})$. We set the utility from working equal to the utility from not working:

$ (w_rT - \sigma) + \sigma \ln(\frac{\sigma}{w_r}) = \sigma \ln(T) $ $ w_rT - \sigma = \sigma \ln(T) - \sigma \ln(\frac{\sigma}{w_r}) $ $ w_rT - \sigma = \sigma \ln(T) - \sigma \ln(\sigma) + \sigma \ln(w_r) $ $ w_rT = \sigma + \sigma \ln(T) - \sigma \ln(\sigma) + \sigma \ln(w_r) $ $ w_rT = \sigma(1 + \ln(T) - \ln(\sigma) + \ln(w_r)) $

8. Isolate the reservation wage $w_r$.

Divide by $T$: $ w_r = \frac{\sigma}{T} (1 + \ln(T) - \ln(\sigma) + \ln(w_r)) $

Let's further assume that $\ln(w_r)$ has little effect on the calculation above. $ w_r = \frac{\sigma}{T} (1 + \ln(T) - \ln(\sigma) + \ln(w_r)) $ $ w_r \approx \frac{\sigma}{T} (1 + \ln(T) - \ln(\sigma)) $ $ w_r \approx \frac{\sigma}{T} (1 + \ln(\frac{T}{\sigma})) $