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In this static model with no savings/intertemporal aspect, labor supply does not depend on the wage. Using the equality relations, we can maximize over $L^s$ without constraints

$$\max_{L^s} U = \left(\frac{L^s\cdot W}{P}\right)^a\cdot (T-L^s)^{1-a}$$

$$\frac {\partial U}{\partial L^s} = \frac a{L^s}\cdot U - \frac {1-a}{T-L^s}\cdot U$$

and setting this equal to zero we get the optimal $L^s$

$$\frac a{L^s}= \frac {1-a}{T-L^s} \implies L^s = aT$$

One can verify that, at the optimum, the second derivative of the Utility function is negative so we do have a maximum.

In this static model with no savings/intertemporal aspect, labor supply does not depend on the wage. Using the equality relations, we can maximize over $L^s$ without constraints

$$\max_{L^s} U = \left(\frac{L^s\cdot W}{P}\right)^a\cdot (T-L^s)^{1-a}$$

$$\frac {\partial U}{\partial L^s} = \frac a{L^s}\cdot U - \frac {1-a}{T-L^s}\cdot U$$

and setting this equal to zero we get the optimal $L^s$

$$\frac a{L^s}= \frac {1-a}{T-L^s} \implies L^s = aT$$

One can verify that the optimum the second derivative of the Utility function is negative so we do have a maximum.

In this static model with no savings/intertemporal aspect, labor supply does not depend on the wage. Using the equality relations, we can maximize over $L^s$ without constraints

$$\max_{L^s} U = \left(\frac{L^s\cdot W}{P}\right)^a\cdot (T-L^s)^{1-a}$$

$$\frac {\partial U}{\partial L^s} = \frac a{L^s}\cdot U - \frac {1-a}{T-L^s}\cdot U$$

and setting this equal to zero we get the optimal $L^s$

$$\frac a{L^s}= \frac {1-a}{T-L^s} \implies L^s = aT$$

One can verify that, at the optimum, the second derivative of the Utility function is negative so we do have a maximum.

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In this static model with no savings/intertemporal aspect, labor supply does not depend on the wage. Using the equality relations, we can maximize over $L^s$ without constraints

$$\max_{L^s} U = \left(\frac{L^s\cdot W}{P}\right)^a\cdot (T-L^s)^{1-a}$$

$$\frac {\partial U}{\partial L^s} = \frac a{L^s}\cdot U - \frac {1-a}{T-L^s}\cdot U$$

and setting this equal to zero we get the optimal $L^s$

$$\frac a{L^s}= \frac {1-a}{T-L^s} \implies L^s = aT$$

One can verify that the optimum the second derivative of the Utility function is negative so we do have a maximum.