The question:

A price-taking farmer produces a crop with labor L as the only input. His production function is:$$F(L) = 10L^{1/2} − 2L$$

He has 4 units of labor in his family and he cannot hire labor from the wage labor market. He does not face any cost of employing family labor.

a. Find out his equilibrium level of output.

b. Suppose that the government imposes an income tax at the rate of 10 per cent. How does this affect his equilibrium output?

c. Suppose an alternative production technology given by: $$F(L) = 11 L^{1/2} − L −15$$ is available. Will the farmer adopt this alternative technology?

My approach:

a. As the cost is 0 and the farmer is price taker, he will try to maximize his production, by FOC differentiating F(L) we L=6.25 where function is maximum. But he can employ L=4, hence his equilibrium level of output will be 4

b. Profit when he is taxed $$R(P,L)=PF(L)-0.1PF(L)=.9PF(L)$$ again he will maximize his production and by FOC L=6.25 but he will employ only L=4 because it's his maximum limit for L ,so equilibrium out will not change even after tax.

c. he will not adopt new technology because if we put L=4 in new production function then F(L)=3 which is less than production from his earlier technology.

Is my approach correct for all three parts?

up vote 3 down vote accepted

Given the production function \begin{eqnarray*} F(L) = 10 \sqrt{L} - 2L \end{eqnarray*} marginal product is \begin{eqnarray*} F'(L) = \frac{5}{ \sqrt{L}} - 2 \end{eqnarray*} Since there is no cost for the family for using its own labor and it cannot hire from outside, family will maximize its production subject to the constraint that it has 4 units of labor. \begin{eqnarray*} \max_{L\geq 0} && 10 \sqrt{L} - 2L \\ \text{s.t.} && L \leq 4 \end{eqnarray*} Given that $F'(L)$ is falling in $L$ and is positive at $L = 4$ $(F'(4) = 0.5)$, family will use all 4 units in optimum. The corresponding output is $10\sqrt{4} - 2(4) = 12$.

With $10\%$ income tax, the objective of the firm will be to maximize total output net of tax which is equal to $0.9 F(L)$. Solution to this problem will stay the same as before because we are just multiplying the objective by a positive constant.

\begin{eqnarray*} \max_{L\geq 0} && 0.9(10 \sqrt{L} - 2L) \\ \text{s.t.} && L \leq 4 \end{eqnarray*}

Therefore, the equilibrium output is still $10\sqrt{4} - 2(4) = 12$ out of which 1.2 is paid as taxes and 10.8 is consumed.

With new technology the problem is \begin{eqnarray*} \max_{L\geq 0} && 11 \sqrt{L} - L - 15\\ \text{s.t.} && L \leq 4 \end{eqnarray*}

The above problem again give $L = 4$ as the solution and the corresponding equilibrium output is $3$ which is way less than what the family had earlier. Therefore, it will not use this technology.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.