1. A firm has the following production function: \(Q=2K^{0.2} L^{0.8}\)
Where Q is output, K is capital, L is labor and the firm is subject to the following isocost:
\(C=P_k K +P_1 L\), where: \(P_k=80, P_1=12, C=200,000\)
Find the values of K and L that maximize Q. Having found the solution, set up the problem as one of cost minimization, i.e. what values of K and L minimize cost for the Q you found. What is the interpretation of lambda?
\(200,000=80K+12L\)
Setting up the Lagrangian:
\(\mathcal{L}=2K^{0.2} L^{0.8}+\lambda[200,000-80K-12L]\)
\(\frac{\partial \mathcal{L}}{\partial K}=0=0.4(\frac{L}{K})^{0.8}-80\lambda\)
\(80\lambda = 0.4(\frac{L}{K})^{0.8}\)
\(\frac{\partial \mathcal{L}}{\partial K}=0=1.6(\frac{K}{L})^{0.2}-12\lambda\)
\(12\lambda = 1.6(\frac{K}{L})^{0.2}\)
\(\frac{80\lambda}{12\lambda} = \frac{0.4(\frac{L}{K})^{0.8}}{1.6(\frac{K}{L})^{0.2}}\)
\(\frac{20}{3} = \frac{1}{4}(\frac{L}{K})\)
\(\frac{L}{K} = \frac{80}{3}\)
\(L = \frac{80}{3}K\)
We substitute this result into the constraint: \(200,000=80K+12L\)
\(200,000=80K+12[\frac{80}{3}K]\)
\(400K = 200,000\)
\(K = 500\)
We substitute this result in the constraint to find L:
\(200,000=80*500+12L\)
\(12L = 160,000\)
\(L = \frac{40,000}{3} \approx 13,333.33\)
We plug our results in the production function \(Q=2K^{0.2} L^{0.8}\) to find for Q:
\(Q=2*500^{0.2} (\frac{40,000}{3})^{0.8} \approx 13,828.497\)
We can set the problem as one of cost minimization as follows:
\(\mathcal{L}= 80K+12L+\lambda[Q-2K^{0.2}L^{0.8}]\)
As we know that \(Q \approx 13,828.497\) we can rewrite the Lagrangian as follows (for simplicity we will consider the approximate value of Q as an exact value when we set the Lagrangian):
\(\mathcal{L}= 80K+12L+\lambda[13,828.497-2K^{0.2}L^{0.8}]\)
\(\frac{\partial \mathcal{L}}{\partial K}=0=80-\lambda0.4K^{-0.8}L^{0.8}\)
\(\lambda = 200(\frac{K}{L})^{0.8}\)
\(\frac{\partial \mathcal{L}}{\partial L}=0=12-\lambda1.6K^{0.2}L^{-0.2}\)
\(\lambda = \frac{15}{2}(\frac{K}{L})^{-0.2}\)
\(200(\frac{K}{L})^{0.8} = \frac{15}{2}(\frac{K}{L})^{-0.2}\)
\(\frac{K}{L}= \frac{3}{80}\)
\(K = \frac{3}{80}L\)
Substituting our result in the constraint \(Q=2K^{0.2}L^{0.8}\):
\(Q = 2*500^{0.2} (\frac{40,000}{3})^{0.8} = 2(\frac{3}{80}L)^{0.2}L^{0.8}\)
\(L = (\frac{500}{\frac{3}{80}})^{0.2} (\frac{40,000}{3})^{0.8}\)
\(L = (\frac{40,000}{3})^{0.2} (\frac{40,000}{3})^{0.8}=\frac{40,000}{3}\)
We substitute this result again in the constraint:
\(2*500^{0.2} (\frac{40,000}{3})^{0.8} = 2K^{0.2}(\frac{40,000}{3})^{0.8}\)
\(K=500\)
The values of K and L are the same when we minimize for the Q we found.
Lambda, in the maximization case, is the increase in output given a one-unit relaxation of the budget constraint. In the minimization case, lambda is the increase in cost given a one-unit increase in output.
2. You are given two pieces of land: plot A and plot B. On plot A you raise artichokes for $4.00 per pound and on plot B you raise beans that sell for $2.00 per pound. The production functions for each plot in terms of output per day are:
\(A=10L_a-L_a^2\)
\(B=15L_b-L_b ^2\)
(where \(L_a\) is the amount of labor on plot A and \(L_b\) is the amount of labor on plot B. A represents output of artichokes and B represents beans, both in pounds.)
It is assumed that you work from dawn to dusk, with a 1 hour break, and the average number of daylight hours in the region is 12. Distribute the labor optimally between plots. If labor is paid the value of its marginal product on both plots, what is the hourly wage? Assuming that the wage share of the total output does not exhaust the total product, the remainder is paid on rent. What rent is paid to plot A and plot B?
You work 11 hours a day so \(L_a + L_b = 11\).
We want to maximize the monetary value of production. We can now set the Lagrangian:
\(\mathcal{L} = 4[10L_a-L_a^2]+2[15L_b-L_b^2] +\lambda[11-L_a - L_b]\)
\(\frac{\partial \mathcal{L}}{\partial L_a} = 0 =40 -8L_a -\lambda\)
\(\frac{\partial \mathcal{L}}{\partial L_a} = 0 = 30 -4L_b -\lambda\)
\(40-8L_a = \lambda = 30-4L_b\)
\(L_a = \frac{5}{4} + \frac{1}{2}L_b\)
Substituting this result in the constraint:
\(\frac{5}{4} + \frac{1}{2}L_a + L_a = 11\)
\(1.5L_a = \frac{39}{4}\)
\(L_a = \frac{9}{2}\)
\(\frac{9}{2}+L_b=11\)
\(L_b=\frac{13}{2}\)
Labor is paid its marginal productivity so we first compute the physical marginal product:
\(\frac{\partial A}{\partial L_a} = 10 - 2L_a\)
\(\frac{\partial B}{\partial L_b} = 15 - 2L_b\)
Hence at the optimum the two marginal physical products are :
\(\frac{\partial A}{\partial L_a} = 10 - 2*\frac{9}{2}=1\)
\(\frac{\partial B}{\partial L_b} = 15 - 2*\frac{13}{2} = 2\)
Since artichokes are sold for $4.00 and beans for 2.00$ than the wage received for \(L_a\) is:
\(W_a=4*\frac{\partial A}{\partial L_a} = 4*1 = 4\)
\(W_b=2*\frac{\partial B}{\partial L_b} = 2*2 = 4\)
This is coherent with pure verbal logical. If the marginal productivities were not equal and that therefore the two wages were not equal, the worker could benefit by shifting part of his time from the activity were the marginal productivity is lower to the activity were the marginal productivity is higher.
On total, the worker is paid \(TW=4*\frac{9}{2}+4*\frac{13}{2} = 18 + 26 = 44\).
The total output in monetary terms is equal to:
\(TR = 4[10L_a-L_a^2]+2[15L_b-L_b^2]\)
\(TR = 4[10*\frac{9}{2}-(\frac{9}{2})^2]+2[15*\frac{13}{2}-(\frac{13}{2})^2]\)
\(TR = 180-81+195-\frac{195}{2} = 209.5\)
Total rent is equal to \(TR = TR-TW = 209.5-44=165.5\).
3) A simple economy produces just one product, Q, with two factors of production M (men) and A (acres). There are 100 identical firms, each with production function
\(Q=10M^{\frac{1}{3}}A^{\frac{2}{3}}\)
To the economy as a whole, M is in absolutely inelastic supply at M=900, and A is in absolutely inelastic supply at A=600.
a) Taking the product Q as a numéraire, find the following in equilibrium:
i) The mean and acres employed per firm
Since we have 100 identical firms and since the supply for labor is purely inelastic, the number of men per firm is equal to \(\frac{900}{100} = 9\).
Since we have 100 identical firms and since the supply for acres is purely inelastic, the number of men per firm is equal to \(\frac{600}{100} = 6\).
ii) The marginal product of men and the marginal product of acres.
\(\frac{\partial Q}{\partial M} = \frac{10}{3}M^{-\frac{2}{3}}A^{\frac{2}{3}}\)