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?
\(200,000=80K+12L\) \(\mathcal{L}=2K^{0.2} L^{0.8}+\lambda(200,000-80K-12L)\)
\(\frac{\partial \mathcal{L}}{\partial \lambda}=200,000-80K-12L=0\) \(\frac{\partial \mathcal{L}}{\partial K}=0.4K^{-0.8}L^{0.8}-80\lambda=0\) \(\frac{\partial \mathcal{L}}{\partial L}=1.6K^{0.2}L^{-0.2}-12\lambda=0\)
\(200,000=80K+12L\) \(0.4K^{-0.8}L^{0.8}=80\lambda\) \(1.6K^{0.2}L^{-0.2}=12\lambda\)
\(K=-\frac{3}{20} L+2500\) \(\frac{1}{200}K^{-0.8}L^{0.8}=\lambda\) \(\frac{2}{15}K^{0.2}L^{-0.2}=\lambda\)
\(\frac{1}{200}K^{-0.8}L^{0.8}=\frac{2}{15}K^{0.2}L^{-0.2}\) \(\frac{15}{2}L=200K\) \(15L=400K\) \(K=\frac{3}{80}L\)
\(\frac{3}{80} L=-\frac{3}{20} L+2500\) \(\frac{3}{16} L=2500\) \(L\approx13333.3\) \(K\approx500\) \(Q=13828.49719\)
\(\mathcal{L}=80K +12L+\lambda(13828.49719-2K^{0.2} L^{0.8})\)
\(\frac{\partial C}{\partial \mathcal{L}}=13828.49719-2K^{0.2} L^{0.8}=0\) \(\frac{\partial C}{\partial K}=80+-0.4K^{-0.8} L^{0.8} \lambda=0\) \(\frac{\partial C}{\partial L}=12+-1.6K^{0.2} L^{-0.2} \lambda=0\)
\(\) \(13828.49719=2K^{0.2}L^{0.8}\) \(80=0.4K^{-0.8} L^{0.8} \lambda\) \(12=1.6K^{0.2} L^{-0.2} \lambda\)
\(K^{0.2}L^{0.8}=6914.248595\) \(\frac{80}{0.4K^{-0.8} L^{0.8}}= \lambda\) \(\frac{12}{1.6K^{0.2} L^{-0.2}}= \lambda\)
\(\frac{80}{0.4K^{-0.8} L^{0.8}}=\frac{12}{1.6K^{0.2} L^{-0.2}}\) \(\frac{24}{5}K^{-0.8} L^{0.8}=128K^{0.2} L^{-0.2}\) \(L=\frac{80}{3}K\) \(K^{0.2} (\frac{80}{3} K)^{0.8}=6914.248595\)
\(K=500\) \(L\approx13333.33\) Lambda is the marginal cost of adding an additional unit of 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?
\(A=10L_a-L_a^2\) \(B=15L_b-L_b ^2\) \(11=L_a+L_b\)
\(\mathcal{L}=4(10L_a-L_a^2)+2(15L_b-L_b^2)+\lambda(11-L_a-L_b)\)
\(\frac{\partial \mathcal{L}}{\partial \lambda}=11-L_a-L_b=0\) \(\frac{\partial \mathcal{L}}{\partial L_a}=40-8L_a -\lambda=0\) \(\frac{\partial \mathcal{L}}{\partial L_b}=30-4L_b-\lambda=0\)
\(11=L_a+L_b\) \(\lambda=-8L_a+40\) \(\lambda=30-4L_b\)
\(-8L_a+40=30-4L_b\) \(L_a=\frac{5}{4}+\frac{L_b}{2}\) \(11=\frac{5}{4}+\frac{1}{2} L_b+L_b\)
\(L_b=\frac{13}{2}=6.5\) \(L_a=\frac{9}{2}=4.5\)
\(w=P_aMP_{L_{a}}=P_bMP_{L_{b}}=w_a=w_b\) \(\frac{dA}{dL_a}=10-2L_a=MP_{L_{a}}\) \(\frac{dB}{dL_b}=15-2L_b=MP_{L_b}\)
\(w_a=4(10-2(4.5))=4\) \(w_b=2(15-2(6.5))=4\) The wage rate is $4 per hour.
\(TP_A=Q_A P_A=4(10(4.5)-(4.5)^2)=99\) \(TP_A=w L_a+r_a\) \(99=4(4.5)+r_a\) \(r_a=81\)
\(TP_B=Q_B P_B=2(15(6.5)-(6.5)^2)=110.5\) \(TP_B=w L_b+r_b\) \(110.5=4(6.5)+r_b\) \(r_b=84.5\)
\(r_a +r_b =81+84.5=165.5\) Rent spent on plot A is $81 and rent spent on plot B is $84.5.
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 numeraire, find the following in equilibrium:
i. The men and acres employed per firm.
\(M\) is the total number of men employed in the market. For an individual identical firm, the number of men will be \(\frac{M}{Number of firms}=\frac{900}{100}=m=9\)
\(A\) is the total number of acres used in the market. For an individual identical firm, the number of acres will be \(\frac{A}{Number of firms}=\frac{600}{100}=a=6\)
ii. The marginal product of men and the marginal product of acres.
\(MP_M=\frac{10}{3} M^{- \frac{2}{3}} A^{\frac{2}{3}}=\frac{10}{3} (\frac{2}{3})^\frac{2}{3} \approx 2.5438\) \(MP_A=\frac{20}{3}M^{\frac{1}{3}}A^{- \frac{1}{3}}=\frac{20}{3} (\frac{3}{2})^\frac{1}{3} \approx 7.6314\)
iii. The wage per man and the rent per acre.
\(\frac {W}{M}=w=MP_M Q=\frac{10}{3} \frac{A}{M} ^\frac{2}{3} Q =\frac{10}{3} (\frac{2}{3})^\frac{2}{3} Q \approx 2.5438 Q\) \(\frac{R}{A}=r=MP_A Q=\frac{20}{3} \frac{M}{A} ^\frac{1}{3} Q =\frac{20}{3} (\frac{3}{2})^\frac{1}{3} Q \approx 7.6314 Q\)
iv. The marginal cost.
\(\frac{w}{MP_M}=\frac{r}{MP_A}=\lambda\) \(\frac{w}{MP_A}=\frac{\frac{10}{3} (\frac{2}{3})^\frac{2}{3} Q}{\frac{10}{3} (\frac{2}{3})^\frac{2}{3} Q}=1\) \(\frac{r}{MP_A}=\frac{\frac{20}{3} (\frac{3}{2})^\frac{1}{3} Q}{\frac{20}{3} (\frac{3}{2})^\frac{1}{3} Q}=1\)
The marginal cost is 1 which means that the cost of producing an additional 1Q requires another 1Q.
v. The absolute shares of the two factors.
\(TP=\frac{10}{3} (\frac{2}{3})^\frac{2}{3} Q_M +\frac{20}{3} (\frac{3}{2})^\frac{1}{3} Q_A=\frac{10}{3} (\frac{2}{3})^\frac{2}{3} (900)+\frac{20}{3} (\frac{3}{2})^\frac{1}{3} (600) \approx 2289.4285+4578.8570\approx 6868.2855\)
This shows that the share paid to men is 2289.4285 and the share paid to acres is 4578.8570.
b. Do the same, on the assumption that a minimum wage equal to 5 units of Q has been legally established and enforced. Explain.
i. The men and acres employed per firm.
A minimum wage that is legally established and enforced (therefore no black market wages) ensures that firms will hire workers up to the point where marginal product equals the minimum wage: \(MP_M=w_{min}\) \(w_{min}=5=\frac{10}{3} \frac{A}{M} ^\frac{2}{3} \) \((\frac{3}{2})^\frac{3}{2} M=A\) Individual firms will still rent 6 acres because this number of acres maximizes profit. \(A=600=(\frac{3}{2})^\frac{3}{2} M\) \(M\approx 326.5986\) This means that there will be about 3.2660 men hired per firm or 326 men employed and 574 men unemployed.
ii. The marginal product of men and the marginal product of acres.
\(MP_M=\frac{10}{3} \frac{A}{M} ^\frac{2}{3} Q=\frac{10}{3} \frac{600}{326.5986} ^\frac{2}{3} Q \approx 5Q\) \(MP_A=\frac{20}{3} M^\frac{1}{3} A^{- \frac{1}{3}} Q=\frac{20}{3} (326.5986)^{\frac{1}{3}} (600)^{- \frac{1}{3}} Q\approx 5.443Q\)
iii. The wage per man and rent per acre.
Men will be paid 5Q due to the minimum wage and rent will be 5.443Q because of the lower marginal product of an acre.
iv. The marginal cost.
\(\frac{w}{MP_M}=\frac{r}{MP_A}=\lambda\) \(\frac{w_{min}}{MP_M}=\frac{5Q}{5Q}=1\) \(\frac{r_{new}}{MP_A}=\frac{5.443Q}{5.443Q}=1\) The marginal cost is 1 which means that the cost of producing an additional 1Q requires 1Q.
v. The absolute shares of the two factors.
\(TP=w(Q_M)+r(Q_A)=5(326.5986)+5.443(600)\approx4898.79\) This shows that the share amount paid to men is about 1632.993 and the share amount paid to acres is 3265.8.
When \(\frac{MP}{TP}=share\) \(\frac{MP_M}{TP}=\frac{1632.993}{4898.79}=\frac{1}{3}\) which is the fractional share of the total product paid to men and \(\frac{MP_A}{TP}=\frac{3265.8}{4898.79}=\frac{2}{3}\) is the fractional share of the total product paid to acres.
Because of the minimum wage the total product is less than that without the price floor as both the share paid to men and to acres is lower.
c. What would be the consequences of establishing a maximum wage equal to 3 units of Q? What if the maximum wage were 2? Explain.
If the maximum wage was equal to 3, this price ceiling would not cause a change in the number of men and acres employed since this wage is above the equilibrium wage found in part 3. a. i. which is 2.5438. Thus the answers in sections i.-v. in part a. satisfy the question. The remaining below answers if the maximum wage is 2, a wage below the equilibrium wage rate, causing a shortage of labor.
i. The men and acres per firm.
The quantity of men will still be the point where marginal product equals the maximum wage, that is: \(MP_M=w_{max}=2Q\)
\(2Q=\frac{10}{3} \frac{A}{M} ^\frac{2}{3} Q\) \(M=(\frac{5}{3})^\frac{3}{2} A\) When \(A=600\), \(M=(\frac{5}{3})^\frac{3}{2} (600)\approx1290.994\) The number of men desired per firm is 12, however the market can only provide 9 per firm. Since \(900=(\frac{5}{3})^\frac{3}{2} A\), \(A=418.282\) and each firm will employ 4.183 acres.
ii. The marginal product of men and the marginal product of acres.
\(MP_M=\frac{10}{3} \frac{A}{M} ^\frac{2}{3} Q=\frac{10}{3} \frac{\frac{900}{(\frac{5}{3})^\frac{3}{2}}}{900} ^\frac{2}{3} Q=2Q\) \(MP_A=\frac{20}{3}M^{\frac{1}{3}}A^{- \frac{1}{3}} Q=\frac{20}{3} (\frac{900}{\frac{900}{(\frac{5}{3})^\frac{3}{2}}})^\frac{1}{3} Q \approx 8.607Q\)
iii. The wage per man and rent per acre.
Men will be paid 2Q due to the maximum wage and 8.607Q will go to each acre.
iv. The marginal cost.
\(\frac{w}{MP_M}=\frac{r}{MP_A}=\lambda\) \(\frac{w_{max}}{MP_M}=\frac{2Q}{2Q}=1\) \(\frac{r_{new}}{MP_A}=\frac{8.607Q}{8.607Q}=1\)
v. The absolute shares of the two factors.
\(TP=w(Q_M)+r(Q_A)=2(900)+\frac{20}{3} (\frac{900}{\frac{900}{(\frac{5}{3})^\frac{3}{2}}})^\frac{1}{3} \frac{900}{(\frac{5}{3})^\frac{3}{2}}=5400\) \(share=\frac{factor}{TP}\)
\(\frac{w(Q_M)}{TP}=\frac{1800}{5400}=\frac{1}{3}\) \(\frac{r(Q_A)}{TP}=\frac{3600}{5400}=\frac{2}{3}\) This shows that the amount paid to men is 1800 which is \(\frac{1}{3}\) of the total product and the amount paid to acres is 3600 which is \(\frac{2}{3}\) of the total product.