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\begin{document}
\title{Problem Set II (Williams)}
\author[1]{Clara Jace}%
\affil[1]{George Mason University}%
\vspace{-1em}
\date{\today}
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\section*{Question 1}
{\label{740685}}
Finding the values that maximize quantity (Q):
\begin{quote}
\(Q = 2K^{0.2} L^{0.8}\)
\(200,000 = 80K + 12L\)
\end{quote}
Using the Lagrangian:
\begin{quote}
\(\mathcal{L} = 2K^{0.2} L^{0.8} + \lambda [200,000 - 80K - 12L]\)
\end{quote}
The first-order conditions are:
\begin{quote}
\(\frac{d \mathcal{L}}{dK} = 0.4K^{-0.8} L^{0.8} + \lambda (-80) = 0\)
\(\frac{d \mathcal{L}}{dL} = 1.6K^{0.2}L^{-0.2} + \lambda (-12) = 0\)
\(\frac{d \mathcal{L}}{d \lambda} = 200,000-80K -12L = 0\)
\end{quote}
Setting both equal to~\(\lambda\)~so that they will be equal to
each other, we divide the first equation by 80 and the second equation
by 12:
\begin{quote}
\(\frac{0.005L^{0.8}}{K^{0.8} } = \frac{0.133K^{0.2}}{L^{0.2}}\)
\end{quote}
Now cross-multiplying and dividing we find:
\begin{quote}
\(0.005L = 0.133K\)
\(L = 26.7K\)
\end{quote}
Substituting this into the last equation yields:
\begin{quote}
\(200,000 - 80K - 12(26.7K) = 0\)
\(200,000 = 400K \Longrightarrow K = 500\)
\end{quote}
Placing this into the isocost constraint to solve for L:
\begin{quote}
\(200,000 - 80(500) - 12L = 0\)
\(12L = 160,000 \Longrightarrow L = 13,333\)
\end{quote}
Thus the total quantity produced would be
\begin{quote}
\(Q = 2(500)^{0.2} (13,333)^{0.8} \Longrightarrow Q = 13,828\)
\end{quote}
Finding the values that minimize cost (C) should be the same. Now, the
objective function becomes the isocost and the constraint is the
production function:
\begin{quote}
\(13,828 = 2K^{0.2} L^{0.8}\)
\(C = 80K + 12L\)
\(\mathcal{L} = 80K + 12L + \lambda [13,828 - 2K^{0.2} L^{0.8}]\)
\end{quote}
Finding the first-order conditions:
\begin{quote}
\(\frac{d \mathcal{L}}{dK} = 80 + \lambda(-0.4K^{-0.8} L^{0.8}) = 0\)
\(\frac{d \mathcal{L}}{dL} = 12 + \lambda (-1.6K^{0.2}L^{-0.2}) = 0\)
\(\frac{d \mathcal{L}}{d \lambda} = 13,828 - 2K^{0.2} L^{0.8} = 0\)
\end{quote}
Setting the first and second equations equal to~\(\lambda\)~and
then to each other yields:
\begin{quote}
\(\frac{200K^{0.8}}{L^{0.8}} = \frac{7.5L^{0.2}}{K^{0.2}}\)
\end{quote}
Cross-multiplying and dividing will give us:
\begin{quote}
\(200K = 7.5 L\)
\(L = 26.7K\)
\end{quote}
Substituting this into the quantity equation yields:
\begin{quote}
\(13,828 = 2K^{0.2} (26.7K)^{0.8}\)
\(13,828 = 27.68K \Longrightarrow K \approx 500\)
\end{quote}
Then to find the quantity of labor:
\begin{quote}
\(L = 26.7(500) \Longrightarrow L \approx 13,333\)
\end{quote}
(The numbers are not exact because of the compounding from rounding.)
The Lagrange multiplier approximates the marginal impact caused by a
small change in the price of the inputs, capital (K) and labor (L). This
is why the Lagrange multiplier is often called the~\emph{shadow price}.~
\par\null
\section*{Question 2}
{\label{825551}}
\textbf{~}The number of working hours per day will be 11. The production
functions for plot A and plot B are:
\begin{quote}
\(A = 10L_a - L_a^2\)
\(B = 15L_b - L_b^2\)
\end{quote}
Giving us a time-cost function of
\begin{quote}
\(11 = L_a + L_b\)
\end{quote}
The profit function to optimize will be:
\begin{quote}
\(\pi = 4A + 2B - wA - wB\)
\end{quote}
Finding the marginal production of each plot:
\begin{quote}
\(A' = 10 - 2 L_a\)
\(B' = 15 - 2 L_b\)
\end{quote}
When the plots are optimized, their marginal productivity will be equal
to each other. Setting the equations equal and multiplying by their
respective prices gives:
\begin{quote}
\(4(10 - 2L_a) = 2(15 - 2L_b)\)
\(L_b = -2.5 + 2L_a\)
\end{quote}
Then putting it back into the time-cost function we have:
\begin{quote}
\(11= L_a + 2L_a - 2.5\)
\(13.5 = 3L_a\)
\(L_a = 4.5\)
\(L_b = 6.5\)
\end{quote}
Since hourly wage is equal to the marginal product of labor on both
plots,~
\begin{quote}
\(A = 4(10 - 2(4.5)) \Longrightarrow w_a = 4\) dollars per hour
\(B = 2(15 - 2(6.5)) \Longrightarrow w_b = 4\) dollars per hour
\end{quote}
We can find the total product of each plot by multiplying the quantity
produced by the price of each product, then subtracting the wages paid:~
\begin{quote}
\(TP_A = 4(10(4.5) - (4.5)^2) = 99\)
\(99 - 4.5(4) = r_a = 81\) dollars spent for rent on Plot A
\(TP_B = 2(15(6.5) - (6.5)^2) = 110.5\)
\(110.5 - 6.5(4) = r_b = 84.5\) dollars spent for rent on Plot B
\end{quote}
The total rent would be
\begin{quote}
\(\pi = 81 + 84.5 = 165.5\) dollars.
\end{quote}
\section*{}
{\label{725218}}
\section*{Question 3}
{\label{725218}}
\(Q = 10M^{1/3} A^{2/3}\)
\(M = 900\) and~\(A = 600\)
\subsection*{Part A~}
{\label{700419}}
\begin{quote}
\textbf{i)~}If there are 100 identical firms, this would mean the
resources are divided from the total number of men or acres over the
total number of firms. Thus 9 men per firm and 6 acres per firm:~
\begin{quote}
\(\frac{900}{100} = 9 , \frac{600}{100} = 6\)
\end{quote}
\textbf{ii)~}The marginal product of men is equal to the derivative of
their total product:
\begin{quote}
\(\frac{dQ}{dM} = 3.33(9)^{-2/3}(6)^{2/3} \approx ~2.5437\)~units of Q
\end{quote}
The marginal product of acres is equal to:
\begin{quote}
\(\frac{dQ}{dA} = 10(9)^{1/3}(\frac{2}{3})(6)^{-1/3} \approx 7.6314\) units of Q
\end{quote}
\textbf{iii)~}In equilibrium, factors are paid their marginal product.
Since Q is numeraire, the wage and rent will be equal the marginal
product of men and acres, about 2.5 and 7.6 units of Q respectively.~
\textbf{iv)~}Marginal cost is found by taking the derivative of the
total cost function. Since we know the simple economy is optimizing, we
can assume the marginal costs of the two factors are identical and take
it just with respect to M:
\begin{quote}
\(TC = 2.5M(Q) + 7.6A(Q)\)
~\(MC = \frac{\partial TC}{\partial Q} = \frac{\partial TC}{\partial M} \cdot \frac{\partial M}{\partial Q} = 2.5Q \cdot \frac{2}{5Q} = 1\) unit of Q is the marginal cost of producing 1 unit
of Q.~
\end{quote}
This is confirmed because MC should also be equal to the price, 1 unit
of Q.~~
\textbf{v)~}The absolute shares of each factor can be found by:
\begin{quote}
\(TP = 10(900^{1/3}) (600^{2/3}) = 6868.29\)
\end{quote}
The share paid to men:~
\begin{quote}
\(2.5437(900) = 2289.33\)
\end{quote}
The share paid to the acreage:
\begin{quote}
\(7.6314(600) = 4578.84\)
\end{quote}
Please note, the sum of the shares (6868.17) is a little off since the
marginal product of each input was rounded to the nearest decimal
place.~~
\end{quote}
\subsection*{Part B}
{\label{529596}}
If there was a minimum wage imposed equal to 5 units of Q, the following
would change as such:
\begin{quote}
\textbf{i)~}Men would be hired up to the point where~\(MP_M = w\),
that is, at 5 units of Q. Individual firms would still rent 6 acres of
land since this is the number that maximizes profit, thus the full
number of acres would be employed. The marginal product of M and A would
still be such that:
\begin{quote}
\(\frac{dQ}{dM} = 3.33(M)^{-2/3}(600)^{2/3} = 5 \)
\(M ^{-2/3} = 0.0211 \Longrightarrow M \approx 326\)~
\end{quote}
With the enforced minimum wage, only 326 men would be employed, thus
about 3.26 men per firm. The amount of acres would stay the same (6).~ ~
\textbf{ii)~}Now to find the marginal product of men and of acres, we
can put the new figures into our previous equations:
\begin{quote}
\(\frac{dQ}{dM} = 3.33(3.26)^{-2/3}(6)^{2/3} \approx ~5\) units of Q
\(\frac{dQ}{dA} = 10(3.26)^{1/3}(\frac{2}{3})(6)^{-1/3} \approx 5.44\) units of Q
\end{quote}
\textbf{iii)~}Again, we know that men and acres are paid their marginal
product, thus the wage is 5 (the minimum wage) and the rent is 5.44
units of Q, lower than before because of the artificially high minimum
wage.~~
\textbf{iv)~}Marginal cost should be the same as before, but we can also
calculate it differently by dividing the wage/rent by the respective
marginal product of the input:
\begin{quote}
\(\frac{w}{MP_M} = {5}{5} = 1\) and likewise~\(\frac{r}{MP_A} = \frac{5.44}{5.44} = 1\)
\end{quote}
As previously seen, the marginal cost is 1 unit of Q, the amount it
takes to produce an additional unit of Q.~
\textbf{v)~}The total production is now~
\begin{quote}
~\(TP = 5(326) + 5.44(600) \approx 4894\)
\end{quote}
Using this, we can see that the share amount paid to men is roughly 1630
while the share amount paid to acres is 3264. The fractional shares are
equal to their exponents, 1/3 for men and 2/3 for acres. Thus, due to
the minimum wage, the total product has been reduced as well as the
shares paid to the factors of production.
\end{quote}
\subsection*{Part C}
{\label{781512}}
There would be no impact of establishing a maximum wage equal to 3 units
of Q, as the marginal product of men is below that ceiling at 2.56.
Thus, all the answers would be the same as in Part A.
However, if the maximum wage were 2, the answers would adjust as such:
\begin{quote}
\textbf{i)~}The quantity of men employed is still equal to the point
where the marginal product equals the maximum wage:~
\begin{quote}
\(\frac{dQ}{dM} = 3.33(M)^{-2/3}(600)^{2/3} = 2\)
\(M^{-2/3} = 0.0084 \Longrightarrow M \approx 1289\)
\end{quote}
This would mean that the number of men desired per firm is roughly 12.
Therefore, since the market can only employ 900 men, the amount of acres
each firm employs will change:
\begin{quote}
\((1.67)^{3/2}(A) = 900 \approx 417.03\)~
\end{quote}
So each firm will employ 9 men and 4.17 acres.~
\textbf{ii)~}Finding the marginal product of men and of acres:
\begin{quote}
\(\frac{dQ}{dM} = 3.33(9)^{-2/3}(4.17)^{2/3} \approx ~2\) units of Q
\(\frac{dQ}{dA} = 10(9)^{1/3}(\frac{2}{3})(4.17)^{-1/3} \approx 8.6\) units of Q
\end{quote}
\textbf{iii)~}The wage is equal to the marginal product of men, being 2
units of Q. The rent will be equal to the marginal product of each
acres, 8.6 units of Q.~
\textbf{iv)~}Finding the marginal cost of men and acreage will the the
same as previously:
\begin{quote}
\(\frac{w}{MP_M} = \frac{2}{2} = 1\) and likewise~\(\frac{r}{MP_A} = \frac{8.6}{8.6} = 1\) unit of Q.
\end{quote}
\textbf{v)~}To find the absolute shares, we first calculate the total
product:
\begin{quote}
\(TP = 2(900) + 8.6(417) \approx 5386\)
\end{quote}
Therefore, we know the share of men is 1800, which is a third of the
total product and acres are paid 3586, which is two thirds of the total
product. Due to a maximum wage, there is a lower total product and
product paid to each share than in the absence of market controls.~
\end{quote}
\par\null
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