Question 1

Finding the values that maximize quantity (Q):
\(Q = 2K^{0.2} L^{0.8}\)
\(200,000 = 80K + 12L\)
Using the Lagrangian:
\(\mathcal{L} = 2K^{0.2} L^{0.8} + \lambda [200,000 - 80K - 12L]\)
The first-order conditions are:
\(\frac{d \mathcal{L}}{dK} = 0.4K^{-0.8} L^{0.8} + \lambda (-80) = 0\)
\(\frac{d \mathcal{L}}{dL} = 1.6K^{0.2}L^{-0.2} + \lambda (-12) = 0\)
\(\frac{d \mathcal{L}}{d \lambda} = 200,000-80K -12L = 0\)
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:
\(\frac{0.005L^{0.8}}{K^{0.8} } = \frac{0.133K^{0.2}}{L^{0.2}}\)
Now cross-multiplying and dividing we find:
\(0.005L = 0.133K\)
\(L = 26.7K\)
Substituting this into the last equation yields:
\(200,000 - 80K - 12(26.7K) = 0\)
\(200,000 = 400K \Longrightarrow K = 500\)
Placing this into the isocost constraint to solve for L:
\(200,000 - 80(500) - 12L = 0\)
\(12L = 160,000 \Longrightarrow L = 13,333\)
Thus the total quantity produced would be
\(Q = 2(500)^{0.2} (13,333)^{0.8} \Longrightarrow Q = 13,828\)
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:
\(13,828 = 2K^{0.2} L^{0.8}\)
\(C = 80K + 12L\)
\(\mathcal{L} = 80K + 12L + \lambda [13,828 - 2K^{0.2} L^{0.8}]\)
Finding the first-order conditions:
\(\frac{d \mathcal{L}}{dK} = 80 + \lambda(-0.4K^{-0.8} L^{0.8}) = 0\)
\(\frac{d \mathcal{L}}{dL} = 12 + \lambda (-1.6K^{0.2}L^{-0.2}) = 0\)
\(\frac{d \mathcal{L}}{d \lambda} = 13,828 - 2K^{0.2} L^{0.8} = 0\)
Setting the first and second equations equal to \(\lambda\) and then to each other yields:
\(\frac{200K^{0.8}}{L^{0.8}} = \frac{7.5L^{0.2}}{K^{0.2}}\)
Cross-multiplying and dividing will give us:
\(200K = 7.5 L\)
\(L = 26.7K\)
Substituting this into the quantity equation yields:
\(13,828 = 2K^{0.2} (26.7K)^{0.8}\)
\(13,828 = 27.68K \Longrightarrow K \approx 500\)
Then to find the quantity of labor:
\(L = 26.7(500) \Longrightarrow L \approx 13,333\)
(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 shadow price

Question 2

 The number of working hours per day will be 11. The production functions for plot A and plot B are:
\(A = 10L_a - L_a^2\)
\(B = 15L_b - L_b^2\)
Giving us a time-cost function of
\(11 = L_a + L_b\)
The profit function to optimize will be:
\(\pi = 4A + 2B - wA - wB\)
Finding the marginal production of each plot:
\(A' = 10 - 2 L_a\)
\(B' = 15 - 2 L_b\)
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:
\(4(10 - 2L_a) = 2(15 - 2L_b)\)
\(L_b = -2.5 + 2L_a\)
Then putting it back into the time-cost function we have:
\(11= L_a + 2L_a - 2.5\)
\(13.5 = 3L_a\)
\(L_a = 4.5\)
\(L_b = 6.5\)
Since hourly wage is equal to the marginal product of labor on both plots, 
\(A = 4(10 - 2(4.5)) \Longrightarrow w_a = 4\) dollars per hour
\(B = 2(15 - 2(6.5)) \Longrightarrow w_b = 4\) dollars per hour
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: 
\(TP_A = 4(10(4.5) - (4.5)^2) = 99\)
\(99 - 4.5(4) = r_a = 81\) 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
The total rent would be
\(\pi = 81 + 84.5 = 165.5\) dollars.

Question 3

\(Q = 10M^{1/3} A^{2/3}\)
\(M = 900\) and \(A = 600\)

Part A