## Exam 1

1. #### Question

A firm has the following production function:
 $F\left(K,L\right)=K{L}^{3}.$

The price for one unit of capital is ${p}_{K}=20$ and the price for one unit of labor is ${p}_{L}=11$. Minimize the costs of the firm considering its production function and given a target production output of 730 units.
How high are in this case the minimal costs?

#### Solution

Step 1: Formulating the minimization problem.
 $\begin{array}{ccc}\multicolumn{1}{c}{\underset{K,L}{min}C\left(K,L\right)}& =\hfill & {p}_{K}K+{p}_{L}L\hfill \\ \multicolumn{1}{c}{}& =\hfill & 20K+11L\hfill \\ \multicolumn{1}{c}{\text{subject to:}}& \hfill & F\left(K,L\right)=Q\hfill \\ \multicolumn{1}{c}{}& \hfill & K{L}^{3}=730\hfill \end{array}$

Step 2: Lagrange function.
 $\begin{array}{ccc}\multicolumn{1}{c}{L\left(K,L,\lambda \right)}& =\hfill & C\left(K,L\right)-\lambda \left(F\left(K,L\right)-Q\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & 20K+11L-\lambda \left(K{L}^{3}-730\right)\hfill \end{array}$

Step 3: First order conditions.
 $\begin{array}{cccc}\multicolumn{1}{c}{\frac{\partial L}{\partial K}}& =\hfill & 20-\lambda {L}^{3}=0\hfill & \hfill \left(1\right)\\ \multicolumn{1}{c}{\frac{\partial L}{\partial L}}& =\hfill & 11-3\lambda K{L}^{3-1}=0\hfill & \hfill \left(2\right)\\ \multicolumn{1}{c}{\frac{\partial L}{\partial \lambda }}& =\hfill & -\left(K{L}^{3}-730\right)=0\hfill & \hfill \left(3\right)\end{array}$

Step 4: Solve the system of equations for $K$, $L$, and $\lambda$.
Equating Equations (1) and (2) after solving for $\lambda$ gives:
 $\begin{array}{ccc}\multicolumn{1}{c}{\frac{20}{{L}^{3}}}& =\hfill & \frac{11}{3K{L}^{3-1}}\hfill \\ \multicolumn{1}{c}{K}& =\hfill & \frac{11}{3·20}·{L}^{3-\left(3-1\right)}\hfill \\ \multicolumn{1}{c}{K}& =\hfill & \frac{11}{60}·L\hfill \end{array}$

Substituting this in the optimization constraint gives:

The minimal costs can be obtained by substituting the optimal factor combination in the objective function:
 $\begin{array}{ccc}\multicolumn{1}{c}{C\left(K,L\right)}& =\hfill & 20K+11L\hfill \\ \multicolumn{1}{c}{}& =\hfill & 29.126734+87.380201\hfill \\ \multicolumn{1}{c}{}& =\hfill & 116.506935\approx 116.51\hfill \end{array}$

Given the target output, the minimal costs are $116.51$. 