<<echo=FALSE, results=hide>>=
fmt <- function(x, digits = 2) {
   x <- round(x, digits = digits)
   if(digits <= 3L) {
     format(x, nsmall = digits, scientific = FALSE, digits = 12)
   } else {
     format(x, scientific = FALSE, digits = 12)
   }
}
@

<<echo=FALSE, results=hide>>=
## DATA GENERATION
ok <- FALSE

while(!ok){
L_multipli <- sample(1:3, 1)
L_multipli_print <- if(L_multipli != 1) L_multipli else "" 
Q <- sample(seq(100, 1000, by=10), 1)
p_K <- sample(2:30, 1)
p_L <- sample(2:30, 1)

## QUESTION/ANSWER GENERATION
L <- round((L_multipli * (p_K/p_L) * Q)^(1/(L_multipli+1)), digits = 8)
K <- round((1/L_multipli) * (p_L/p_K) * L, digits = 8)
cost <- (p_L*L + p_K*K)
lambda <- round(p_K/(L^L_multipli),digits=6)
ratio <- round((p_L)/(p_K * L_multipli), digits=6)

type <- sample(c("labor", "capital", "costs"), 1)

switch(type,
"labor" = {
  question <- "What is the amount of the input factor \\\\\\emph{labor} in this minimum?"
  solution <- paste0("Given the target output, the optimal amount of the input factor \\\\\\emph{labor} is $L = ", fmt(L), "$.")
  sol <- L
  com <- "%"
},
"capital" = {
  question <- "What is the amount of the input factor \\\\\\emph{capital} in this minimum?"
  solution <- paste0("Given the target output, the optimal amount of the input factor \\\\\\emph{capital} is $K = ", fmt(K), "$.")
  sol <- K
  com <- "%"
},
"costs" = {
  question <- "How high are in this case the minimal costs?"
  solution <- paste0("Given the target output, the minimal costs are $", fmt(cost), "$.")
  sol <- cost
  com <- ""
})

ok <- L > 1 & K > 1 & cost > 0
}
@

\begin{question}
A firm has the following production function: \[F(K,L)= K L^{\Sexpr{L_multipli_print}}.\]
The price for one unit of \emph{capital} is $p_K = \Sexpr{p_K}$ 
and the price for one unit of \emph{labor} is $p_L = \Sexpr{p_L}$.
Minimize the costs of the firm considering its production function and given a target production output of \Sexpr{Q} units.

\Sexpr{question}
\end{question}

\begin{solution}
\emph{Step~1}: Formulating the minimization problem.
\begin{eqnarray*}
\min_{K,L} C(K,L) &=& p_K K + p_L L\\
 &=& \Sexpr{p_K} K + \Sexpr{p_L} L\\
\mbox{subject to:} &&  F(K,L) = Q \\
&& K L^{\Sexpr{L_multipli_print}} = \Sexpr{Q} 
\end{eqnarray*}

\emph{Step~2}: Lagrange function.
\begin{eqnarray*}
\mathcal{L}(K, L, \lambda) &=& C(K, L)-\lambda (F(K, L) - Q) \\
&=& \Sexpr{p_K} K + \Sexpr{p_L} L - \lambda (K L^{\Sexpr{L_multipli_print}} -\Sexpr{Q})
\end{eqnarray*}

\emph{Step~3}: First order conditions.
\begin{eqnarray}
\frac{\partial {\mathcal {L}}}{\partial K} & = & \Sexpr{p_K} - \lambda L^{\Sexpr{L_multipli_print}} = 0\\
\frac{\partial {\mathcal {L}}}{\partial L} & = & \Sexpr{p_L} - {\Sexpr{L_multipli}} \lambda K L^{\Sexpr{L_multipli}-1} = 0 \\
\frac{\partial {\mathcal {L}}}{\partial \lambda} & = & -(K L^{\Sexpr{L_multipli_print}}-\Sexpr{Q}) = 0
\end{eqnarray}

\emph{Step~4}: Solve the system of equations for $K$, $L$, and $\lambda$.

Equating Equations~(1) and (2) after solving for $\lambda$ gives:
\begin{eqnarray*}
\frac{\Sexpr{p_K}}{L^{\Sexpr{L_multipli_print}}} & = & \frac{\Sexpr{p_L}}{{\Sexpr{L_multipli}} K L^{\Sexpr{L_multipli}-1}}\\
K & = & \frac{\Sexpr{p_L}}{\Sexpr{L_multipli} \cdot \Sexpr{p_K}} \cdot L^{\Sexpr{L_multipli} - (\Sexpr{L_multipli} -1)}\\
K & = & \frac{\Sexpr{p_L}}{\Sexpr{p_K * L_multipli}} \cdot L
\end{eqnarray*}

Substituting this in the optimization constraint gives:
\begin{eqnarray*}
K L^{\Sexpr{L_multipli_print}} & = & \Sexpr{Q}\\
\left(\frac{\Sexpr{p_L}}{\Sexpr{p_K * L_multipli}}\cdot L \right) L^{\Sexpr{L_multipli_print}} & = & \Sexpr{Q}\\
\frac{\Sexpr{p_L}}{\Sexpr{p_K * L_multipli}} L^{\Sexpr{L_multipli + 1}} & = & \Sexpr{Q}\\
L & = & \left(\frac{\Sexpr{L_multipli*p_K}}{\Sexpr{p_L}} \cdot 
\Sexpr{Q}\right)^{\frac{1}{\Sexpr{L_multipli + 1}}} = \Sexpr{L}~\approx~\Sexpr{fmt(L)}\\
K & = & \frac{\Sexpr{p_L}}{\Sexpr{p_K * L_multipli}} \cdot L = \Sexpr{K}~\approx~\Sexpr{fmt(K)}
\end{eqnarray*}

\Sexpr{com} The minimal costs can be obtained by substituting the optimal factor combination in the objective function:
\Sexpr{com} \begin{eqnarray*}
\Sexpr{com} C(K, L) & = & \Sexpr{p_K} K + \Sexpr{p_L} L\\
\Sexpr{com}         & = & \Sexpr{fmt(p_K * K, 6)} + \Sexpr{fmt(p_L * L, 6)} \\
\Sexpr{com}         & = & \Sexpr{fmt(cost, 6)} \approx \Sexpr{fmt(cost)}
\Sexpr{com} \end{eqnarray*}
\Sexpr{solution}

<<echo=FALSE, fig=TRUE>>=
costfunction <- function(x1, x2) (p_L * x1 + p_K * x2)
prodfunction <- function(x) (Q/x^(L_multipli))

x1 <- seq(0, L * 3, length = L * 10)
x2 <- seq(0, K * 4, length = K * 10)
y <- outer(x1, x2, costfunction)

contour(x1, x2, y, xaxs = "i", yaxs = "i", xlab = "L", ylab = "K", col = "gray")
plot(prodfunction, 0, L * 10, add = TRUE, lty = 2)
contour(x1, x2, y, add = TRUE, xaxs = "i", yaxs = "i",
  levels = costfunction(L, K), labcex = 0.8, lwd = 1.5)
lines(c(L,L),c(0,K), lty=3)
lines(c(0,L),c(K,K), lty=3)
points(L, K, pch = 19, col = "red")
@
\end{solution}


%% META INFORMATION
%% \extype{num}
%% \exsolution{\Sexpr{fmt(sol)}}
%% \exname{Lagrange cost minimization}
%% \extol{0.01}