<>= 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) } } @ <>= ## 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} <>= 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}