{r data generation, 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 _labor_ in this minimum?" solution <- paste0("Given the target output, the optimal amount of the input factor _labor_ is $L = ", fmt(L), "$.") sol <- L com1 <- "" }, "capital" = { question <- "What is the amount of the input factor _capital_ in this minimum?" solution <- paste0("Given the target output, the optimal amount of the input factor _capital_ is $K = ", fmt(K), "$.") sol <- K com1 <- "" }, "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 com1 <- "" com2 <- "" }) ok <- L > 1 & K > 1 & cost > 0 }  Question ======== A firm has the following production function: $$F(K,L)= K L^{r L_multipli_print}.$$ The price for one unit of _capital_ is $p_K = r p_K$ and the price for one unit of _labor_ is $p_L = r p_L$. Minimize the costs of the firm considering its production function and given a target production output of r Q units. r question Solution ======== _Step 1_: Formulating the minimization problem. \begin{aligned} \min_{K,L} C(K,L) & = p_K K + p_L L\\ & = r p_K K + r p_L L\\ \mbox{subject~to:} & F(K,L) = Q \\ & K L^{r L_multipli_print} = r Q \end{aligned} _Step 2_: Lagrange function. \begin{aligned} \mathcal{L}(K, L, \lambda) & = C(K, L) - \lambda (F(K, L) - Q) \\ & = r p_K K + r p_L L - \lambda (K L^{r L_multipli_print} -r Q) \end{aligned} _Step 3_: First order conditions. \begin{aligned} \frac{\partial {\mathcal {L}}}{\partial K} & = r p_K - \lambda L^{r L_multipli_print} = 0\\ \frac{\partial {\mathcal {L}}}{\partial L} & = r p_L - {r L_multipli} \lambda K L^{r L_multipli - 1} = 0 \\ \frac{\partial {\mathcal {L}}}{\partial \lambda} & = -(K L^{r L_multipli_print}-r Q) = 0 \end{aligned} _Step 4_: Solve the system of equations for $K$, $L$, and $\lambda$. Solving the first two equations for $\lambda$ and equating them gives: \begin{aligned} \frac{r p_K}{L^{r L_multipli_print}} & = \frac{r p_L}{{r L_multipli} K L^{r L_multipli - 1}}\\ K & = \frac{r p_L}{r L_multipli \cdot r p_K} \cdot L^{r L_multipli - (r L_multipli - 1)}\\ K & = \frac{r p_L}{r p_K * L_multipli} \cdot L \end{aligned} Substituting this in the optimization constraint gives: \begin{aligned} K L^{r L_multipli_print} & = r Q\\ \left(\frac{r p_L}{r p_K * L_multipli}\cdot L \right) L^{r L_multipli_print} & = r Q\\ \frac{r p_L}{r p_K * L_multipli} L^{r L_multipli + 1} & = r Q\\ L & = \left(\frac{r L_multipli*p_K}{r p_L} \cdot r Q\right)^{\frac{1}{r L_multipli + 1}} = r L \approx r fmt(L)\\ K & = \frac{r p_L}{r p_K * L_multipli} \cdot L = r K \approx r fmt(K) \end{aligned} r com1 The minimal costs can be obtained by substituting the optimal factor combination in the objective function: \begin{aligned} C(K, L) & = r p_K K + r p_L L\\ & = r fmt(p_K * K, 6) + r fmt(p_L * L, 6) \\ & = r fmt(cost, 6) \approx r fmt(cost) \end{aligned} r com2 r solution \ {r contourplot, echo = FALSE, fig.height = 5, fig.width = 5, fig.path = "", fig.cap = ""} 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")  Meta-information ================ extype: num exsolution: r fmt(sol) exname: Lagrange cost minimization extol: 0.01