In exams: Automatic Generation of Exams in R

## 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 <- "<!--"
  com2 <- "-->"
},
"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 <- "<!--"
  com2 <- "-->"
},
"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 \

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")