## 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 }
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
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")
extype: num
exsolution: r fmt(sol)
exname: Lagrange cost minimization
extol: 0.01
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