A producer is a price-taker on both the market for input factors labor and capital, and the market for end products. The cost of one unit of labor equals $w=9$, the cost of one unit of capital equals $r=81$, while the selling price of the end products equals $p=27$. The production function of this producer is given by $Y(L,K)=L^{\frac{1}{3}}K^{\frac{1}{3}}$. Determine the maximum profit.
Antwoord 1 correct
Correct
Antwoord 2 optie
$0$
Antwoord 2 correct
Fout
Antwoord 3 optie
$100$
Antwoord 3 correct
Fout
Antwoord 4 optie
$9$
Antwoord 4 correct
Fout
Antwoord 1 optie
$1$
Antwoord 1 feedback
Correct: The revenue function is given by $R(L,K)=pY(L,K)=27L^{\frac{1}{3}}K^{\frac{1}{3}}$ and the cost function by $C(L,K)=wL+rK=9L+81K$, which results in the profit function
\[
\pi (L,K)=27L^{\frac{1}{3}}K^{\frac{1}{3}}-9L-81K.
\]
Since the partial derivatives of $\pi(L,K)$ equal
$\pi'_{L}(L,K) =9L^{-\frac{2}{3}}K^{\frac{1}{3}}-9$ and
$\pi'_{K}(L,K) =9L^{\frac{1}{3}}K^{-\frac{2}{3}}-81$,
the stationary points of $\pi(L,K)$ are solutions of the following system:
$$\begin{align*}
9L^{-\frac{2}{3}}K^{\frac{1}{2}}-9&=0,\\
9L^{\frac{1}{3}}K^{-\frac{2}{3}}-81&=0.
\end{align*}$$
Hence, $K^{\frac{1}{3}}=L^{\frac{2}{8}}$ and therefore, $K=L^2$.
Consequently, $L^{\frac{1}{3}}(L^2)^{-\frac{2}{3}}=9$, which gives $L^{-1}=9$. Thus, $L=\frac{1}{9}$, which gives $K=\frac{1}{81}$.
Hence, $(L,K)=(\frac{1}{9},\frac{1}{81})$ is the only stationary point. By the use of the criterion function we investigate whether or not this point is a maximum location. It holds that $\pi''_{LL}(L,K)=-6L^{-\frac{5}{3}}K^{\frac{1}{3}}$, $\pi''_{KK}(L,K)=-6L^{\frac{1}{3}}K^{-\frac{5}{3}}$ and $\pi''_{LK}(L,K)=3L^{-\frac{2}{3}}K^{-\frac{2}{3}}$, which implies that the criterion function is given by
$$\begin{align*}
C(L,K) &= \pi''_{LL}(L,K)\pi''_{KK}(L,K)-(\pi''_{LK}(L,K))^{2}\\
&=(-6L^{-\frac{5}{3}}K^{\frac{1}{3}}) \cdot (-6L^{\frac{1}{3}}K^{-\frac{5}{3}})-(3L^{-\frac{2}{3}}K^{-\frac{2}{3}})^2\\
&=36L^{-\frac{4}{3}}K^{-\frac{4}{3}}-9L^{-\frac{4}{3}}K^{-\frac{4}{3}}\\
&=27L^{-\frac{4}{3}}K^{-\frac{4}{3}}.\\
\end{align*}$$
Hence, as $C(\frac{1}{9},\frac{1}{81})>0$ and $\pi''_{LL}(\frac{1}{9},\frac{1}{81})<0$ it follows that $\pi (L,K)$ has a maximum at $(L,K)=(\frac{1}{9},\frac{1}{81})$, with value $\pi(\frac{1}{9},\frac{1}{81})=1.$
Go on.
\[
\pi (L,K)=27L^{\frac{1}{3}}K^{\frac{1}{3}}-9L-81K.
\]
Since the partial derivatives of $\pi(L,K)$ equal
$\pi'_{L}(L,K) =9L^{-\frac{2}{3}}K^{\frac{1}{3}}-9$ and
$\pi'_{K}(L,K) =9L^{\frac{1}{3}}K^{-\frac{2}{3}}-81$,
the stationary points of $\pi(L,K)$ are solutions of the following system:
$$\begin{align*}
9L^{-\frac{2}{3}}K^{\frac{1}{2}}-9&=0,\\
9L^{\frac{1}{3}}K^{-\frac{2}{3}}-81&=0.
\end{align*}$$
Hence, $K^{\frac{1}{3}}=L^{\frac{2}{8}}$ and therefore, $K=L^2$.
Consequently, $L^{\frac{1}{3}}(L^2)^{-\frac{2}{3}}=9$, which gives $L^{-1}=9$. Thus, $L=\frac{1}{9}$, which gives $K=\frac{1}{81}$.
Hence, $(L,K)=(\frac{1}{9},\frac{1}{81})$ is the only stationary point. By the use of the criterion function we investigate whether or not this point is a maximum location. It holds that $\pi''_{LL}(L,K)=-6L^{-\frac{5}{3}}K^{\frac{1}{3}}$, $\pi''_{KK}(L,K)=-6L^{\frac{1}{3}}K^{-\frac{5}{3}}$ and $\pi''_{LK}(L,K)=3L^{-\frac{2}{3}}K^{-\frac{2}{3}}$, which implies that the criterion function is given by
$$\begin{align*}
C(L,K) &= \pi''_{LL}(L,K)\pi''_{KK}(L,K)-(\pi''_{LK}(L,K))^{2}\\
&=(-6L^{-\frac{5}{3}}K^{\frac{1}{3}}) \cdot (-6L^{\frac{1}{3}}K^{-\frac{5}{3}})-(3L^{-\frac{2}{3}}K^{-\frac{2}{3}})^2\\
&=36L^{-\frac{4}{3}}K^{-\frac{4}{3}}-9L^{-\frac{4}{3}}K^{-\frac{4}{3}}\\
&=27L^{-\frac{4}{3}}K^{-\frac{4}{3}}.\\
\end{align*}$$
Hence, as $C(\frac{1}{9},\frac{1}{81})>0$ and $\pi''_{LL}(\frac{1}{9},\frac{1}{81})<0$ it follows that $\pi (L,K)$ has a maximum at $(L,K)=(\frac{1}{9},\frac{1}{81})$, with value $\pi(\frac{1}{9},\frac{1}{81})=1.$
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