経済学で出る数学

ワークブックでじっくり攻める：問 3.43

1. $L=\left(\dfrac{r{\alpha}}{w{\beta}}\right)^{\frac{\beta}{{\alpha}+{\beta}}}x^{\frac{1}{{\alpha}+{\beta}}}$, $K=\left(\dfrac{w{\beta}}{r{\alpha}}\right)^{\frac{\alpha}{{\alpha}+{\beta}}}x^{\frac{1}{{\alpha}+{\beta}}} $ に対して， \begin{align} L^{\alpha}K^{\beta}&= \left(\dfrac{r{\alpha}}{w{\beta}}\right)^{\frac{\alpha\beta}{{\alpha}+{\beta}}}x^{\frac{\alpha}{{\alpha}+{\beta}}} \left(\dfrac{w{\beta}}{r{\alpha}}\right)^{\frac{\alpha\beta}{{\alpha}+{\beta}}}x^{\frac{\beta}{{\alpha}+{\beta}}}\\ &=\left(\dfrac{r{\alpha}}{w{\beta}}\right)^{\frac{\alpha\beta}{{\alpha}+{\beta}}} \left(\dfrac{r{\alpha}}{w{\beta}}\right)^{-\frac{\alpha\beta}{{\alpha}+{\beta}}} x^{\frac{\alpha}{{\alpha}+{\beta}}}x^{\frac{\beta}{{\alpha}+{\beta}}}\qquad 第2項を逆数にするのがポイント\\ &=\left(\dfrac{r{\alpha}}{w{\beta}}\right)^{\frac{\alpha\beta}{{\alpha}+{\beta}}-\frac{\alpha\beta}{{\alpha}+{\beta}}}x^{\frac{\alpha +\beta}{{\alpha}+{\beta}}}\\ &=\left(\dfrac{r{\alpha}}{w{\beta}}\right)^0x=x \end{align}


2. $\left\{ \begin{array}{c@{\;}c@{\;}l} w&=&\displaystyle \frac{p}{3}L^{-\frac{2}{3}}K^{\frac{1}{3}}\\[2mm] r&=&\displaystyle \frac{p}{3}L^{\frac{1}{3}}K^{-\frac{2}{3}} \end{array} \right.$ に対して， \begin{align} w^2r&=\left(\frac{p^2}{3^2}L^{-\frac{4}{3}}K^{\frac{2}{3}}\right)\left(\frac{p}{3}L^{\frac{1}{3}}K^{-\frac{2}{3}}\right)\\ &=\frac{p^3}{3^3}L^{-\frac{4}{3}+\frac{1}{3}}K^{\frac{2}{3}-\frac{2}{3}}\\ &=\frac{p^3}{3^3}L^{-1} \end{align} \begin{align} wr^2&=\left(\frac{p}{3}L^{-\frac{2}{3}}K^{\frac{1}{3}}\right)\left(\frac{p^2}{3^2}L^{\frac{2}{3}}K^{-\frac{4}{3}}\right)\\ &=\frac{p^3}{3^3}L^{-\frac{2}{3}+\frac{2}{3}}K^{\frac{1}{3}-\frac{4}{3}}\\ &=\frac{p^3}{3^3}K^{-1} \end{align}


3. 例題7.6の解答例参照．