経済学で出る数学

ワークブックでじっくり攻める：応用問題

オイラーの定理の逆（作成：2015.11.24）

【問】　以下の$2$変数関数について $ \dfrac{\partial f}{\partial x_1}(x_1,x_2)x_1+ \dfrac{\partial f}{\partial x_2}(x_1,x_2)x_2 $ を計算しなさい．
１．　$\displaystyle f(x_1,x_2)=x_1^{\frac{1}{3}}x_2^{\frac{2}{3}}$
２．　$\displaystyle f(x_1,x_2)=x_1^{\frac{3}{4}}x_2^{\frac{5}{4}}$
３．　$\displaystyle f(x_1,x_2)=x_1^{\alpha}x_2^{\beta}$
４．　$\displaystyle f(x_1,x_2)=x_1^3+x_1^2x_2+2x_1x_2^2$

1. $\dfrac{\partial f}{\partial x_1}(x_1,x_2) =\dfrac{1}{3}x_1^{-\frac{2}{3}}x_2^{\frac{2}{3}}， \dfrac{\partial f}{\partial x_1}(x_1,x_2) =\dfrac{2}{3}x_1^{\frac{1}{3}}x_2^{-\frac{2}{3}}$ なので， $\dfrac{\partial f}{\partial x_1}(x_1,x_2)x_1+ \dfrac{\partial f}{\partial x_2}(x_1,x_2)x_2 =\dfrac{1}{3}x_1^{\frac{1}{3}}x_2^{\frac{2}{3}} +\dfrac{2}{3}x_1^{\frac{1}{3}}x_2^{\frac{2}{3}} =x_1^{\frac{1}{3}}x_2^{\frac{2}{3}}=f(x_1,x_2)$：1次同次関数


2. $\dfrac{\partial f}{\partial x_1}(x_1,x_2) =\dfrac{3}{4}x_1^{-\frac{1}{4}}x_2^{\frac{5}{4}}， \dfrac{\partial f}{\partial x_1}(x_1,x_2) =\dfrac{5}{4}x_1^{\frac{3}{4}}x_2^{\frac{1}{4}}$ なので， $\dfrac{\partial f}{\partial x_1}(x_1,x_2)x_1+ \dfrac{\partial f}{\partial x_2}(x_1,x_2)x_2 =\dfrac{3}{4}x_1^{\frac{3}{4}}x_2^{\frac{5}{4}} +\dfrac{5}{4}x_1^{\frac{3}{4}}x_2^{\frac{5}{4}} =2x_1^{\frac{3}{4}}x_2^{\frac{5}{4}}=2f(x_1,x_2)$：2次同次関数


3. $\dfrac{\partial f}{\partial x_1}(x_1,x_2) =\alpha x_1^{\alpha -1}x_2^{\beta}， \dfrac{\partial f}{\partial x_1}(x_1,x_2) =\beta x_1^{\alpha}x_2^{\beta -1}$ なので， $\dfrac{\partial f}{\partial x_1}(x_1,x_2)x_1+ \dfrac{\partial f}{\partial x_2}(x_1,x_2)x_2 =\alpha x_1^{\alpha}x_2^{\beta} +\beta x_1^{\alpha}x_2^{\beta} =(\alpha +\beta )(x_1^{\alpha}x_2^{\beta})=(\alpha +\beta )f(x_1,x_2)$： $(\alpha +\beta )$次同次関数


4. $\dfrac{\partial f}{\partial x_1}(x_1,x_2) =3x_1^2+2x_1x_2+x^2_2， \dfrac{\partial f}{\partial x_1}(x_1,x_2) =x_1^2x_2+4x_1x_2$ なので， $\dfrac{\partial f}{\partial x_1}(x_1,x_2)x_1+ \dfrac{\partial f}{\partial x_2}(x_1,x_2)x_2 =3x_1^3+3x_1^2x_2+6x_1x_2^2=3f(x_1,x_2)$： $3$次同次関数