経済学で出る数学

ワークブックでじっくり攻める：問7.19解答例

• 【Step1】ラグランジュ関数を作ると， $ {\cal{L}}(x_1,x_2,x_3,\lambda)=(x_1)^{\alpha}(x_2)^{\beta}(x_3)^{\gamma}+\lambda\left(I-p_1x_1-p_2x_2-p_3x_3\right). $


• 【Step2】 各変数で偏微分してイコールゼロとおくと， \[ \left\{ \begin{array}{lll} 0=\displaystyle \frac{\partial\cal{L}}{\partial x_1} ={\alpha}(x_1)^{\alpha -1}(x_2)^{\beta}(x_3)^{\gamma}-{\lambda}p_1 & \rightarrow {\lambda}p_1={\alpha}(x_1)^{\alpha -1}(x_2)^{\beta}(x_3)^{\gamma} &\qquad (1)\\[2ex] 0=\displaystyle \frac{\partial\cal{L}}{\partial x_2} ={\beta}(x_1)^{\alpha}(x_2)^{\beta -1}(x_3)^{\gamma}-{\lambda}p_2 & \rightarrow {\lambda}p_2= {\beta}(x_1)^{\alpha}(x_2)^{\beta -1}(x_3)^{\gamma} &\qquad (2)\\[2ex] 0=\displaystyle \frac{\partial\cal{L}}{\partial x_3} ={\gamma}(x_1)^{\alpha}(x_2)^{\beta}(x_3)^{\gamma -1}-{\lambda}p_3 & \rightarrow {\lambda}p_3= {\gamma}(x_1)^{\alpha}(x_2)^{\beta}(x_3)^{\gamma -1} &\qquad (3)\\[2ex] 0=I-p_1x_1-p_2x_2-p_3x_3& & \qquad (4) \end{array} \right. \]


• 【Step3】 あとは工夫して解く．
  • $(1) \div (2)$ から $\dfrac{\alpha}{\beta}x_1^{-1}x_2=\dfrac{p_1}{p_2} \Longrightarrow x_2=\dfrac{\beta p_1}{\alpha p_2}x_1$
  
  
  • $(1) \div (3)$ から $\dfrac{\alpha}{\gamma}x_1^{-1}x_3=\dfrac{p_1}{p_3} \Longrightarrow x_3=\dfrac{\gamma p_1}{\alpha p_3}x_1$
  
  
  • $(4)$ に代入すると， $p_1x_1+p_2\dfrac{\beta p_1}{\alpha p_2}x_1+p_3\dfrac{\gamma p_1}{\alpha p_3}x_1=I \Longrightarrow \dfrac{\alpha + \beta +\gamma}{\alpha}p_1x_1=I$
  
  
  • $x_1=\dfrac{\alpha}{\alpha + \beta +\gamma}\dfrac{I}{p_1}$
  
  
  • $x_2= \dfrac{\beta p_1}{\alpha p_2}\dfrac{\alpha}{\alpha + \beta +\gamma}\dfrac{I}{p_1} =\dfrac{\beta}{\alpha + \beta +\gamma}\dfrac{I}{p_2}$
  
  
  • $x_3= \dfrac{\gamma p_1}{\alpha p_3}\dfrac{\alpha}{\alpha + \beta +\gamma}\dfrac{I}{p_1} =\dfrac{\gamma}{\alpha + \beta +\gamma}\dfrac{I}{p_3}$
  
  
  ゆえに最適解は，$(x_1,x_2,x_3)=\left( \dfrac{\alpha}{\alpha + \beta +\gamma}\dfrac{I}{p_1}, \dfrac{\beta}{\alpha + \beta +\gamma}\dfrac{I}{p_2}, \dfrac{\gamma}{\alpha + \beta +\gamma}\dfrac{I}{p_3} \right)$．