ラグランジュの未定乗数法

拘束条件がある関数の極値を調べるにはどうすればいいだろうか? 単純に考えれば,実数値関数の微分をとって0になる点を求めれば極値である. しかし,拘束条件がある場合では単純ではない. 拘束条件のもと,調べたい関数の全微分を取って極値を調べる.

2変数の場合.

2つの微分可能な連続実数値関数\(f(x,y),g(x,y)\)とし, 以下のような微分可能な連続実数値関数を用意する. \[ {\mathscr L}(x,y,\lambda) = f(x,y) - \lambda g(x,y) \] 上記の偏微分が\((x,y,\lambda) = (a, b, l)\)のとき, \[ \left.\frac{\partial {\mathscr L}(x,y,\lambda)}{\partial x}\right|_{(a,b,l)} = \left.\frac{\partial {\mathscr L}(x,y,\lambda)}{\partial y}\right| _{(a,b,l)} = \left. \frac{\partial {\mathscr L}(x,y,\lambda)}{\partial \lambda} \right| _{(a,b,l)} = 0 \] が成立する場合,\((x,y,\lambda) = (a, b, l)\)で\(f(x,y)\)は,\(g(x,y) = 0\)の拘束条件で極値を取る.

  • 証明
    連立方程式は, \[ \left\{ \begin{array}{ll} \left. \displaystyle \frac{\partial {\mathscr L}(x,y,\lambda)}{\partial \lambda} \right| _{(a,b,l)} &= g(a,b) = 0 \\ \left. \displaystyle \frac{\partial {\mathscr L}(x,y,\lambda)}{\partial x} \right| _{(a,b,l)} &= \left. \displaystyle \frac{\partial f(x,y)}{\partial x} \right| _{(a,b,l)} - \lambda \left. \displaystyle \frac{\partial g(x,y)}{\partial x} \right| _{(a,b,l)} = 0 \\ \left. \displaystyle \frac{\partial {\mathscr L}(x,y,\lambda)}{\partial y} \right| _{(a,b,l)} &= \left. \displaystyle \frac{\partial f(x,y)}{\partial y} \right| _{(a,b,l)} - \lambda \left. \displaystyle \frac{\partial g(x,y)}{\partial y} \right| _{(a,b,l)} = 0 \end{array} \right. \] 拘束条件は,第1式が満たしている.\((\frac{\partial g(x,y)}{\partial x},\frac{\partial g(x,y)}{\partial y}) = (0,0)\)ならば,\(g(x,y)\)が定数を表しているので,\(x,y\)を拘束せず, 拘束条件のない極値を求める問題となる.

    \(\left(\left. \frac{\partial g(x,y)}{\partial x}\right|_{(a,b)}, \left.\frac{\partial g(x,y)}{\partial y}\right| _{(a,b)} \right) \not = (0,0)\)の場合を考える. \(\left.\frac{\partial g(x,y)}{\partial y}\right| _{(a,b)} \not = 0\)とする. 前述の仮定と陰関数定理より,\(a,b\)の近傍で微分可能な連続関数\(y=h(x)\)が存在する. \(h\)は微分可能で,\(g(x,y) = 0\)なので, \[ \frac{d g(x,y)}{d x} = \frac{d g(x,h(x))}{d x} = \frac{\partial g}{\partial x} + \frac{\partial g}{\partial y}\frac{d h}{d x} = 0 \] から, \[ \frac{d h(x)}{d x} = -\frac{\frac{\partial g(x,y)}{\partial x}}{\frac{\partial g(x,y)}{\partial y}} \] また,\(f(x,y) = f(x,h(x))\)と書くことができ, \[ \frac{d f(x,y)}{d x} = \frac{d f(x,h(x))}{d x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{d h}{d x} \] これが,\((x,y) = (a,b)\)で極値を取るなら, \[ \left. \frac{\partial f(x,y)}{\partial x} \right| _{(a,b)} + \left. \frac{\partial f(x,y)}{\partial y} \frac{d h(x)}{d x} \right| _{(a,b)} = 0 \] \(\displaystyle \frac{d h(x)}{d x} = -\frac{\frac{\partial g(x,y)}{\partial x}}{\frac{\partial g(x,y)}{\partial y}}\)から, \[ \left. \frac{\partial f(x,y)}{\partial x} \right| _{(a,b)} - \left. \frac{\partial f(x,y)}{\partial y} \frac{\frac{\partial g(x,y)}{\partial x}}{\frac{\partial g(x,y)}{\partial y}} \right| _{(a,b)} = \left. \frac{\partial f(x,y)}{\partial x} \right| _{(a,b)} - \left. \frac{\frac{\partial f(x,y)}{\partial y}}{\frac{\partial g(x,y)}{\partial y}}\frac{\partial g(x,y)}{\partial x} \right| _{(a,b)} = 0 \] ここで,\(\displaystyle l=\left. \frac{\frac{\partial f(x,y)}{\partial y}}{\frac{\partial g(x,y)}{\partial y}}\right| _{(a,b)}\)とおくと, \[ \left. \frac{\partial f(x,y)}{\partial x} \right| _{(a,b)} - \left. l \frac{\partial g(x,y)}{\partial x} \right| _{(a,b)} = \left. \frac{\partial f(x,y)}{\partial x} \right| _{(a,b,l)} - \lambda \left. \frac{\partial g(x,y)}{\partial x} \right| _{(a,b,l)} = 0 \] 第2式が成立し,第3式は, \[ \begin{align} \left. \frac{\partial f(x,y)}{\partial y} \right| _{(a,b,l)} - \lambda \left. \frac{\partial g(x,y)}{\partial y} \right| _{(a,b,l)} &= \left. \frac{\partial f(x,y)}{\partial y} \right| _{(a,b)} - l \left. \frac{\partial g(x,y)}{\partial y} \right| _{(a,b)} \\ &= \left. \frac{\partial f(x,y)}{\partial y} \right| _{(a,b)} - \left. \frac{\frac{\partial f(x,y)}{\partial y}}{\frac{\partial g(x,y)}{\partial y}} \frac{\partial g(x,y)}{\partial y} \right| _{(a,b)} = 0 \end{align} \] となる. \(\left.\frac{\partial g(x,y)}{\partial x}\right| _{(a,b)} \not = 0\)のときも同様.