2変数関数の極値

■問題

次の関数の極値を求めよ．

${\displaystyle f\left(x,y\right)=xy+\frac{1}{x}+\frac{1}{y}}$

■答

点 ${\displaystyle \left(1,1\right)}$ で極小値 ${\displaystyle 3}$をとる．

■ヒント

2変数関数の極値の定理1を使用する．

与えられた関数を ${\displaystyle x,y}$ でそれぞれ偏微分し，連立方程式

${\displaystyle \{\begin{array}{l}{\displaystyle \frac{\partial }{\partial x}f\left(x,y\right)=0}\\ {\displaystyle \frac{\partial }{\partial y}f\left(x,y\right)=0}\end{array}}$

とし，その解${\displaystyle \left(x,y\right)=\left(a,b\right)}$を求める．

更に

${\displaystyle \frac{{\partial }^2}{\partial x^2}f\left(x,y\right)=f_{xx}\left(x,y\right)}$ ，${\displaystyle \frac{{\partial }^2}{\partial y^2}f\left(x,y\right)=f_{yy}\left(x,y\right)}$ ，${\displaystyle \frac{{\partial }^2}{\partial y\partial x}f\left(x,y\right)=f_{xy}\left(x,y\right)}$

をそれぞれ求め

${\displaystyle A=f_{xx}\left(a,b\right),}$ ${\displaystyle D={\left\{f_{xy}\left(a,b\right)\right\}}^2-f_{xx}\left(a,b\right)·f_{yy}\left(a,b\right)}$

を計算して極値を判定する．

■解答

与式を ${\displaystyle x}$ で偏微分（偏導関数の定義より， ${\displaystyle y}$ を定数とみなして ${\displaystyle x}$ で微分）すると

${\displaystyle \frac{\partial }{\partial x}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(xy+\frac{1}{x}+\frac{1}{y}\right)}$ ${\displaystyle =y-\frac{1}{x^2}}$

次に ${\displaystyle f\left(x,y\right)}$ を ${\displaystyle y}$ で偏微分（偏導関数の定義より， ${\displaystyle x}$ を定数とみなして ${\displaystyle y}$ で微分）すると

${\displaystyle \frac{\partial }{\partial y}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(xy+\frac{1}{x}+\frac{1}{y}\right)}$ ${\displaystyle =x-\frac{1}{y^2}}$

両者を連立させる．

${\displaystyle \{\begin{array}{l}{\displaystyle y-\frac{1}{x^2}=0\cdots \cdots \left(1\right)}\\ {\displaystyle x-\frac{1}{y^2}=0\cdots \cdots \left(2\right)}\end{array}}$

(1)から

${\displaystyle y-\frac{1}{x^2}=0}$，${\displaystyle y=\frac{1}{x^2}}$　･･････(3)

これを(2)に代入する．

${\displaystyle x-\frac{1}{\frac{1}{x^2}}=0}$，${\displaystyle x-\frac{1\times x^2}{\frac{1}{x^2}\times x^2}=0}$（分数の性質を用いる．），${\displaystyle x-\frac{x^2}{1}=0}$，${\displaystyle x-x^2=0}$，${\displaystyle x^2-x=0}$，${\displaystyle x\left(x-1\right)=0}$

${\displaystyle \{\begin{array}{l}x=0\\ x-1=0\cdots \cdots \left(4\right)\end{array}}$

(4)から

${\displaystyle x-1=0}$，${\displaystyle x=1}$

また，与式より${\displaystyle x\ne 0}$ である．

よって

${\displaystyle x=1}$

${\displaystyle x=1}$ を(3)に代入する．

${\displaystyle y=\frac{1}{1^2}}$${\displaystyle =1}$

以上から極値をとる候補は ${\displaystyle \left(1,1\right)}$ となる．

次に

${\displaystyle \frac{{\partial }^2}{\partial x^2}f\left(x,y\right)=f_{xx}\left(x,y\right)}$ ，${\displaystyle \frac{{\partial }^2}{\partial y^2}f\left(x,y\right)=f_{yy}\left(x,y\right)}$ ，${\displaystyle \frac{{\partial }^2}{\partial y\partial x}f\left(x,y\right)=f_{xy}\left(x,y\right)}$

をそれぞれ求める．

${\displaystyle f_{xx}\left(x,y\right)=}$${\displaystyle \frac{{\partial }^2}{\partial x^2}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(y-\frac{1}{x^2}\right)}$ ${\displaystyle =\frac{2}{x^3}}$

${\displaystyle f_{yy}\left(x,y\right)=}$ ${\displaystyle \frac{{\partial }^2}{\partial y^2}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(x-\frac{1}{y^2}\right)}$ ${\displaystyle =\frac{2}{y^3}}$

${\displaystyle f_{xy}\left(x,y\right)=}$ ${\displaystyle \frac{{\partial }^2}{\partial y\partial x}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(y-\frac{1}{x^2}\right)}$ ${\displaystyle =1}$

$\left(x,y\right)=\left(a,b\right)$における$A$，$D$の値は

${\displaystyle A}$${\displaystyle =f_{xx}\left(a,b\right)}$$$ {\displaystyle =\frac{2}{a^3}}$$

${\displaystyle D}$${\displaystyle ={\left\{f_{xy}\left(a,b\right)\right\}}^2-f_{xx}\left(a,b\right)·f_{yy}\left(a,b\right)}$${\displaystyle =1^2-\frac{2}{a^3}·\frac{2}{b^3}}$${\displaystyle =1-\frac{4}{a^3{b}^3}}$

これを元に各点における ${\displaystyle A,D}$ を求める．

●点 ${\displaystyle \left(1,1\right)}$ においては

${\displaystyle A}$${\displaystyle =\frac{2}{1^3}}$${\displaystyle =\frac{2}{1}}$${\displaystyle =2}$

${\displaystyle D}$${\displaystyle =1-\frac{4}{1^3·1^3}}$${\displaystyle =1-\frac{4}{1·1}}$${\displaystyle =1-\frac{4}{1}}$${\displaystyle =1-4}$${\displaystyle =-3}$

以上より， ${\displaystyle A>0,D<0}$ より，点 ${\displaystyle \left(1,1\right)}$ で極小となる．

この点での値は

${\displaystyle f\left(1,1\right)}$${\displaystyle =1·1+\frac{1}{1}+\frac{1}{1}}$${\displaystyle =1+1+1}$${\displaystyle =3}$

従って，この関数は点 ${\displaystyle \left(1,1\right)}$ で極小値 ${\displaystyle 3}$ をとる．

　

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最終更新日： 2023年9月21日