2変数関数の極値

■問題

次の関数の極値を求めよ．

${\displaystyle f\left(x,y\right)=x^2-2xy+3y^2-4x+5y}$

■答

${\displaystyle \left(\frac{7}{4},-\frac{1}{4}\right)}$ で極小値 ${\displaystyle -\frac{33}{8}}$ 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．

■解説

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(x^2-2xy+3y^2-4x+5y\right)}$${\displaystyle =2x-2y-4}$

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(x^2-2xy+3y^2-4x+5y\right)}$${\displaystyle =-2x+6y+5}$

両者を連立させる．

${\displaystyle \{\begin{array}{l}2x-2y-4=0\cdots \cdots \left(1\right)\\ -2x+6y+5=0\cdots \cdots \left(2\right)\end{array}}$

(1)から

${\displaystyle 2x-2y-4}$${\displaystyle =0}$

${\displaystyle x-y-2}$${\displaystyle =0}$

${\displaystyle x}$${\displaystyle =y+2}$　･･････(3)

これを(2)に代入する．

${\displaystyle -2\left(y+2\right)+6y+5}$${\displaystyle =0}$

${\displaystyle -2y-4+6y+5}$${\displaystyle =0}$

${\displaystyle 4y+1}$${\displaystyle =0}$

${\displaystyle 4y}$${\displaystyle =-1}$

${\displaystyle y}$${\displaystyle =-\frac{1}{4}}$

求めた${\displaystyle y}$ を(3)に代入する．

${\displaystyle x}$${\displaystyle =-\frac{1}{4}+2}$${\displaystyle =\frac{-1+8}{4}}$${\displaystyle =\frac{7}{4}}$

以上から極値をとる候補は ${\displaystyle \left(\frac{7}{4},-\frac{1}{4}\right)}$ 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)=}$${\displaystyle \frac{{\partial }^2}{\partial x^2}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(2x-2y-4\right)}$${\displaystyle =2}$

${\displaystyle f_{yy}\left(x,y\right)=}$${\displaystyle \frac{{\partial }^2}{\partial y^2}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(-2x+6y+5\right)}$${\displaystyle =6}$

${\displaystyle f_{xy}\left(x,y\right)=}$${\displaystyle \frac{{\partial }^2}{\partial y\partial x}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(2x-2y-4\right)}$${\displaystyle =-2}$

以上から ${\displaystyle A,D}$ を求めると

${\displaystyle A}$${\displaystyle =f_{xx}\left(\frac{7}{4},\frac{1}{4}\right)}$ ${\displaystyle =2}$

${\displaystyle D}$${\displaystyle ={\left\{f_{xy}\left(\frac{7}{4},\frac{1}{4}\right)\right\}}^2-f_{xx}\left(\frac{7}{4},\frac{1}{4}\right)·f_{yy}\left(\frac{7}{4},\frac{1}{4}\right)}$ ${\displaystyle ={\left(-2\right)}^2-2·6}$${\displaystyle =4-12}$${\displaystyle =-8}$

となる．

${\displaystyle A>0,D<0}$ より，点 ${\displaystyle \left(\frac{7}{4},-\frac{1}{4}\right)}$ で極小となる．

この点での値は

${\displaystyle f\left(\frac{7}{4},-\frac{1}{4}\right)}$${\displaystyle ={\left(\frac{7}{4}\right)}^2-2·\frac{7}{4}·\left(-\frac{1}{4}\right)+3·{\left(-\frac{1}{4}\right)}^2-4·\frac{7}{4}+5·\left(-\frac{1}{4}\right)}$

${\displaystyle =\frac{49}{16}+\frac{14}{16}+\frac{3}{16}-7-\frac{5}{4}}$

${\displaystyle =\frac{49+14+3-112-20}{16}}$

${\displaystyle =-\frac{66}{16}}$

${\displaystyle =-\frac{33}{8}}$

従って，この関数は点 ${\displaystyle \left(\frac{7}{4},-\frac{1}{4}\right)}$ で極小値 ${\displaystyle -\frac{33}{8}}$ をとる．

　

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最終更新日： 2023年9月21日