2変数関数の極値

■問題

次の関数の極値を求めよ．

${\displaystyle f\left(x,y\right)=x^2+2y^2+10x}$

■答

点${\displaystyle \left(-5,0\right)}$ で極小値 ${\displaystyle -25}$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

${\displaystyle \frac{\partial }{\partial x}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(x^2+2y^2+10x\right)}$ ${\displaystyle =2x+10}$

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+2y^2+10x\right)}$${\displaystyle =4y}$

両者を連立させる．

${\displaystyle \{\begin{array}{l}2x+10=0\cdots \cdots \left(1\right)\\ 4y=0\cdots \cdots \left(2\right)\end{array}}$

(1)から

${\displaystyle 2x}$ ${\displaystyle =-10}$

${\displaystyle x}$ ${\displaystyle =-5}$

(2)から

${\displaystyle y=0}$

以上から極値をとる候補は ${\displaystyle \left(-5,0\right)}$ 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+10\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(4y\right)}$${\displaystyle =4}$

${\displaystyle f_{xy}\left(x,y\right)=}$ ${\displaystyle \frac{{\partial }^2}{\partial y\partial x}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(2x+10\right)}$${\displaystyle =0}$

以上から ${\displaystyle A,D}$ を求めると

${\displaystyle A}$ ${\displaystyle =f_{xx}\left(5,0\right)}$ ${\displaystyle =2}$

${\displaystyle D}$ ${\displaystyle ={\left\{f_{xy}\left(5,0\right)\right\}}^2-f_{xx}\left(5,0\right)·f_{yy}\left(5,0\right)}$ ${\displaystyle =0^2-2·4}$${\displaystyle =-8}$

となる．

${\displaystyle A>0,D<0}$ より，点 ${\displaystyle \left(-5,0\right)}$ で極小となる．

この点での値は

${\displaystyle f\left(-5,0\right)}$ ${\displaystyle ={\left(-5\right)}^2+2·0^2+10·\left(-5\right)}$ ${\displaystyle =25+0-50}$ ${\displaystyle =-25}$

したがって，この関数は点 ${\displaystyle \left(-5,0\right)}$ で極小値 ${\displaystyle -25}$ をとる．

　

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最終更新日： 2023年9月21日