2変数関数の極値

■問題

次の関数の極値を求めよ．

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

■答

${\displaystyle \left(1,-\frac{1}{2}\right)}$ で極大値 ${\displaystyle \frac{1}{2}}$ 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

与式を展開する．

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

これを ${\displaystyle x}$ 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

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

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

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

両者を連立させる．

${\displaystyle \{\begin{array}{l}2xy-2y^2-3y=0\cdots \cdots \left(1\right)\\ x^2-4xy-3x=0\cdots \cdots \left(2\right)\end{array}}$

(1)は

${\displaystyle 2xy-2y^2-3y}$${\displaystyle =0}$

${\displaystyle y\left(2x-2y-3\right)}$${\displaystyle =0}$

となり， ${\displaystyle y=0}$ または ${\displaystyle 2x-2y-3=0}$ を満たせば良い．

同様に(2)も，

${\displaystyle x^2-4xy-3x}$${\displaystyle =0}$

${\displaystyle x\left(x-4y-3\right)}$${\displaystyle =0}$

となり， ${\displaystyle x=0}$ または ${\displaystyle x-4y-3=0}$ を満たせば良い．

よってこの連立方程式の解は

(i)${\displaystyle \{\begin{array}{l}y=0\\ x=0\end{array}}$

(ii)${\displaystyle \{\begin{array}{l}y=0\\ x-4y-3=0\cdots \cdots \left(3\right)\end{array}}$

(iii)${\displaystyle \{\begin{array}{l}2x-2y-3=0\cdots \cdots \left(4\right)\\ x=0\end{array}}$

(iv)${\displaystyle \{\begin{array}{l}2x-2y-3=0\cdots \cdots \left(4\right)\\ x-4y-3=0\cdots \cdots \left(3\right)\end{array}}$

これら4組の連立方程式の解となる．

(i)の連立方程式の解は${\displaystyle \left(x,y\right)=}$ ${\displaystyle \left(0,0\right)}$

(ii)の連立方程式の解は ${\displaystyle y=0}$ を(3)に代入して

${\displaystyle x-4·0-3=0}$，${\displaystyle x-3=0}$，${\displaystyle x=3}$

よって，${\displaystyle \left(x,y\right)=}$ ${\displaystyle \left(3,0\right)}$

(iii)の連立方程式の解は ${\displaystyle x=0}$ を(4)に代入して

${\displaystyle 2·0-2y-3=0}$，${\displaystyle -2y-3=0}$，${\displaystyle -2y=3}$，${\displaystyle y=-\frac{3}{2}}$

よって， ${\displaystyle \left(x,y\right)=}$${\displaystyle \left(0,-\frac{3}{2}\right)}$

(iv)の連立方程式の解は(4)から

${\displaystyle 2x-2y-3=0}$，${\displaystyle 2x=2y+3}$，${\displaystyle x=y+\frac{3}{2}}$　･･････(5)

これを(3)に代入して

${\displaystyle y+\frac{3}{2}-4y-3=0}$，${\displaystyle -3y+\frac{3-6}{2}=0}$，${\displaystyle -3y-\frac{3}{2}=0}$，${\displaystyle -3y=\frac{3}{2}}$，${\displaystyle y=-\frac{1}{2}}$

求めた${\displaystyle y}$ を(5)に代入して

${\displaystyle x}$${\displaystyle =-\frac{1}{2}+\frac{3}{2}}$${\displaystyle =\frac{-1+3}{2}}$${\displaystyle =\frac{2}{2}}$${\displaystyle =1}$

よって，(i)，(ii)，(iii)，(iv)より極値をとる候補は ${\displaystyle \left(0,0\right)}$，${\displaystyle \left(3,0\right)}$，${\displaystyle \left(0,-\frac{3}{2}\right)}$，${\displaystyle \left(1,-\frac{1}{2}\right)}$ の4点となる．

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

${\displaystyle f_{yy}\left(x,y\right)=}$${\displaystyle \frac{{\partial }^2}{\partial y^2}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(x^2-4xy-3x\right)}$ ${\displaystyle =-4x}$

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

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

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

${\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 ={\left(2a-4b-3\right)}^2-2b·\left(-4a\right)}$

${\displaystyle ={\left(2a-4b-3\right)}^2+8ab}$

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

●点 ${\displaystyle \left(0,0\right)}$ においては

${\displaystyle A}$${\displaystyle =2·0}$${\displaystyle =0}$

${\displaystyle D}$${\displaystyle ={\left(2·0-4·0-3\right)}^2+8·0·0}$${\displaystyle ={\left(-3\right)}^2+0}$${\displaystyle =9}$

${\displaystyle D>0}$ となり ${\displaystyle \left(0,0\right)}$ は極値ではない．

●点 ${\displaystyle \left(3,0\right)}$ においては

${\displaystyle A}$${\displaystyle =2·0}$${\displaystyle =0}$

${\displaystyle D}$${\displaystyle ={\left(2·3-4·0-3\right)}^2+8·3·0}$${\displaystyle ={\left(6-3\right)}^2+0}$${\displaystyle =3^2}$${\displaystyle =9}$

${\displaystyle D>0}$ となり ${\displaystyle \left(3,0\right)}$ は極値ではない．

●点 ${\displaystyle \left(0,-\frac{3}{2}\right)}$ においては

${\displaystyle A}$${\displaystyle =2·\left(-\frac{3}{2}\right)}$${\displaystyle =-3}$

${\displaystyle D>0}$ となり ${\displaystyle \left(0,-\frac{3}{2}\right)}$ は極値ではない．

●点 ${\displaystyle \left(1,-\frac{1}{2}\right)}$ においては

${\displaystyle A}$${\displaystyle =2·\left(-\frac{1}{2}\right)}$${\displaystyle =-1}$

となり， ${\displaystyle A<0,D<0}$ より点 ${\displaystyle \left(1,-\frac{1}{2}\right)}$ で極大となる．

この点での値は

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

以上からこの関数は点 ${\displaystyle \left(1,-\frac{1}{2}\right)}$ で極大値 ${\displaystyle \frac{1}{2}}$ をとる．

　

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最終更新日： 2023年9月21日