2変数関数の極値

■問題

次の関数の極値を求めよ．

${\displaystyle f\left(x,y\right)=x^3+y^3+6xy-24}$

■答

${\displaystyle \left(-2,-2\right)}$ で極大値 ${\displaystyle -16}$ 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

${\displaystyle \frac{\partial }{\partial x}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(x^3+y^3+6xy-24\right)}$ ${\displaystyle =3x^2+6y}$

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

${\displaystyle \frac{\partial }{\partial y}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(x^3+y^3+6xy-24\right)}$ ${\displaystyle =3y^2+6x}$

両者を連立させる．

${\displaystyle \{\begin{array}{l}3x^2+6y=0\cdots \cdots \left(1\right)\\ 3y^2+6x=0\cdots \cdots \left(2\right)\end{array}}$

(1)から

${\displaystyle 3x^2+6y=0}$，${\displaystyle x^2+2y=0}$，${\displaystyle 2y=-x^2}$，${\displaystyle y=-\frac{1}{2}{x}^2}$　･･････(3)

これを(2)に代入する．

${\displaystyle 3{\left(-\frac{1}{2}{x}^2\right)}^2+6x=0}$，${\displaystyle {\left(-\frac{1}{2}{x}^2\right)}^2+2x=0}$，${\displaystyle \frac{1}{4}{x}^4+2x=0}$，${\displaystyle x^4+8x=0}$，${\displaystyle x\left(x^3+8\right)=0}$

${\displaystyle \{\begin{array}{l}x=0\\ x^3+8=0\cdots \cdots \left(4\right)\end{array}}$

(4)から

${\displaystyle x^3+8=0}$，${\displaystyle x^3=-8}$，${\displaystyle x^3}$${\displaystyle ={\left(-2\right)}^3}$，${\displaystyle x=-2}$

(3)に ${\displaystyle x=0}$ を代入する．

${\displaystyle y}$${\displaystyle =-\frac{1}{2}·0^2}$${\displaystyle =0}$

(3)に ${\displaystyle x=-2}$ を代入する．

${\displaystyle y}$${\displaystyle =-\frac{1}{2}·{\left(-2\right)}^2}$${\displaystyle =-\frac{1}{2}·4}$${\displaystyle =-2}$

以上から極値をとる候補は ${\displaystyle \left(0,0\right)}$，${\displaystyle \left(-2,-2\right)}$ の2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)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(3x^2+6y\right)}$ ${\displaystyle =6x}$

${\displaystyle f_{yy}\left(x,y\right)=}$${\displaystyle \frac{{\partial }^2}{\partial y^2}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(3y^2+6x\right)}$ ${\displaystyle =6y}$

${\displaystyle f_{xy}\left(x,y\right)=}$${\displaystyle \frac{{\partial }^2}{\partial y\partial x}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(3x^2+6y\right)}$ ${\displaystyle =6}$

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

${\displaystyle A}$${\displaystyle =f_{xx}\left(a,b\right)}$${\displaystyle =6a}$

${\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 =6^2-6a·6b}$${\displaystyle =36-36ab}$${\displaystyle =36\left(1-ab\right)}$

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}$${\displaystyle =6·0}$${\displaystyle =0}$

${\displaystyle D}$${\displaystyle =36\left(1-0·0\right)}$${\displaystyle =36·1}$${\displaystyle =36}$

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

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

${\displaystyle A}$${\displaystyle =6·\left(-2\right)}$ ${\displaystyle =-12}$

${\displaystyle D}$${\displaystyle =36\left\{1-\left(-2\right)·\left(-2\right)\right\}}$${\displaystyle =36·\left(1-4\right)}$${\displaystyle =36·\left(-3\right)}$${\displaystyle =-108}$

となり， ${\displaystyle A<0,D<0}$ から点 ${\displaystyle \left(-2,-2\right)}$ で極大となる．

この点での値は，

${\displaystyle f\left(-2,-2\right)}$${\displaystyle ={\left(-2\right)}^3+{\left(-2\right)}^3+6·\left(-2\right)·\left(-2\right)-24}$${\displaystyle =-8-8+24-24}$${\displaystyle =-16}$

以上からこの関数は点 ${\displaystyle \left(-2,-2\right)}$ で極大値 ${\displaystyle -16}$ をとる．

　

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最終更新日： 2023年9月21日