2変数関数の極値

■問題

次の関数の極値を求めよ．

${\displaystyle f\left(x,y\right)=cosx+cosy+cos\left(x+y\right)}$　${\displaystyle \left(0<x<\pi ,0<y<\pi \right)}$

■答

${\displaystyle \left(\frac{2}{3}\pi ,\frac{2}{3}\pi \right)}$ で極小値 ${\displaystyle -\frac{3}{2}}$ 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 ${\displaystyle x}$ 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

${\displaystyle \frac{\partial }{\partial x}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(cosx+cosy+cos\left(x+y\right)\right)}$ ${\displaystyle =-sinx-sin\left(x+y\right)·1}$ ${\displaystyle =-sinx-sin\left(x+y\right)}$

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

${\displaystyle \frac{\partial }{\partial y}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial y}\left(cosx+cosy+cos\left(x+y\right)\right)}$ ${\displaystyle =-siny-sin\left(x+y\right)·1}$ ${\displaystyle =-siny-sin\left(x+y\right)}$

両者を連立させる．

${\displaystyle \{\begin{array}{l}-sinx-sin\left(x+y\right)=0\cdots \cdots \left(1\right)\\ -siny-sin\left(x+y\right)=0\cdots \cdots \left(2\right)\end{array}}$

(1)から

${\displaystyle -sinx-sin\left(x+y\right)=0}$，${\displaystyle sin\left(x+y\right)=-sinx}$

これを(2)に代入する．

${\displaystyle -siny-\left(-sinx\right)=0}$，${\displaystyle -siny+sinx=0}$，${\displaystyle siny=sinx}$

${\displaystyle 0<x<\pi }$ ，${\displaystyle 0<y<\pi }$より

${\displaystyle y=x}$，${\displaystyle y=\pi -x}$ 　（∵${\displaystyle \sin \theta =\sin \left(\pi -\theta \right)}$ 　ここを参照）

◇${\displaystyle y=x}$ の時

この関係を(1)に代入して

${\displaystyle -sinx-sin\left(x+x\right)=0}$，${\displaystyle sinx+sin2x=0}$，${\displaystyle sinx+2sinxcosx=0}$（2倍角の公式を使用する），${\displaystyle sinx\left(1+2cosx\right)=0}$

${\displaystyle 0<x<\pi }$ より${\displaystyle \sin x\ne 0}$ ，よって

${\displaystyle 1+2\cos x=0}$

この方程式を解く．

${\displaystyle 1+2cosx=0}$，${\displaystyle 2cosx=-1}$，${\displaystyle cosx=-\frac{1}{2}}$

${\displaystyle 0<x<\pi }$ の条件から，これを満たす ${\displaystyle x}$ は

${\displaystyle x=\frac{2}{3}\pi }$　ここを参照

${\displaystyle y=x}$ の関係から，極値をとる候補は ${\displaystyle \left(\frac{2}{3}\pi ,\frac{2}{3}\pi \right)}$ の1点となる．

◇${\displaystyle y=\pi -x}$ の時

この関係を(1)に代入する.

${\displaystyle -\sin x-\sin \left(x+\pi -x\right)=0}$ ，${\displaystyle -\sin x-\sin \pi =0}$ ，${\displaystyle -\sin x=0}$ ，${\displaystyle \sin x=0}$

${\displaystyle \sin x\ne 0}$ より，${\displaystyle y=\pi -x}$ のとき解なし．

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)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(-sinx-sin\left(x+y\right)\right)}$

${\displaystyle =-cosx-cos\left(x+y\right)}$

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

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

${\displaystyle =-cosy-cos\left(x+y\right)}$

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

${\displaystyle A}$${\displaystyle =f_{xx}\left(a,b\right)}$${\displaystyle =-cosa-cos\left(a+b\right)}$

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

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

${\displaystyle A}$${\displaystyle =-cosx-cos\left(x+y\right)}$${\displaystyle =-cos\frac{2}{3}\pi -cos\left(\frac{2}{3}\pi +\frac{2}{3}\pi \right)}$

${\displaystyle =-cos\frac{2}{3}\pi -cos\frac{4}{3}\pi }$${\displaystyle =-\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)}$${\displaystyle =\frac{1}{2}+\frac{1}{2}}$${\displaystyle =1}$

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

${\displaystyle ={\left(\frac{1}{2}\right)}^2-\left(-1\right)·\left(-1\right)}$${\displaystyle =\frac{1}{4}-1}$${\displaystyle =-\frac{3}{4}}$

以上より, ${\displaystyle A>0,D<0}$ から点 ${\displaystyle \left(\frac{2}{3}\pi ,\frac{2}{3}\pi \right)}$ で極小となる．

この点での値は

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

${\displaystyle =cos\frac{2}{3}\pi +cos\frac{2}{3}\pi +cos\frac{4}{3}\pi }$${\displaystyle =-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}}$${\displaystyle =-\frac{3}{2}}$

従って，この関数は点 ${\displaystyle \left(\frac{2}{3}\pi ,\frac{2}{3}\pi \right)}$ で極小値 ${\displaystyle -\frac{3}{2}}$ をとる．

　

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最終更新日： 2023年9月22日