2変数関数の極値

■問題

次の関数の極値を求めよ．

$3$\displaystyle{f $ x,y $ =\cos x+\cos y+\cos $ x+y $ }$ 　 $3$\displaystyle{ \left({ 0< x< {\pi},0< y< {\pi} }\right) }$

■答

$3$\displaystyle{ $ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi},\frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi} $ }$ で極小値 $3$\displaystyle{- \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{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 $3$\displaystyle{x,y}$ でそれぞれ偏微分し，連立方程式

$3$\displaystyle{ \{\begin{array}{lll}\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}f $ x,y $ =0}& \vspace{6}\\ \displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}f $ x,y $ =0}& \vspace{6}\\ \end{array} }$

とし，その解 $3$\displaystyle{\left({ x,y }\right)=\left({ a,b }\right)}$ を求める．

更に

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2} \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}f\left({ x,y }\right)=\hspace{1}{f}_{ xx }\left({ x,y }\right)}$ ， $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2} \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}f\left({ x,y }\right)=\hspace{1}{f}_{ yy }\left({ x,y }\right)}$ ， $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2} \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}f\left({ x,y }\right)=\hspace{1}{f}_{ xy }\left({ x,y }\right)}$

をそれぞれ求め

$3$\displaystyle{ A=\hspace{1}{f}_{ xx } $ a,b $ \text{\hspace{1}},\text{\hspace{1}} }$ $3$\displaystyle{ D={ \{ \hspace{1}{f}_{ xy } $ a,b $ \} }^{2}- \hspace{1}{f}_{ xx } $ a,b $ \cdot \hspace{1}{f}_{ yy } $ a,b $ }$

を計算して極値を判定する．

■解説

与式を $3$\displaystyle{x}$ で偏微分（偏導関数の定義より， $3$\displaystyle{y}$ を定数とみなして $3$\displaystyle{x}$ で微分）すると

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}f $ x,y $ }$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \cos x+\cos y+\cos \( x+y $ \) }$ $3$\displaystyle{=- \sin x- \sin $ x+y $ \cdot 1}$ $3$\displaystyle{=- \sin x- \sin $ x+y $ }$

次に $3$\displaystyle{f $ x,y $ }$ を $3$\displaystyle{y}$ で偏微分（偏導関数の定義より， $3$\displaystyle{x}$ を定数とみなして $3$\displaystyle{y}$ で微分）すると

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}f $ x,y $ }$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \cos x+\cos y+\cos \( x+y $ \) }$ $3$\displaystyle{=- \sin y- \sin $ x+y $ \cdot 1}$ $3$\displaystyle{=- \sin y- \sin $ x+y $ }$

両者を連立させる．

$3$\displaystyle{ \{\begin{array}{lll}- \sin x- \sin $ x+y $ =0\text{\hspace{5}}\cdots \cdots $1$ & \vspace{6}\\ - \sin y- \sin $ x+y $ =0\text{\hspace{5}}\cdots \cdots $2$ & \vspace{6}\\ \end{array} }$

(1)から

$3$\displaystyle{- \sin x- \sin $ x+y $ =0}$ ， $3$\displaystyle{\sin $ x+y $ =- \sin x}$

これを(2)に代入する．

$3$\displaystyle{- \sin y- $ - \sin x $ =0}$ ， $3$\displaystyle{- \sin y+\sin x=0}$ ， $3$\displaystyle{\sin y=\sin x}$

$3$\displaystyle{ 0< x< {\pi} }$ ， $3$\displaystyle{ 0< y< {\pi} }$ より

$3$\displaystyle{y=x}$ ， $3$\displaystyle{y={\pi}- x}$ 　（∵ $3$\displaystyle{\sin {\theta}=\sin \left({ {\pi}- {\theta} }\right)}$ 　ここを参照）

◇ $3$\displaystyle{y=x}$ の時

この関係を(1)に代入して

$3$\displaystyle{- \sin x- \sin $ x+x $ =0}$ ， $3$\displaystyle{\sin x+\sin 2x=0}$ ， $3$\displaystyle{\sin x+2\sin x\cos x=0}$ （2倍角の公式を使用する）， $3$\displaystyle{\sin x $ 1+2\cos x $ =0}$

$3$\displaystyle{ 0< x< {\pi} }$ より $3$\displaystyle{\sin x\neq 0}$ ，よって

$3$\displaystyle{ 1+2\cos x=0 }$

この方程式を解く．

$3$\displaystyle{1+2\cos x=0}$ ， $3$\displaystyle{2\cos x=- 1}$ ， $3$\displaystyle{\cos x=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}}$

$3$\displaystyle{0< x< {\pi}}$ の条件から，これを満たす $3$\displaystyle{x}$ は

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

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

◇ $3$\displaystyle{y={\pi}- x}$ の時

この関係を(1)に代入する.

$3$\displaystyle{- \sin x- \sin \left({ x+{\pi}- x }\right)=0}$ ， $3$\displaystyle{- \sin x- \sin {\pi}=0}$ ， $3$\displaystyle{- \sin x=0}$ ， $3$\displaystyle{\sin x=0}$

$3$\displaystyle{\sin x\neq 0}$ より， $3$\displaystyle{y={\pi}- x}$ のとき解なし．

次に

をそれぞれ求める．

$3$\displaystyle{ \hspace{1}{f}_{ xx }\left({ x,y }\right)= }$ $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2} \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}f $ x,y $ }$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}f \( x,y $ \) }$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ - \sin x- \sin \( x+y $ \) }$

$3$\displaystyle{=- \cos x- \cos $ x+y $ }$

$3$\displaystyle{ \hspace{1}{f}_{ xy }\left({ x,y }\right)= }$ $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2} \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}f $ x,y $ }$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}f \( x,y $ \) }$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ - \sin x- \sin \( x+y $ \) }$ $3$\displaystyle{=- \cos $ x+y $ }$

$3$\displaystyle{ \hspace{1}{f}_{ yy }\left({ x,y }\right)= }$ $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2} \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}f $ x,y $ }$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}f \( x,y $ \) }$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ - \sin y- \sin \( x+y $ \) }$

$3$\displaystyle{=- \cos y- \cos $ x+y $ }$

$3$\left({ x,y }\right)=\left({ a,b }\right)$ における $3$A$ ， $3$D$ の値は

$3$\displaystyle{A}$ $3$\displaystyle{=\hspace{1}{f}_{ xx } $ a,b $ }$ $3$\displaystyle{=- \cos a- \cos $ a+b $ }$

$3$\displaystyle{D}$ $3$\displaystyle{={ \{ \hspace{1}{f}_{ xy } $ a,b $ \} }^{2}- \hspace{1}{f}_{ xx } $ a,b $ \cdot \hspace{1}{f}_{ yy } $ a,b $ }$

$3$\displaystyle{={ \{ - \cos $ a+b $ \} }^{2}}$ $3$\displaystyle{- $ - \cos a- \cos \( a+b $ \) \cdot $ - \cos b- \cos \( a+b $ \) }$

$3$\displaystyle{={ \{ - \cos $ a+b $ \} }^{2}}$ $3$\displaystyle{- \{ \cos a+\cos $ a+b $ \} \cdot \{ \cos b+\cos $ a+b $ \} }$

これを元に各点における $3$\displaystyle{A,D}$ を求める．

●点 $3$\displaystyle{ $ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi},\frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi} $ }$ においては

$3$\displaystyle{A}$ $3$\displaystyle{=- \cos x- \cos $ x+y $ }$ $3$\displaystyle{=- \cos \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}- \cos $ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi} $ }$

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

$3$\displaystyle{D}$ $3$\displaystyle{={ \{ - \cos $ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi} $ \} }^{2}}$

$3$\displaystyle{- \{ \cos \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\cos $ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi} $ \} \cdot \{ \cos \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\cos $ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi} $ \} }$

$3$\displaystyle{={ $ - \cos \frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}{\pi} $ }^{2}}$ $3$\displaystyle{- $ \cos \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\cos \frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}{\pi} $ \cdot $ \cos \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\cos \frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}{\pi} $ }$

$3$\displaystyle{={ \{ - $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ \} }^{2}}$ $3$\displaystyle{- $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ \cdot $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ }$

$3$\displaystyle{={ $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ }^{2}- $ - 1 $ \cdot $ - 1 $ }$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}- 1}$ $3$\displaystyle{=- \frac{\hspace{2}3\hspace{2}}{\hspace{2}4\hspace{2}}}$

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

この点での値は

$3$\displaystyle{f $ 1,1 $ }$ $3$\displaystyle{=\cos \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\cos \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\cos $ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}{\pi} $ }$

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

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

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最終更新日： 2023年9月22日