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 $ =xy+\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}y\hspace{2}}}$

■答

点 $3$\displaystyle{ $ 1,1 $ }$ で極小値 $3$\displaystyle{3}$ 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．

■解答

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}} $ xy+\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}y\hspace{2}} $ }$ $3$\displaystyle{=y- \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}}}$

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}} $ xy+\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}y\hspace{2}} $ }$ $3$\displaystyle{=x- \frac{\hspace{2}1\hspace{2}}{\hspace{2} {y}^{2} \hspace{2}}}$

両者を連立させる．

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

(1)から

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

これを(2)に代入する．

$3$\displaystyle{x- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}} \hspace{2}}=0}$ ， $3$\displaystyle{x- \frac{\hspace{2} 1\times {x}^{2} \hspace{2}}{\hspace{2} \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}}\times {x}^{2} \hspace{2}}=0}$ （分数の性質を用いる．）， $3$\displaystyle{x- \frac{\hspace{2} {x}^{2} \hspace{2}}{\hspace{2}1\hspace{2}}=0}$ ， $3$\displaystyle{x- {x}^{2}=0}$ ， $3$\displaystyle{{x}^{2}- x=0}$ ， $3$\displaystyle{x $ x- 1 $ =0}$

$3$\displaystyle{ \{\begin{array}{lll}x=0& \vspace{6}\\ x- 1=0\text{\hspace{5}}\cdots \cdots $4$ & \vspace{6}\\ \end{array} }$

(4)から

$3$\displaystyle{x- 1=0}$ ， $3$\displaystyle{x=1}$

また，与式より $3$\displaystyle{x\neq 0}$ である．

よって

$3$\displaystyle{x=1}$

$3$\displaystyle{x=1}$ を(3)に代入する．

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

以上から極値をとる候補は $3$\displaystyle{ $ 1,1 $ }$ となる．

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

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

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

$3$\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$$ = 2 a 3

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

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

●点 $3$\displaystyle{ $ 1,1 $ }$ においては

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

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

以上より， $3$\displaystyle{A> 0,D< 0}$ より，点 $3$\displaystyle{ $ 1,1 $ }$ で極小となる．

この点での値は

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

従って，この関数は点 $3$\displaystyle{ $ 1,1 $ }$ で極小値 $3$\displaystyle{3}$ をとる．

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最終更新日： 2023年9月21日