# 2変数関数の極値

## ■問題

$f\left(x,y\right)={x}^{2}-2xy+3{y}^{2}-4x+5y$

## ■答

$\left(\frac{7}{4},-\frac{1}{4}\right)$ で極小値 $-\frac{33}{8}$ をとる．

## ■ヒント

2変数関数の極値の定理1を使用する．

$\left\{\begin{array}{l}\frac{\partial }{\partial x}f\left(x,y\right)=0\\ \frac{\partial }{\partial y}f\left(x,y\right)=0\end{array}$

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

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

$A={f}_{xx}\left(a,b\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}$ $D={\left\{{f}_{xy}\left(a,b\right)\right\}}^{2}-{f}_{xx}\left(a,b\right)·{f}_{yy}\left(a,b\right)$

を計算して極値を判定する．

## ■解説

$\frac{\partial }{\partial x}f\left(x,y\right)$ $=\frac{\partial }{\partial x}\left({x}^{2}-2xy+3{y}^{2}-4x+5y\right)$$=2x-2y-4$

$\frac{\partial }{\partial y}f\left(x,y\right)$ $=\frac{\partial }{\partial y}\left({x}^{2}-2xy+3{y}^{2}-4x+5y\right)$$=-2x+6y+5$

$\left\{\begin{array}{l}2x-2y-4=0\text{ }\cdots \cdots \left(1\right)\\ -2x+6y+5=0\text{ }\cdots \cdots \left(2\right)\end{array}$

(1)から

$2x-2y-4$$=0$

$x-y-2$$=0$

$x$$=y+2$　･･････(3)

これを(2)に代入する．

$-2\left(y+2\right)+6y+5$$=0$

$-2y-4+6y+5$$=0$

$4y+1$$=0$

$4y$$=-1$

$y$$=-\frac{1}{4}$

$x$$=-\frac{1}{4}+2$$=\frac{-1+8}{4}$$=\frac{7}{4}$

$\frac{{\partial }^{2}}{\partial {x}^{2}}f\left(x,y\right)={f}_{xx}\left(x,y\right)$$\frac{{\partial }^{2}}{\partial {y}^{2}}f\left(x,y\right)={f}_{yy}\left(x,y\right)$$\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．

${f}_{xx}\left(x,y\right)=$$\frac{{\partial }^{2}}{\partial {x}^{2}}f\left(x,y\right)$ $=\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)$ $=\frac{\partial }{\partial x}\left(2x-2y-4\right)$$=2$

${f}_{yy}\left(x,y\right)=$$\frac{{\partial }^{2}}{\partial {y}^{2}}f\left(x,y\right)$ $=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f\left(x,y\right)\right)$ $=\frac{\partial }{\partial y}\left(-2x+6y+5\right)$$=6$

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

$A$$={f}_{xx}\left(\frac{7}{4},\frac{1}{4}\right)$ $=2$

$D$$={\left\{{f}_{xy}\left(\frac{7}{4},\frac{1}{4}\right)\right\}}^{2}-{f}_{xx}\left(\frac{7}{4},\frac{1}{4}\right)·{f}_{yy}\left(\frac{7}{4},\frac{1}{4}\right)$ $={\left(-2\right)}^{2}-2·6$$=4-12$$=-8$

となる．

$A>0,D<0$ より，点 $\left(\frac{7}{4},-\frac{1}{4}\right)$ で極小となる．

この点での値は

$f\left(\frac{7}{4},-\frac{1}{4}\right)$$={\left(\frac{7}{4}\right)}^{2}-2·\frac{7}{4}·\left(-\frac{1}{4}\right)+3·{\left(-\frac{1}{4}\right)}^{2}-4·\frac{7}{4}+5·\left(-\frac{1}{4}\right)$

$=\frac{49}{16}+\frac{14}{16}+\frac{3}{16}-7-\frac{5}{4}$

$=\frac{49+14+3-112-20}{16}$

$=-\frac{66}{16}$

$=-\frac{33}{8}$

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