# 2変数関数の極値

## ■問題

$f\left(x,y\right)={x}^{2}+2{y}^{2}+10x$

## ■答

$\left(-5,0\right)$ で極小値 $-25$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

$\frac{\partial }{\partial x}f\left(x,y\right)$ $=\frac{\partial }{\partial x}\left({x}^{2}+2{y}^{2}+10x\right)$ $=2x+10$

$\frac{\partial }{\partial y}f\left(x,y\right)$ $=\frac{\partial }{\partial y}\left({x}^{2}+2{y}^{2}+10x\right)$$=4y$

$\left\{\begin{array}{l}2x+10=0\text{ }\cdots \cdots \left(1\right)\\ 4y=0\text{ }\cdots \cdots \left(2\right)\end{array}$

(1)から

$2x$ $=-10$

$x$ $=-5$

(2)から

$y=0$

$\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+10\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(4y\right)$$=4$

${f}_{xy}\left(x,y\right)=$ $\frac{{\partial }^{2}}{\partial y\partial x}f\left(x,y\right)$ $=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)$ $=\frac{\partial }{\partial y}\left(2x+10\right)$$=0$

$A$ $={f}_{xx}\left(5,0\right)$ $=2$

$D$ $={\left\{{f}_{xy}\left(5,0\right)\right\}}^{2}-{f}_{xx}\left(5,0\right)·{f}_{yy}\left(5,0\right)$ $={0}^{2}-2·4$$=-8$

となる．

$A>0,D<0$ より，点 $\left(-5,0\right)$ で極小となる．

この点での値は

$f\left(-5,0\right)$ $={\left(-5\right)}^{2}+2·{0}^{2}+10·\left(-5\right)$ $=25+0-50$ $=-25$

したがって，この関数は点 $\left(-5,0\right)$ で極小値 $-25$ をとる．

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