# 2変数関数の極値

## ■問題

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

## ■答

$\left(0,-2\right)$ で極小値 $-4$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

$\frac{\partial }{\partial x}f\left(x,y\right)$ $=\frac{\partial }{\partial x}\left(4{x}^{2}+2xy+{y}^{2}+4x+4y\right)$ $=8x+2y+4$

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

$\left\{\begin{array}{l}8x+2y+4=0\text{ }\cdots \cdots \left(1\right)\\ 2x+2y+4=0\text{ }\cdots \cdots \left(2\right)\end{array}$

(2)から

$2x+2y+4$$=0$

$x+y+2$$=0$

$x$$=-y-2$　･･････(3)

これを(1)に代入する．

$8\left(-y-2\right)+2y+4$$=0$

$-8y-16+2y+4$$=0$

$-6y-12$$=0$

$-6y$$=12$

$y$$=-2$

$x$$=-\left(-2\right)-2$$=2-2$$=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)$ $=\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)$ $=\frac{\partial }{\partial x}\left(8x+2y+4\right)$ $=8$

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

${f}_{xy}\left(x,y\right)=$$\frac{{\partial }^{2}}{\partial y\partial x}f\left(x,y\right)$ $=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}f\left(x,y\right)\right)$ $=\frac{\partial }{\partial y}\left(8x+2y+4\right)$ $=2$

$A$$={f}_{xx}\left(0,-2\right)$ $=8$

$D$$={\left\{{f}_{xy}\left(0,-2\right)\right\}}^{2}-{f}_{xx}\left(0,-2\right)·{f}_{yy}\left(0,-2\right)$ $={2}^{2}-8·2$$=4-16$$=-12$

となる．

$A>0,D<0$ より，点 $\left(0,-2\right)$ で極小となる．

この点での値は

$f\left(0,-2\right)$$=4·{0}^{2}-2·0·\left(-2\right)+{\left(-2\right)}^{2}+4·0$$+4·\left(-2\right)$

$=0+0+4+0-8$

$=-4$

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