第7章 7.24

$x^2 + y^2 = 1 \quad \text{条件}$

$g = x^2 + y^2 - 1 \text{ とおく}$

$f = x^2 + 4\sqrt{2}xy + 3y^2$

$g_x = 2x \qquad g_y = 2y \qquad f_x = 2x + 4\sqrt{2}y \qquad f_y = 4\sqrt{2}x + 6y$

ラグランジュの乗数法から

$\begin{cases} 2x + 4\sqrt{2}y - \lambda 2x = 0 & ① \\ 4\sqrt{2}x + 6y - \lambda 2y = 0 & ② \end{cases}$

①から $x = \dfrac{4\sqrt{2}y}{2(\lambda - 1)}$

②に代入して$\lambda$について解くと $(\lambda - 5)(\lambda + 1) = 0$ $\therefore \lambda = 5, -1$

$\lambda = 5$のとき①から $y = \sqrt{2}x$

条件式に代入して $x = \pm\dfrac{1}{\sqrt{3}} \qquad y = \pm\dfrac{\sqrt{2}}{\sqrt{3}}$

$\lambda = -1$のとき①から $x = -\sqrt{2}y$

条件式に代入して $y = \pm\dfrac{1}{\sqrt{3}} \qquad x = \mp\dfrac{\sqrt{2}}{\sqrt{3}}$

$f_{\left(\pm\frac{1}{\sqrt{3}}, \pm\frac{\sqrt{2}}{\sqrt{3}}\right)} = 5 \qquad f_{\left(\pm\frac{\sqrt{2}}{\sqrt{3}}, \mp\frac{1}{\sqrt{3}}\right)} = -1$

よって$\left(\pm\dfrac{1}{\sqrt{3}}, \pm\dfrac{\sqrt{2}}{\sqrt{3}}\right)$のとき最大値$5$ $\left(\pm\dfrac{\sqrt{2}}{\sqrt{3}}, \mp\dfrac{1}{\sqrt{3}}\right)$のとき最小値$-1$