第7章 7.17

(1)

$S = x\sqrt{r^2 - \left(\frac{x}{2}\right)^2}$

$S' = \sqrt{r^2 - \left(\frac{x}{2}\right)^2} + x\frac{-\frac{1}{2}x}{2\sqrt{r^2 - \left(\frac{x}{2}\right)^2}}$

$= \sqrt{r^2 - \left(\frac{x}{2}\right)^2} + x\frac{-\frac{1}{2}x}{2\sqrt{r^2 - \left(\frac{x}{2}\right)^2}} = \frac{4\left(r^2 - \frac{x^2}{4}\right) - x^2}{4\sqrt{r^2 - \left(\frac{x}{2}\right)^2}} = \frac{4r^2 - 2x^2}{4\sqrt{r^2 - \left(\frac{x}{2}\right)^2}}$

$S'=0$のとき $4r^2-2x^2=0$ $x=\pm\sqrt{2}r$ $x>0$より $x=\sqrt{2}r$

これから２辺は $\sqrt{2}r, \dfrac{1}{\sqrt{2}}r$

(2)

$\frac{l}{2} = x + \sqrt{r^2 - \frac{x^2}{4}}$

$\left(\frac{l}{2}\right)' = 1 + \frac{-\frac{1}{2}x}{2\sqrt{r^2-\frac{x^2}{4}}} = \frac{4\sqrt{r^2-\frac{x^2}{4}} - x}{4\sqrt{r^2-\frac{x^2}{4}}}$

$\left(\dfrac{l}{2}\right)'=0$のとき $4\sqrt{r^2-\dfrac{x^2}{4}} - x = 0$ $r^2 - \dfrac{x^2}{4} = \dfrac{1}{16}x^2$ $16r^2-4x^2=x^2$ $x=\dfrac{4}{\sqrt{5}}r$

これから２辺は $\dfrac{4}{\sqrt{5}}r, \dfrac{1}{\sqrt{5}}r$