Optimization 3.2

$x = 2r\sin(\dfrac{\theta}{2})$ and $h^{*} = r\cos(\dfrac{\theta}{2})$ $\implies A_{\triangle{ABC}} = \dfrac{1}{2}r^2\sin(\theta) = a \implies r = \sqrt{\dfrac{2a}{\sin(\theta)}}$ .

$rH = 1 \implies H = \sqrt{\dfrac{\sin(\theta)}{2a}} \implies D(\theta) = d^2(\theta) = \dfrac{\sin(\theta)}{2a} + \dfrac{2a}{\sin(\theta)} \implies \dfrac{dD}{d\theta} = \dfrac{\cos(\theta)}{2a} - \dfrac{2a\cos(\theta)}{\sin^2(\theta)} =$

$\cos(\theta)(\dfrac{\sin^2(\theta) - 4a^2}{2a\sin^2(\theta)}) = 0.$

For $(0 < \theta < \pi)$ $\:\ \sin(\theta) > 0 \implies \sin(\theta) = 2a \implies$ $(0 < a = \dfrac{\sin(\theta)}{2} \leq \dfrac{1}{2})$ .

$\dfrac{d^2\theta}{d\theta^2} = 2\cot^2(\theta) > 0$ for $\theta \neq \dfrac{\pi}{2}$ .

For $\theta = \dfrac{\pi}{2}$ and $a = \dfrac{1}{2}$ :

On $(0,\dfrac{\pi}{2}) \implies \dfrac{dD}{d\theta} < 0$ and on $(\dfrac{\pi}{2},\pi) \implies \dfrac{dD}{d\theta} > 0$ $\implies d$ is minimized wherever $(0 < a \leq \dfrac{1}{2})$ .

(1) $a_{max} = \dfrac{1}{2}$

(2) $d_{min} = \sqrt{2}$

(3) $r = 1$ and $a_{max} = \dfrac{1}{2} \implies \theta = \dfrac{\pi}{2} \implies x = 2\sin(\dfrac{\pi}{4}) = \sqrt{2}$ and $h^{*} = \dfrac{1}{\sqrt{2}}$ and converting to degrees $\theta = 90^{\circ}$

Let $\lambda = m\angle{FAE}$ .

(4) Clearly $\lambda = \boxed{60^{\circ}}$ , since $\triangle{FAE}$ is an equilateral triangle and $H = 1$ .

Using the law of cosines $\implies 4(1 - \cos(\lambda)) = 2 \implies \cos(\lambda) = \dfrac{1}{2} \implies \lambda = 60^{\circ}$

$\therefore d = \sqrt{2} = x, r = 1, h^{*} = \dfrac{1}{\sqrt{2}},\theta = 90^{\circ}, H = 1$ and $\lambda = 60^{\circ} \implies \lfloor{d}\rfloor$ + $\lfloor{x}\rfloor$ + $\lfloor{r}\rfloor$ + $\lfloor{h^{*}}\rfloor$ + $\lfloor{\theta}\rfloor$ + $\lfloor{H}\rfloor$ + $\lfloor{\lambda}\rfloor = \boxed{154}$