Oblique Cylinder

$V = \pi r^2 H = 1 \implies H = \dfrac{1}{\pi r^2}$ and $A = 2\pi rs = 1 \implies s = \dfrac{1}{2\pi r}$

$\implies Y(r) = y^2(r) = s^2(r) - H^2(r) = \dfrac{1}{4\pi^2 r^2 } - \dfrac{1}{\pi^2 r^4} \implies \dfrac{dY}{dr} = \dfrac{1}{\pi^2}(\dfrac{8 - r^2}{2r^5})$ $\:\ r > 0 \implies r = 2\sqrt{2} \implies$ $s = \dfrac{1}{4\sqrt{2}\pi}$ and $H = \dfrac{1}{8\pi}$ and $y = \sqrt{s^2 - H^2} = \dfrac{1}{8\pi} = H$ .

$\dfrac{d^2Y}{dr^2}|_{r = 2\sqrt{2}} = \dfrac{-1}{(8\pi)^2} < 0 \implies$ max at $r = 2\sqrt{2}$ .

Let $m\angle{PTR} = \lambda \implies \cos(\lambda) = \dfrac{y(r)}{s(r)} = \dfrac{1}{\sqrt{2}}$ and $\theta = 180 - \lambda \implies \cos(\theta) = \cos(180 - \theta) = -\cos(\lambda) = -\dfrac{1}{\sqrt{2}} \implies \boxed{\theta = 135^\circ}$

$\implies d^2 = s^2 + 4r^2 + 4rs \cos(\lambda) = \dfrac{1}{32\pi^2} + 32 + \dfrac{\sqrt{2}}{\pi} =$ $\dfrac{32^2\pi^2 + 32\sqrt{2}\pi + 1}{32\pi^2} \implies$ $d = \dfrac{\sqrt{32^2\pi^2 + 32\sqrt{2}\pi + 1}}{4\sqrt{2}\pi} \approx \boxed{5.69678}$

$\therefore \lfloor{d + \theta}\rfloor = \boxed{140}$