Solution:

Place the rectangle with corners $(0, 0),$ $(11, 0),$ $(11, 10),$ $(0, 10).$ A maximal equilateral triangle can be enlarged unless it is pinned by the rectangle, and the extremal position has one vertex at a corner, say the origin, with the other two vertices $s(\cos\theta, \sin\theta)$ and $s\bigl(\cos(\theta + 60^\circ), \sin(\theta + 60^\circ)\bigr)$ touching the far sides $x = 11$ and $y = 10:$ $$s\cos\theta = 11, \qquad s\sin(\theta + 60^\circ) = 10.$$

Dividing, $11\sin(\theta + 60^\circ) = 10\cos\theta,$ and expanding the left side gives $\frac{11}{2}\sin\theta + \frac{11\sqrt{3}}{2}\cos\theta = 10\cos\theta,$ so $\tan\theta = \frac{20 - 11\sqrt{3}}{11}$ (about $4.9^\circ,$ a legal tilt). Then $$s^2 = \frac{121}{\cos^2\theta} = 121\left(1 + \tan^2\theta\right) = 121 + \left(20 - 11\sqrt{3}\right)^2 = 884 - 440\sqrt{3}.$$

The area is $\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}\left(884 - 440\sqrt{3}\right) = 221\sqrt{3} - 330 \approx 52.8,$ which indeed beats the untilted triangle of side $10.$ Thus $p + q + r = 221 + 3 + 330 = 554.$