Minimum of perimeter (part 2)

By the perimeter equation, $P = a + b + c = 5$ , and by Heron's area formula, $A = \sqrt{s(s - a)(s - b)(s - c)} = 1$ .

Using $s = \frac{1}{2}P = \frac{5}{2}$ and combining equations gives $\frac{5}{2}(\frac{5}{2} - b)(\frac{5}{2} - c)(b + c - \frac{5}{2}) = 1$ , which rearranges to $40b^2c - 100b^2 + 40bc^2 - 300bc + 500b - 100c^2 + 500c - 641 = 0$ .

By implicit differentiation, $80bc + 40b^2 \frac{dc}{db} - 200b + 40c^2 + 80bc\frac{dc}{db} - 300c - 300b\frac{dc}{db} + 500 - 200c\frac{dc}{db} + 500\frac{dc}{db} = 0$ . Since the maximum $c$ occurs when $\frac{dc}{db} = 0$ , this equation simplifies to $8bc - 20b + 4c^2 - 30c + 50 = 0$ .

The two equations numerically solve to $(b, c) \approx (1.42215, 2.15569)$ or $(b, c) \approx (1.97465, 1.05071)$ , and $c \approx 2.15569$ is the maximum of the two.

Therefore, $\lfloor 10000M \rfloor = \boxed{21556}$ .