Just The Right Amount of Constraints

We first note that

(i) $(a + b + c) + (3a + b - c) = 50 + 70 \Longrightarrow 4a + 2b = 120 \Longrightarrow 2a + b = 60 \Longrightarrow b = 60 - 2a$ ,

(ii) $(3a + b - c) - (a + b + c) = 70 - 50 \Longrightarrow 2a - 2c = 20 \Longrightarrow a - c = 10 \Longrightarrow c = a - 10$ .

Thus $5a + 4b + 2c = 5a + 4(60 - 2a) + 2(a - 10) = -a + 220$ .

Now as $b \gt 0$ we can infer from $2a + b = 60$ that $a \lt 30$ , and as $c \gt 0$ we can infer from $a - c = 10$ that $a \gt 10$ . So $10 \lt a \lt 30 \Longrightarrow 190 \lt -a + 220 \lt 210$ , and thus $m + n = 190 + 210 = \boxed{400}$ .

Interpret the two given equations as planes in the 3D-Cartesian system. The allowed values would be a line segment (because the coordinates has to be positive). Call this line segment $l$ . Solving the given equations along with:

(i) $a=0$ gives $a=0,b=60,c=-10$

(ii) $b=0$ gives $a=30,b=0,c=20$

(iii) $c=0$ gives $a=10,b=40,c=0$

Case (i) involves negative values, so it is rejected. Hence the endpoints of $l$ are given by (ii) and (iii). $5a+4b+2c=k$ is also describes a plane $P$ for some constant $k$ , that is always perpendicular to the vector $(5,4,2)$ . Since all relationships are linear, the extreme values of $k$ can be found relating it with the endpoints found before, i.e. (ii) $5*30+4*0+2*20=190$ and (iii) $5*10+4*40+2*0=210$ . (This can be interpreted as sliding a point along $l$ at constant speed, and also sliding the plane $p$ with the restriction that $p$ has to contain the point, then $k$ should vary linearly as the point slides from one endpoint of $l$ to the other.) Hence the answer is $190+210=\boxed{400}$ .