Choice A is the correct answer. Find vertices of the feasible region.

1. Intersection of lines: $2x + y = 10$ and $x + 2y = 8$.
   • Multiply second by -2: $-2x - 4y = -16$
   • Add to first: $-3y = -6 \rightarrow y = 2$
   • Find x: $x + 2(2) = 8 \rightarrow x = 4$. Vertex (4, 2).
2. Intercepts:
   • $2x + y = 10 \rightarrow (5, 0)$ and $(0, 10)$.
   • $x + 2y = 8 \rightarrow (8, 0)$ and $(0, 4)$.
   • Feasible region vertices: $(0,0), (5,0)$ is OUT (fails $x+2y \leq 8$), $(4,0)$ is IN? No, wait.
   • Check intercepts against other inequalities:
     • $(5,0)$: $5 + 2(0) = 5 \leq 8$ (Valid). So $(5,0)$ is a vertex? Wait. $2(5)+0=10$. Yes.
     • $(0,4)$: $2(0) + 4 = 4 \leq 10$ (Valid). So $(0,4)$ is a vertex.
3. Vertices: $(0,0), (5,0), (4,2), (0,4)$.
4. Evaluate P:
   • $(0,0): 0$
   • $(5,0): 4(5) + 3(0) = 20$
   • $(4,2): 4(4) + 3(2) = 16 + 6 = 22$
   • $(0,4): 4(0) + 3(4) = 12$
5. Maximum: 22.

?�� Strategic Tip: The intersection of the boundary lines is often the optimal point.

Choice B is incorrect.Choice C is incorrect.Choice D is incorrect.