Set 13: Linear Inequalities

Choice A is correct. Choice A is the correct answer. Find vertices of the feasible region. 1. Intersection of lines: $2x + y = 10$ and $x + 2 y = 8$. - Multiply second by -2: $-2 x - 4 y = -16$ - Add to first: $-3 y = -6 \rightarrow y = 2$ - Find x: $x + 2(2) = 8 \rightarrow x = 4$. Vertex (4, 2). 2. Intercepts: - $2x + y = 10 \rightarrow (5, 0)$ and $(0, 10)$. - $x + 2 y = 8 \rightarrow (8, 0)$ and $(0, 4)$. - Feasible region vertices: $(0,0), (5,0)$ is OUT (fails $x+2 y \leq 8$), $(4,0)$ 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.