Set 15: Linear Inequalities

Choice A is correct. Choice A is the correct answer. Find vertices and evaluate. 1. Intersection 1: $y = -x + 4$ and $y = x + 2$. $x + 2 = -x + 4 \rightarrow 2 x = 2 \rightarrow x = 1, y = 3$. Vertex $(1, 3)$. 2. Intersection 2: $y = -x + 4$ and $y = 0$. $x = 4$. Vertex $(4, 0)$. 3. Intersection 3: $y = x + 2$ and $y = 0$. $x = -2$. Vertex $(-2, 0)$. 4. Evaluate P: - $(1, 3): 1 + 3 = 4$ - $(4, 0): 4 + 0 = 4$ - $(-2, 0): -2 + 0 = -2$ 5. Max: 4 Strategic Tip: Sometimes multiple vertices give the same optimal value (multiple solutions). Choice B is incorrect. Choice C is incorrect. Choice D is incorrect.