Inequalities System

1. Problem 3: Solve the system of inequalities $$4x + 8y \leq 640; 6x + 4y \geq 480; x \geq 0; y \geq 0$$ Step 1: Rewrite inequalities for boundary lines: $$4x + 8y = 640 \Rightarrow y = \frac{640 - 4x}{8} = 80 - 0.5x$$ $$6x + 4y = 480 \Rightarrow y = \frac{480 - 6x}{4} = 120 - 1.5x$$ Step 2: Identify feasible regions based on inequalities: - For $$4x + 8y \leq 640$$, feasible region is below the line $$y = 80 - 0.5x$$. - For $$6x + 4y \geq 480$$, feasible region is above the line $$y = 120 - 1.5x$$. - Also, $$x, y \geq 0$$. Step 3: Find intersection points of the boundary lines: Set $$80 - 0.5x = 120 - 1.5x$$ $$80 - 0.5x = 120 - 1.5x$$ $$-0.5x + 1.5x = 120 - 80$$ $$x = 40$$ Then, $$y = 80 - 0.5(40) = 80 - 20 = 60$$ So, intersection point is $$ (40, 60) $$. Step 4: Find intercepts: - For $$4x + 8y = 640$$: - $$x=0 \Rightarrow y=80$$ - $$y=0 \Rightarrow x=160$$ - For $$6x + 4y = 480$$: - $$x=0 \Rightarrow y=120$$ - $$y=0 \Rightarrow x=80$$ Step 5: Feasible region vertices are: - $$ (0, 80) $$ from first inequality - $$ (40, 60) $$ intersection point - $$ (80, 0) $$ from second inequality - and $$ (0,0) $$ due to non-negativity constraints --- 2. Problem 4: Solve system $$4x + 8y \geq 640; 6x + 4y \geq 480; x \geq 0; y \geq 0$$ Step 1: Boundary lines are same as in problem 3. Step 2: Inequalities define: - above or on line $$y = 80 - 0.5x$$ - above or on line $$y = 120 - 1.5x$$ Step 3: Feasible region is intersection of these two half planes plus non-negative quadrant. Step 4: Find intercepts and intersection point as in problem 3: same points $$ (0, 80), (40,60), (80,0) $$. Step 5: Since inequalities are \(\geq\), feasible region is unbounded above these lines and in first quadrant. --- 3. Problem 5: Solve system $$4x + 6y \geq 640; 8x + 4y \geq 480; x \geq 0; y \geq 0$$ Step 1: Write boundary lines: $$4x + 6y = 640 \Rightarrow y = \frac{640 - 4x}{6} = \frac{320}{3} - \frac{2}{3}x \approx 106.67 - 0.6667x$$ $$8x + 4y = 480 \Rightarrow y = \frac{480 - 8x}{4} = 120 - 2x$$ Step 2: Intersection point: Set equal: $$106.67 - 0.6667x = 120 - 2x$$ $$-0.6667x + 2x = 120 - 106.67$$ $$1.3333x = 13.33$$ $$x = 10$$ Calculate $$y$$: $$y = 120 - 2(10) = 120 - 20 = 100$$ Step 3: Intercepts: - For $$4x + 6y = 640$$: - x-intercept: $$y=0 \Rightarrow 4x=640 \Rightarrow x=160$$ - y-intercept: $$x=0 \Rightarrow 6y=640 \Rightarrow y=106.67$$ - For $$8x + 4y = 480$$: - x-intercept: $$y=0 \Rightarrow 8x=480 \Rightarrow x=60$$ - y-intercept: $$x=0 \Rightarrow 4y=480 \Rightarrow y=120$$ Step 4: Feasible region vertices: $$ (0,106.67), (10, 100), (60, 0), (0, 0)$$ --- 4. Problem 6: Solve system $$4x + 8y \leq 640; 6x + 4y \leq 480; x \geq 0; y \geq 0$$ Step 1: Boundary lines: $$4x + 8y = 640 \Rightarrow y=80 - 0.5x$$ $$6x + 4y = 480 \Rightarrow y=120 - 1.5x$$ Step 2: Inequalities define: - below or on $$y=80 - 0.5x$$ - below or on $$y=120 - 1.5x$$ Step 3: Intersection point (from problem 3) at $$ (40, 60) $$ Step 4: Intercepts: - For $$4x + 8y=640$$: $$x=160$$, $$y=80$$ - For $$6x + 4y=480$$: $$x=80$$, $$y=120$$ Step 5: Feasible region vertices: $$ (0,0), (0,80), (40,60), (80,0) $$ --- Summary: Problems 3 to 6 analyzed as above, plotting and further optimization or shading depends on specific application.