1. We are given the system of inequalities: $$3x + y \geq 15$$ $$x + 2y \geq 10$$ $$x \geq 0$$ $$y \geq 0$$ Our goal is to determine the feasible region defined by these inequalities, typically done by graphing. 2. First, rewrite each inequality as an equation to find boundary lines: - Line 1: $$3x + y = 15$$ - Line 2: $$x + 2y = 10$$ - Line 3: $$x = 0$$ (y-axis) - Line 4: $$y = 0$$ (x-axis) 3. Find intercepts for each line to aid graphing: - For $$3x + y = 15$$: When $$x=0$$, $$y=15$$ When $$y=0$$, $$3x=15 \Rightarrow x=5$$ - For $$x + 2y = 10$$: When $$x=0$$, $$2y=10 \Rightarrow y=5$$ When $$y=0$$, $$x=10$$ 4. Plot boundary lines and consider inequalities: - $$3x + y \geq 15$$ means the region on or above the line $$3x + y=15$$. - $$x + 2y \geq 10$$ means region on or above the line $$x + 2y=10$$. - $$x \geq 0$$ means to the right of or on the y-axis. - $$y \geq 0$$ means above or on the x-axis. 5. To find the feasible region, identify the area that satisfies all these conditions simultaneously. 6. Find intersection points: - Intersection of $$3x + y = 15$$ and $$x + 2y =10$$: From the second equation, express $$x$$ in terms of $$y$$: $$x=10 - 2y$$ Substitute in the first: $$3(10 - 2y) + y =15$$ $$30 - 6y + y = 15$$ $$30 - 5y = 15$$ $$-5y = 15 - 30 = -15$$ $$y = 3$$ Then, $$x = 10 - 2(3) = 10 -6 = 4$$ So, intersection point is $$ (4, 3) $$. - Intersections with axes are already known. 7. The feasible region is bounded by points: - $$ (0, 15) $$ from $$3x + y=15$$ and $$x=0$$ - $$ (5, 0) $$ from $$3x + y=15$$ and $$y=0$$ - $$ (10, 0) $$ from $$x + 2y=10$$ and $$y=0$$ - $$ (0, 5) $$ from $$x + 2y=10$$ and $$x=0$$ 8. Considering $$x \geq 0$$ and $$y \geq 0$$ restricts feasible region to the first quadrant. 9. Final feasible region is the polygon formed by the points $$ (0,15), (4,3), (10,0), (0,5) $$ and portions satisfying the inequalities. Final answer: The feasible region is bounded in the first quadrant by the lines $$3x + y = 15$$, $$x + 2y = 10$$, and axes $$x=0$$ and $$y=0$$, forming a polygon with vertices at $$ (0,15), (4,3), (10,0), (0,5) $$.