1. **Stating the problem:** We need to find the region that satisfies the system of inequalities: $$x \geq 0, \quad y \geq 0, \quad x - 2y \leq 4, \quad 5x + y \geq 10, \quad x + y \leq 8$$ 2. **Understanding the inequalities:** - $x \geq 0$ means the solution is on or to the right of the y-axis. - $y \geq 0$ means the solution is on or above the x-axis. - $x - 2y \leq 4$ can be rewritten as $y \geq \frac{x-4}{2}$. - $5x + y \geq 10$ can be rewritten as $y \geq 10 - 5x$. - $x + y \leq 8$ can be rewritten as $y \leq 8 - x$. 3. **Plotting and analyzing the boundaries:** - The feasible region must be in the first quadrant due to $x \geq 0$ and $y \geq 0$. - The line $y = \frac{x-4}{2}$ has intercepts at $(4,0)$ and extends upward for $x > 4$. - The line $y = 10 - 5x$ intersects the x-axis at $(2,0)$ and decreases steeply. - The line $y = 8 - x$ intersects the axes at $(0,8)$ and $(8,0)$. 4. **Finding the intersection points of the boundary lines to identify the feasible polygon:** - Intersection of $x - 2y = 4$ and $5x + y = 10$: Solve simultaneously: $$x - 2y = 4$$ $$5x + y = 10$$ From first: $x = 4 + 2y$ Substitute into second: $$5(4 + 2y) + y = 10 \Rightarrow 20 + 10y + y = 10 \Rightarrow 11y = -10 \Rightarrow y = -\frac{10}{11}$$ Since $y$ is negative, this point is outside $y \geq 0$ region. - Intersection of $x - 2y = 4$ and $x + y = 8$: From $x + y = 8$, $y = 8 - x$ Substitute into $x - 2y = 4$: $$x - 2(8 - x) = 4 \Rightarrow x - 16 + 2x = 4 \Rightarrow 3x = 20 \Rightarrow x = \frac{20}{3} \approx 6.67$$ Then $y = 8 - 6.67 = 1.33$ - Intersection of $5x + y = 10$ and $x + y = 8$: Subtract second from first: $$5x + y - (x + y) = 10 - 8 \Rightarrow 4x = 2 \Rightarrow x = 0.5$$ Then $y = 8 - 0.5 = 7.5$ 5. **Check which regions satisfy all inequalities:** - The feasible region is bounded by $x \geq 0$, $y \geq 0$, and the three lines. - The polygon formed by points approximately at $(2,0)$, $(6.67,1.33)$, $(0.5,7.5)$, and $(0,4)$ fits the constraints. 6. **Conclusion:** Based on the graph and inequalities, the feasible region corresponds to region **I**. **Final answer:** **A. I**