1. **State the problem:** We need to determine the feasible region and vertices for the system of inequalities: $$\begin{cases} x + 2y < 24 \\ 2x + 4y > 16 \\ x > 0, y > 0 \end{cases}$$ 2. **Rewrite inequalities for clarity:** - From $x + 2y < 24$, the boundary line is $x + 2y = 24$. - From $2x + 4y > 16$, divide both sides by 2: $x + 2y > 8$. - Also, $x > 0$ and $y > 0$ restrict us to the first quadrant. 3. **Find intersection points (vertices):** - Intersection of $x + 2y = 24$ and $x + 2y = 8$ is impossible since they are parallel lines. - Find intersections with axes: - For $x + 2y = 24$: - When $x=0$, $2y=24 \Rightarrow y=12$. - When $y=0$, $x=24$. - For $x + 2y = 8$: - When $x=0$, $2y=8 \Rightarrow y=4$. - When $y=0$, $x=8$. 4. **Vertices of feasible region:** - Since $x + 2y > 8$ and $x + 2y < 24$, the feasible region lies between these two lines. - Along with $x > 0$ and $y > 0$, the vertices are: - $A = (0,4)$ from $x + 2y = 8$ and $x=0$. - $B = (8,0)$ from $x + 2y = 8$ and $y=0$. - $C = (24,0)$ from $x + 2y = 24$ and $y=0$. - $D = (0,12)$ from $x + 2y = 24$ and $x=0$. 5. **Feasible region description:** - The region is the area bounded between the two lines $x + 2y = 8$ and $x + 2y = 24$ in the first quadrant. 6. **Summary:** - The feasible region is the quadrilateral with vertices $A(0,4)$, $B(8,0)$, $C(24,0)$, and $D(0,12)$. - The region satisfies $x + 2y > 8$, $x + 2y < 24$, and $x,y > 0$. --- **Final answer:** Vertices of the feasible region are: $$A(0,4), B(8,0), C(24,0), D(0,12)$$