1. **State the problem:** Find the maximum value of $Z = 4x + 3y$ subject to the constraints: $$2x + y \leq 10,$$ $$x + y \leq 8,$$ $$x \geq 0, y \geq 0.$$ 2. **Graph the constraints and identify feasible region vertices:** - From $2x + y \leq 10$ we find the intercepts: when $x=0$, $y=10$; when $y=0$, $x=5$. - From $x + y \leq 8$ intercepts: when $x=0$, $y=8$; when $y=0$, $x=8$. 3. **Calculate intersection points of boundary lines:** - Intersection of $2x + y = 10$ and $x + y = 8$: $$2x + y = 10,$$ $$x + y = 8.$$ Subtract second from first: $$2x + y - (x + y) = 10 - 8 \, \Rightarrow \, x = 2.$$ Substitute into $x + y = 8$: $$2 + y = 8 \, \Rightarrow \, y = 6.$$ So intersection point is $(2, 6)$. 4. **Identify vertices of the feasible region:** - $(0, 0)$ from non-negativity. - $(0, 8)$ from $x=0$ and $x + y = 8$. - $(2, 6)$ intersection point. - $(5, 0)$ from $y=0$ and $2x + y = 10$. 5. **Evaluate $Z$ at each vertex:** - At $(0,0)$: $Z = 4(0) + 3(0) = 0$. - At $(0,8)$: $Z = 4(0) + 3(8) = 24$. - At $(2,6)$: $Z = 4(2) + 3(6) = 8 + 18 = 26$. - At $(5,0)$: $Z = 4(5) + 3(0) = 20$. 6. **Conclusion:** The maximum value is $Z_{max} = 26$ at point $(2, 6)$.