1. **State the problem:** We want to maximize the objective function $$Z = 6x + 5y$$ subject to the constraints: $$x \geq 0, y \geq 0, x + y \leq 8, x + 2y \leq 10.$$ 2. **Identify feasible region constraints:** - The inequalities $x \geq 0$ and $y \geq 0$ restrict us to the first quadrant. - $x + y \leq 8$ describes the area below the line $y = 8 - x$. - $x + 2y \leq 10$ describes the area below the line $y = \frac{10 - x}{2}$. 3. **Find intersection points of constraints (vertices of feasible region):** - Intersection of $x+y=8$ and $x+2y=10$: Solve the system: $$x + y = 8$$ $$x + 2y = 10$$ Subtract first from second: $$x + 2y - (x + y) = 10 - 8 \Rightarrow y = 2$$ Substitute back: $$x + 2 = 8 \Rightarrow x = 6$$ So point: $(6,2)$ - Intersection of $x + y = 8$ and $y=0$: Substitute $y=0$: $$x + 0 = 8 \Rightarrow x = 8$$ So point: $(8,0)$ - Intersection of $x + 2y = 10$ and $y = 0$: Substitute $y=0$: $$x + 0 = 10 \Rightarrow x = 10$$ But $x=10$ is not feasible because $x + y \leq 8$ limits $x \leq 8$ when $y=0$. So the feasible vertex along $y=0$ is at $(8,0)$. - Intersection of $x = 0$ and $x + 2y = 10$: Substitute $x=0$: $$0 + 2y = 10 \Rightarrow y = 5$$ So point: $(0,5)$ - Intersection of $x=0$ and $x + y = 8$: Substitute $x=0$: $$0 + y = 8 \Rightarrow y=8$$ Check feasibility with $x + 2y \leq 10$: $0 + 2(8) = 16 > 10$, so $(0,8)$ is infeasible. - Intersection of $x=0$ and $y=0$: $(0,0)$ 4. **Evaluate objective function $Z=6x+5y$ at each feasible vertex:** - At $(0,0)$: $Z = 6\times0 + 5\times0 = 0$ - At $(8,0)$: $Z = 6\times8 + 5\times0 = 48$ - At $(6,2)$: $Z = 6\times6 + 5\times2 = 36 + 10 = 46$ - At $(0,5)$: $Z = 6\times0 + 5\times5 = 25$ 5. **Determine maximum:** Among these, the maximum value of $Z$ is $48$ at point $(8,0)$. **Final answer:** $$\boxed{\max Z = 48 \text{ at } (x,y) = (8,0)}.$$