To maximize and minimize $z = 3x + 4y$ subject to the constraints:

1. $2x + y \geq 10$
2. $x + 2y \geq 8$
3. $x, y \geq 0$

Step 1: Identify the Feasible Region

1. Rewrite the inequalities as equalities to find boundary lines:

   • For $2x + y = 10$:
     • When $x = 0$: $y = 10$
     • When $y = 0$: $x = 5$
     • Points: $(0, 10)$ and $(5, 0)$
   • For $x + 2y = 8$:
     • When $x = 0$: $y = 4$
     • When $y = 0$: $x = 8$
     • Points: $(0, 4)$ and $(8, 0)$

2. Plot the lines and determine the feasible region, which is the area where both inequalities $2x + y \geq 10$ and $x + 2y \geq 8$ are satisfied along with $x, y \geq 0$.

3. Find Intersection Points of the boundary lines:

   • Solve $2x + y = 10$ and $x + 2y = 8$ simultaneously:
     • From $2x + y = 10$, we get $y = 10 - 2x$.
     • Substitute into $x + 2(10 - 2x) = 8$: $x + 20 - 4x = 8 \Rightarrow -3x = -12 \Rightarrow x = 4$ $y = 10 - 2(4) = 2$
     • Intersection point: $(4, 2)$

4. Vertices of the feasible region are $(0, 10)$, $(5, 0)$, $(8, 0)$, and $(4, 2)$.

Step 2: Evaluate $z = 3x + 4y$ at Each Vertex

1. At $(0, 10)$: $z = 3(0) + 4(10) = 40$

2. At $(5, 0)$: $z = 3(5) + 4(0) = 15$

3. At $(8, 0)$: $z = 3(8) + 4(0) = 24$

4. At $(4, 2)$: $z = 3(4) + 4(2) = 12 + 8 = 20$

Step 3: Conclusion

• Maximum value of $z = 40$ at $(0, 10)$.
• Minimum value of $z = 15$ at $(5, 0)$.

Would you like a more detailed explanation or have any questions?

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Here are some additional questions for practice:

1. How would the solution change if $z = 5x + 3y$?
2. What would happen if we changed one of the constraints, such as $2x + y \leq 10$?
3. How do we determine if a solution is bounded or unbounded?
4. How would you handle a constraint like $y \leq 6$ in this problem?
5. What if we were asked to only maximize (or only minimize) $z$?

Tip: When working with linear programming, always verify the feasibility of intersection points within the constraints.