To solve this linear programming problem, we will use the method of finding the feasible region defined by the constraints and then evaluating the objective function at each vertex of this region.

Objective Function:

$\text{Maximize } Z = 2.9x + 1.8y$

Constraints:

1. $4y + 5x \leq 20$
2. $8y + 15x \leq 48$
3. $x \geq 0$
4. $y \geq 0$

Step 1: Convert the inequalities to equalities to find the intersection points (vertices).

1. Constraint 1: $4y + 5x = 20$
   $y = \frac{20 - 5x}{4} = 5 - 1.25x$

2. Constraint 2: $8y + 15x = 48$
   $y = \frac{48 - 15x}{8} = 6 - 1.875x$

Step 2: Find the intersection points of the constraints with the axes.

• For $4y + 5x = 20$:

  • When $x = 0$: $y = 5$
  • When $y = 0$: $x = 4$

  Points: $(0, 5)$ and $(4, 0)$

• For $8y + 15x = 48$:

  • When $x = 0$: $y = 6$
  • When $y = 0$: $x = 3.2$

  Points: $(0, 6)$ and $(3.2, 0)$

Step 3: Find the intersection point of the two constraints.

$4y + 5x = 20$ $8y + 15x = 48$

Multiply the first equation by 2: $8y + 10x = 40$

Subtract the first from the second: $8y + 15x - (8y + 10x) = 48 - 40$ $5x = 8$ $x = \frac{8}{5} = 1.6$

Substitute $x = 1.6$ into $4y + 5x = 20$: $4y + 5(1.6) = 20$ $4y + 8 = 20$ $4y = 12$ $y = 3$

Intersection point: $(1.6, 3)$

Step 4: Evaluate the objective function at each vertex.

1. At $(0, 5)$:
   $Z = 2.9(0) + 1.8(5) = 0 + 9 = 9$

2. At $(4, 0)$:
   $Z = 2.9(4) + 1.8(0) = 11.6 + 0 = 11.6$

3. At $(3.2, 0)$:
   $Z = 2.9(3.2) + 1.8(0) = 9.28 + 0 = 9.28$

4. At $(1.6, 3)$:
   $Z = 2.9(1.6) + 1.8(3) = 4.64 + 5.4 = 10.04$

Step 5: Conclusion

The maximum value of $Z$ is 11.6000 at the point $(4, 0)$.

Would you like any additional details or have any questions?

Here are 5 related questions for further exploration:

1. How would the solution change if the objective function was to minimize $2.9x + 1.8y$?
2. What happens to the feasible region if we change the constraint $4y + 5x \leq 20$ to $4y + 5x \leq 15$?
3. How can we solve this problem graphically to verify the solution?
4. What are the implications if there are no feasible solutions for a set of constraints?
5. How would we approach this problem if there were more than two variables?

Tip: Always check for the feasibility of the intersection points with the original constraints, as not all calculated points may be within the feasible region.