To solve this problem, we need to approach it methodically by first addressing part (A) and then analyzing the results for part (B). Let us begin:

(A) Finding all combinations of planes:

We denote:

• $x$: number of 10-passenger planes,
• $y$: number of 15-passenger planes,
• $z$: number of 20-passenger planes.

The constraints are:

1. $x + y + z = 12$ (the total number of planes is 12),
2. $10x + 15y + 20z = 220$ (the total passenger capacity is 220).

Step 1: Simplify the equations

From the first equation: $z = 12 - x - y$ Substitute $z$ into the second equation: $10x + 15y + 20(12 - x - y) = 220$ Simplify: $10x + 15y + 240 - 20x - 20y = 220$ $-10x - 5y + 240 = 220$ -10x - 5y = -20 \quad \text{or} \quad 2x + y = 4 \tag{1}

Step 2: Solve for $x$, $y$, and $z$

Equation (1) gives: y = 4 - 2x \quad \text{(substitute values of x$to find integers for$y$and$z)}.

Step 3: Find all valid combinations

1. $y \geq 0$: This implies $4 - 2x \geq 0$, or $x \leq 2$.
2. $z \geq 0$: This implies $12 - x - y \geq 0$.

Now, calculate valid values:

• For $x = 0$: $y = 4 - 2(0) = 4, \quad z = 12 - 0 - 4 = 8$ Solution: $(x, y, z) = (0, 4, 8)$.

• For $x = 1$: $y = 4 - 2(1) = 2, \quad z = 12 - 1 - 2 = 9$ Solution: $(x, y, z) = (1, 2, 9)$.

• For $x = 2$: $y = 4 - 2(2) = 0, \quad z = 12 - 2 - 0 = 10$ Solution: $(x, y, z) = (2, 0, 10)$.

Thus, all combinations are: $(0, 4, 8), \; (1, 2, 9), \; (2, 0, 10).$

---

(B) Minimizing the leasing cost:

The cost of leasing is: $\text{Cost} = 8000x + 14000y + 16000z$ Substitute the combinations from part (A):

1. For $(x, y, z) = (0, 4, 8)$: $\text{Cost} = 8000(0) + 14000(4) + 16000(8) = 0 + 56000 + 128000 = 184000$

2. For $(x, y, z) = (1, 2, 9)$: $\text{Cost} = 8000(1) + 14000(2) + 16000(9) = 8000 + 28000 + 144000 = 180000$

3. For $(x, y, z) = (2, 0, 10)$: $\text{Cost} = 8000(2) + 14000(0) + 16000(10) = 16000 + 0 + 160000 = 176000$

---

Conclusion:

• All combinations: $(x, y, z) = (0, 4, 8), \; (1, 2, 9), \; (2, 0, 10)$.
• Minimized cost: The least monthly leasing cost is €176,000, achieved with the combination $(x, y, z) = (2, 0, 10)$.

Would you like further clarification or details?

---

Related Questions:

1. What other constraints could be applied to this type of optimization problem?
2. How would the solution change if the total number of planes was different?
3. What is the significance of linear programming in solving similar problems?
4. Could a graphical method be used to solve this system? How?
5. How would this problem differ if we had more plane types?

Tip:

When solving linear equations with constraints, always check for boundary values to identify all valid solutions.