Mr. Squash bought a large parking lot in Utah, which has an area of $600$ square meters. A car needs $6$ square meters of parking space while a bus needs $30$ square meters of parking space. Mr. Squash charges $\$2.50$ per car and $\$7.50$ per bus, but Mr. Squash can only handle at most $60$ vehicles at a time. Find the ordered pair $(a,b)$ where $a$ is the number of cars and $b$ is the number of buses that maximizes the amount of money Mr. Squash makes.

[i]Proposed by Nathan Cho[/i]
To solve this problem, we need to maximize Mr. Squash's revenue given the constraints on the number of vehicles and the area of the parking lot. Let's define the variables:
- \( x \) = number of cars
- \( y \) = number of buses

The constraints are:
1. The total number of vehicles cannot exceed 60:
   \[
   x + y \leq 60
   \]
2. The total area occupied by the vehicles cannot exceed 600 square meters:
   \[
   6x + 30y \leq 600
   \]

The objective function to maximize is the total revenue:
\[
R = 2.50x + 7.50y
\]

We will solve this problem using the method of linear programming. First, we need to find the feasible region defined by the constraints.

1. **Constraint 1: Total number of vehicles**
   \[
   x + y \leq 60
   \]

2. **Constraint 2: Total area occupied**
   \[
   6x + 30y \leq 600
   \]
   Simplifying this constraint by dividing by 6:
   \[
   x + 5y \leq 100
   \]

3. **Non-negativity constraints:**
   \[
   x \geq 0, \quad y \geq 0
   \]

Next, we find the intersection points of the constraints to determine the vertices of the feasible region.

**Intersection of \( x + y = 60 \) and \( x + 5y = 100 \):**
\[
\begin{cases}
x + y = 60 \\
x + 5y = 100
\end{cases}
\]

Subtract the first equation from the second:
\[
(x + 5y) - (x + y) = 100 - 60 \\
4y = 40 \\
y = 10
\]

Substitute \( y = 10 \) into \( x + y = 60 \):
\[
x + 10 = 60 \\
x = 50
\]

So, the intersection point is \( (50, 10) \).

**Vertices of the feasible region:**
1. \( (0, 0) \)
2. \( (0, 20) \) from \( x + 5y = 100 \)
3. \( (50, 10) \) from the intersection
4. \( (60, 0) \) from \( x + y = 60 \)

Now, we evaluate the objective function \( R = 2.50x + 7.50y \) at each vertex:

1. At \( (0, 0) \):
   \[
   R = 2.50(0) + 7.50(0) = 0
   \]

2. At \( (0, 20) \):
   \[
   R = 2.50(0) + 7.50(20) = 150
   \]

3. At \( (50, 10) \):
   \[
   R = 2.50(50) + 7.50(10) = 125 + 75 = 200
   \]

4. At \( (60, 0) \):
   \[
   R = 2.50(60) + 7.50(0) = 150
   \]

The maximum revenue is \( 200 \) dollars, which occurs at the point \( (50, 10) \).

The final answer is \( \boxed{ (50, 10) } \)