A motivational speaker sells regular tickets and vip tickets to their show. In order for the event to be profitable, it must sell a minimum of 80 VIP tickets and a minimum of 120 regular tickets. The speaker makes a profit of $225 for each regular ticket and a profit of $210 for each VIP ticket. The venue has room for 300 people. How many of each ticket should be sold to maximize profits?

This problem involves optimizing profit while considering constraints, which is a typical example of a **linear programming** problem.

### Let's Define Variables:
- Let $ x $ represent the number of **regular tickets** sold.
- Let $ y $ represent the number of **VIP tickets** sold.

### Constraints:
1. **Minimum number of regular tickets**: $ x \geq 120 $
2. **Minimum number of VIP tickets**: $ y \geq 80 $
3. **Venue capacity**: $ x + y \leq 300 $

### Objective Function (Profit):
The objective is to maximize profit $ P $, where:
\[ P = 225x + 210y \]

### Problem Setup:
We need to maximize $ P = 225x + 210y $ subject to:
1. $ x \geq 120 $
2. $ y \geq 80 $
3. $ x + y \leq 300 $

### Graphical Solution:
To find the optimal solution, we graph the constraints and determine the feasible region, then evaluate the profit function at the vertices of this region.

1. **Constraint 1**: $ x \geq 120 $ is a vertical line at $ x = 120 $.
2. **Constraint 2**: $ y \geq 80 $ is a horizontal line at $ y = 80 $.
3. **Constraint 3**: $ x + y \leq 300 $ is a line with intercepts at $ x = 300 $ (when $ y = 0 $) and $ y = 300 $ (when $ x = 0 $).

The feasible region is bounded by these lines, and we will check the profit at the vertices of this region.

### Vertices of the Feasible Region:
The vertices are where the lines intersect:
1. $ (120, 80) $
2. $ (120, 180) $
3. $ (220, 80) $
4. $ (300, 0) $ (this vertex does not satisfy $ y \geq 80 $)

### Calculate Profit at Each Vertex:
1. At $ (120, 80) $: $ P = 225(120) + 210(80) = 27,000 + 16,800 = 43,800 $
2. At $ (120, 180) $: $ P = 225(120) + 210(180) = 27,000 + 37,800 = 64,800 $
3. At $ (220, 80) $: $ P = 225(220) + 210(80) = 49,500 + 16,800 = 66,300 $

### Conclusion:
The maximum profit is $ \boxed{66,300} $, achieved by selling **220 regular tickets** and **80 VIP tickets**.

Would you like more details or have any questions?

### 5 Related Questions:
1. How would the solution change if the profit for VIP tickets increased to $250?
2. What if the venue capacity increased to 350 people?
3. How would additional constraints, like a maximum on ticket prices, affect the solution?
4. How would this problem be solved using simplex methods instead of graphically?
5. What is the significance of vertices in linear programming solutions?

### Tip:
In linear programming, always evaluate the objective function at the vertices of the feasible region since the maximum or minimum values occur there.

Learn how to maximize profits by selling regular and VIP tickets for an event using linear programming. Solve constraints involving minimum ticket sales and venue capacity while maximizing profit. Suitable for high school students and those interested in optimization problems.

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