The Drama Club Is Selling Tickets To Their Play To Raise Money For The Show's Expenses. Each Student Ticket Sells For $ 4.50 \$4.50 $4.50 And Each Adult Ticket Sells For $ 7.50 \$7.50 $7.50 . The Auditorium Can Hold A Maximum Of 150 People. The Drama Club

Mar 12, 2025 by ADMIN 256 views

**The Drama Club's Ticket Sales Dilemma**

Introduction

The drama club is gearing up to put on a spectacular play, but they need to raise some money to cover the expenses. To do this, they have decided to sell tickets to the play. The club has a limited number of tickets available, and they need to determine the optimal pricing strategy to maximize their revenue. In this article, we will explore the problem of ticket sales and how the drama club can use mathematical concepts to make informed decisions.

Problem Statement

The drama club has 150 tickets available for sale, with each student ticket selling for $\$4.50$ and each adult ticket selling for $\$7.50$ . The club wants to maximize their revenue, but they also need to consider the number of tickets sold. If they sell too many tickets, they may not be able to cover the expenses, but if they sell too few tickets, they may not generate enough revenue.

Mathematical Modeling

Let's denote the number of student tickets sold as $x$ and the number of adult tickets sold as $y$ . The total revenue from ticket sales can be represented by the equation:

R = 4.50x + 7.50y

The number of tickets sold is limited by the capacity of the auditorium, which is 150. Therefore, we can write the constraint equation:

x + y \leq 150

We also know that the number of student tickets sold cannot exceed the number of students in the school, and the number of adult tickets sold cannot exceed the number of adults in the school. However, we do not have this information, so we will assume that the number of student tickets sold is less than or equal to the number of adult tickets sold.

Optimization Problem

The drama club wants to maximize their revenue, which is represented by the equation $R = 4.50x + 7.50y$ . However, they are also subject to the constraint equation $x + y \leq 150$ . This is a classic example of a linear programming problem, where we want to maximize a linear objective function subject to a linear constraint.

Graphical Method

To solve this problem graphically, we can plot the constraint equation $x + y \leq 150$ on a coordinate plane. The resulting graph is a line with a slope of -1 and a y-intercept of 150. We can then plot the revenue equation $R = 4.50x + 7.50y$ on the same coordinate plane. The resulting graph is a line with a slope of 4.50 and a y-intercept of 0.

Solving the Problem

To find the optimal solution, we need to find the point of intersection between the two lines. This can be done by solving the system of equations:

x + y = 150

4.50x + 7.50y = R

Substituting the first equation into the second equation, we get:

4.50x + 7.50(150 - x) = R

Simplifying the equation, we get:

4.50x + 1125 - 7.50x = R

Combine like terms:

-3x = R - 1125

Divide both sides by -3:

x = \frac{1125 - R}{3}

Now, substitute the expression for x into the first equation:

\frac{1125 - R}{3} + y = 150

Multiply both sides by 3:

1125 - R + 3y = 450

Add R to both sides:

1125 + 3y = 450 + R

Subtract 1125 from both sides:

3y = R - 1675

Divide both sides by 3:

y = \frac{R - 1675}{3}

Now, substitute the expressions for x and y into the revenue equation:

R = 4.50x + 7.50y

Substitute the expressions for x and y:

R = 4.50\left(\frac{1125 - R}{3}\right) + 7.50\left(\frac{R - 1675}{3}\right)

Simplify the equation:

R = \frac{4500 - 4.50R}{3} + \frac{7.50R - 12637.5}{3}

Multiply both sides by 3:

3R = 4500 - 4.50R + 7.50R - 12637.5

Combine like terms:

3R = 4500 + 3R - 12637.5

Subtract 3R from both sides:

0 = 4500 - 12637.5

Add 12637.5 to both sides:

12637.5 = 4500

This is a contradiction, which means that there is no solution to the problem.

Conclusion

The drama club's ticket sales dilemma is a classic example of a linear programming problem. However, in this case, the problem has no solution. This is because the revenue equation and the constraint equation are inconsistent. The drama club cannot sell a positive number of tickets and still maximize their revenue.

Recommendations

In this case, the drama club should consider alternative pricing strategies or revenue streams. For example, they could offer discounts for bulk ticket purchases or sell concessions during the play. They could also consider partnering with local businesses to sell tickets or promote the play.

Limitations

This problem assumes that the number of student tickets sold is less than or equal to the number of adult tickets sold. However, in reality, the number of student tickets sold may be greater than the number of adult tickets sold. This would require a more complex model that takes into account the different pricing strategies for student and adult tickets.

Future Research

This problem is a classic example of a linear programming problem. However, there are many variations of this problem that could be explored in future research. For example, the problem could be extended to include multiple pricing strategies or revenue streams. The problem could also be solved using different optimization techniques, such as dynamic programming or genetic algorithms.

References

[1] Chvatal, V. (1983). Linear Programming. W.H. Freeman and Company.
[2] Hillier, F. S., & Lieberman, G. J. (2015). Introduction to Operations Research. McGraw-Hill Education.
[3] Winston, W. L. (2018). Operations Research: Applications and Algorithms. Cengage Learning.
The Drama Club's Ticket Sales Dilemma: Q&A =============================================

Introduction

In our previous article, we explored the drama club's ticket sales dilemma and how they can use mathematical concepts to make informed decisions. However, we also found that the problem has no solution. In this article, we will answer some of the most frequently asked questions about the drama club's ticket sales dilemma.

Q: What is the drama club's ticket sales dilemma?

A: The drama club's ticket sales dilemma is a classic example of a linear programming problem. The club wants to maximize their revenue by selling tickets to the play, but they are subject to a constraint that limits the number of tickets they can sell.

Q: Why does the drama club's ticket sales dilemma have no solution?

A: The drama club's ticket sales dilemma has no solution because the revenue equation and the constraint equation are inconsistent. The club cannot sell a positive number of tickets and still maximize their revenue.

Q: What are some alternative pricing strategies that the drama club could use?

A: The drama club could consider alternative pricing strategies such as offering discounts for bulk ticket purchases or selling concessions during the play. They could also consider partnering with local businesses to sell tickets or promote the play.

Q: How can the drama club maximize their revenue?

A: The drama club can maximize their revenue by using a variety of pricing strategies and revenue streams. They could also consider using dynamic pricing, which involves adjusting the price of tickets based on demand.

Q: What are some other ways that the drama club could raise money for the play?

A: The drama club could consider raising money for the play by selling concessions during the play, holding a bake sale or other fundraising event, or seeking donations from local businesses or individuals.

Q: How can the drama club use mathematical concepts to make informed decisions?

A: The drama club can use mathematical concepts such as linear programming and optimization to make informed decisions about ticket pricing and revenue streams. They could also use data analysis and statistical modeling to understand their audience and make data-driven decisions.

Q: What are some common mistakes that the drama club could make when trying to raise money for the play?

A: The drama club could make common mistakes such as underpricing tickets, overpricing tickets, or failing to consider the costs of production. They could also fail to market the play effectively or neglect to engage with their audience.

Q: How can the drama club avoid these mistakes and raise money for the play effectively?

A: The drama club can avoid these mistakes by using a variety of pricing strategies and revenue streams, engaging with their audience, and using data analysis and statistical modeling to make informed decisions. They could also consider seeking advice from a professional fundraiser or using online fundraising platforms.