Maximizing Revenue with Python

Oct 13, 2025 by thesweetdesigns.com 113 views

Introduction

The drama club is facing a challenge in raising funds for their upcoming play. To cover the expenses, they have decided to sell tickets to the event. The club has set a price of $4 for each student ticket and $9 for each adult ticket. However, there is a constraint on the number of tickets that can be sold, as the auditorium can accommodate no more than 110 people. In this article, we will explore the problem of maximizing the revenue generated by the drama club through ticket sales.

Problem Formulation

Let's denote the number of student tickets sold as x and the number of adult tickets sold as y. The revenue generated from the sale of student tickets is $4x, and the revenue generated from the sale of adult tickets is $9y. The total revenue is the sum of these two revenues, which can be expressed as:

R(x, y) = 4x + 9y

However, there are two constraints that need to be considered:

The total number of tickets sold cannot exceed 110, which can be expressed as:

x + y ≤ 110

The number of student tickets sold cannot be negative, which can be expressed as:

x ≥ 0

Mathematical Modeling

To solve this problem, we can use linear programming techniques. The goal is to maximize the revenue R(x, y) subject to the constraints x + y ≤ 110 and x ≥ 0.

Graphical Method

One way to visualize this problem is to use a graphical method. We can plot the constraint x + y ≤ 110 on a coordinate plane, where the x-axis represents the number of student tickets sold and the y-axis represents the number of adult tickets sold.

import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 110, 100)
y = 110 - x
plt.plot(x, y)
plt.xlabel('Number of Student Tickets Sold')
plt.ylabel('Number of Adult Tickets Sold')
plt.title('Constraint x + y ≤ 110')
plt.grid(True)
plt.show()

Solving the Problem

To solve this problem, we need to find the values of x and y that maximize the revenue R(x, y) subject to the constraints x + y ≤ 110 and x ≥ 0.

One way to do this is to use the graphical method. We can plot the revenue function R(x, y) = 4x + 9y on the same coordinate plane as the constraint x + y ≤ 110.

import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 110, 100)
y = 110 - x
R = 4x + 9y
plt.plot(x, R)
plt.xlabel('Number of Student Tickets Sold')
plt.ylabel('Revenue')
plt.title('Revenue Function R(x, y) = 4x + 9y')
plt.grid(True)
plt.show()

Optimal Solution

By analyzing the graphical representation of the problem, we can see that the optimal solution occurs when x = 110 and y = 0. This is because the revenue function R(x, y) = 4x + 9y is maximized when the number of student tickets sold is at its maximum value of 110.

Conclusion

Maximizing Revenue

To maximize the revenue, we need to find the values of x and y that satisfy the constraints x + y ≤ 110 and x ≥ 0. One way to do this is to use the graphical method, where we plot the revenue function R(x, y) = 4x + 9y on the same coordinate plane as the constraint x + y ≤ 110.

import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 110, 100)
y = 110 - x
R = 4x + 9y
plt.plot(x, R)
plt.xlabel('Number of Student Tickets Sold')
plt.ylabel('Revenue')
plt.title('Revenue Function R(x, y) = 4x + 9y')
plt.grid(True)
plt.show()

Optimal Values of x and y

By analyzing the graphical representation of the problem, we can see that the optimal values of x and y are x = 110 and y = 0. This is because the revenue function R(x, y) = 4x + 9y is maximized when the number of student tickets sold is at its maximum value of 110.

Revenue Maximization

import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 110, 100)
y = 110 - x
R = 4x + 9y
plt.plot(x, R)
plt.xlabel('Number of Student Tickets Sold')
plt.ylabel('Revenue')
plt.title('Revenue Function R(x, y) = 4x + 9y')
plt.grid(True)
plt.show()

Conclusion

Maximizing Revenue with Python

To maximize the revenue using Python, we can use the scipy.optimize module to find the optimal values of x and y that satisfy the constraints x + y ≤ 110 and x ≥ 0.

import numpy as np
from scipy.optimize import linprog
c = [-4, -9]  # coefficients of the revenue function
A = [[1, 1]]  # coefficients of the constraint x + y ≤ 110
b = [110]  # right-hand side of the constraint x + y ≤ 110
A_ub = [[0, -1]]  # coefficients of the constraint x ≥ 0
b_ub = [0]  # right-hand side of the constraint x ≥ 0
res = linprog(c, A_ub=A_ub, b_ub=b_ub, method='highs')
print('Optimal values of x and y:', res.x)

Conclusion

Introduction

In our previous article, we explored the problem of maximizing the revenue generated by the drama club through ticket sales. We used linear programming techniques to find the optimal solution that maximizes the revenue subject to the constraints x + y ≤ 110 and x ≥ 0. In this article, we will answer some frequently asked questions related to the drama club's ticket sales dilemma.

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

A: The optimal solution to the drama club's ticket sales dilemma is x = 110 and y = 0. This means that the drama club should sell 110 student tickets and no adult tickets to maximize the revenue.

Q: Why is it optimal to sell 110 student tickets and no adult tickets?

A: It is optimal to sell 110 student tickets and no adult tickets because the revenue function R(x, y) = 4x + 9y is maximized when the number of student tickets sold is at its maximum value of 110. Selling more adult tickets would reduce the revenue, as the revenue function is linear and the slope of the revenue function is negative.

Q: What is the maximum revenue that the drama club can generate through ticket sales?

A: The maximum revenue that the drama club can generate through ticket sales is $440, which is obtained by selling 110 student tickets at $4 each.

Q: How can the drama club use this information to make informed decisions about ticket sales?

A: The drama club can use this information to make informed decisions about ticket sales by considering the optimal solution and the maximum revenue that can be generated. For example, the drama club can set a target revenue of $440 and adjust the number of student and adult tickets sold accordingly to achieve this target.

Q: What are some potential challenges that the drama club may face in implementing this solution?

A: Some potential challenges that the drama club may face in implementing this solution include:

Ensuring that there are enough student tickets available to meet the demand
Managing the logistics of selling tickets, including handling cash and credit card transactions
Communicating the ticket sales information to the audience and ensuring that they understand the pricing and availability of tickets

Q: How can the drama club overcome these challenges and implement the optimal solution?

A: The drama club can overcome these challenges and implement the optimal solution by:

Ensuring that there are enough student tickets available to meet the demand by ordering additional tickets if necessary
Implementing a ticket sales system that can handle cash and credit card transactions efficiently
Communicating the ticket sales information to the audience through various channels, including social media, email, and in-person announcements

Q: What are some potential benefits of implementing this solution?

A: Some potential benefits of implementing this solution include:

Maximizing revenue and ensuring that the drama club has sufficient funds to cover expenses
Providing a clear and transparent ticket sales process that is easy for the audience to understand
Building trust and credibility with the audience by demonstrating a commitment to transparency and fairness in ticket sales

Conclusion

In conclusion, the drama club's ticket sales dilemma can be solved using linear programming techniques. By maximizing the revenue R(x, y) subject to the constraints x + y ≤ 110 and x ≥ 0, we can find the optimal solution that maximizes the revenue generated by the drama club through ticket sales. We hope that this Q&A article has provided valuable insights and information to help the drama club make informed decisions about ticket sales.