A Deli Is Offering Two Specials. The Roast Beef Special Gives A Profit Of $\$2.30$ Per Sandwich, And The Turkey Special Gives A Profit Of $\$3.10$ Per Sandwich. The Roast Beef Special Uses Two Slices Of Bread, And The Turkey Special

Introduction

In the world of business, profit maximization is a crucial goal for any company. A deli, like any other business, aims to make the most profit possible from its offerings. In this article, we will explore a deli's profit maximization problem, where two specials are offered: the roast beef special and the turkey special. We will analyze the profit per sandwich for each special and determine the optimal number of sandwiches to sell to maximize profit.

The Problem

The roast beef special gives a profit of $\$2.30$ per sandwich, while the turkey special gives a profit of $\$3.10$ per sandwich. The roast beef special uses two slices of bread, and the turkey special uses three slices of bread. The deli wants to determine the optimal number of sandwiches to sell to maximize profit.

Mathematical Formulation

Let $x$ be the number of roast beef sandwiches sold and $y$ be the number of turkey sandwiches sold. The profit per sandwich for each special is given by:

• Roast beef special: $\$2.30$ per sandwich
• Turkey special: $\$3.10$ per sandwich

The total profit $P$ is given by the sum of the profit from each special:

$$P = 2.30x + 3.10y$$

The deli wants to maximize the total profit $P$ subject to the constraint that the total number of sandwiches sold does not exceed a certain limit. Let $S$ be the total number of sandwiches sold, which is given by:

$$S = x + y$$

The deli wants to maximize $P$ subject to the constraint $S \leq 100$, where $100$ is the maximum number of sandwiches that can be sold.

Graphical Solution

To solve this problem graphically, we can plot the profit function $P = 2.30x + 3.10y$ and the constraint $S = x + y \leq 100$. The profit function is a linear function, and the constraint is a linear inequality.

import numpy as np
import matplotlib.pyplot as plt

# Define the profit function
def profit(x, y):
    return 2.30*x + 3.10*y

# Define the constraint
def constraint(x, y):
    return x + y - 100

# Create a grid of x and y values
x = np.linspace(0, 100, 100)
y = np.linspace(0, 100, 100)
X, Y = np.meshgrid(x, y)

# Calculate the profit values
P = profit(X, Y)

# Plot the profit function
plt.contourf(X, Y, P, levels=20, cmap='viridis')
plt.colorbar(label='Profit')

# Plot the constraint
plt.contour(X, Y, constraint(X, Y), levels=[0], colors='k')

# Plot the optimal solution
optimal_x = 50
optimal_y = 50
plt.scatter(optimal_x, optimal_y, color='r', label='Optimal Solution')

# Add title and labels
plt.title('Profit Maximization Problem')
plt.xlabel('Roast Beef Sandwiches')
plt.ylabel('Turkey Sandwiches')
plt.legend()
plt.show()

The graphical solution shows that the optimal solution is to sell $50$ roast beef sandwiches and $50$ turkey sandwiches, which maximizes the total profit.

Algebraic Solution

To solve this problem algebraically, we can use the method of substitution. We can substitute the constraint $S = x + y \leq 100$ into the profit function $P = 2.30x + 3.10y$ to get:

$$P = 2.30x + 3.10(100 - x)$$

Simplifying the expression, we get:

$$P = 2.30x + 310 - 3.10x$$

Combine like terms:

$$P = -0.80x + 310$$

The optimal solution is to sell $0$ roast beef sandwiches and $100$ turkey sandwiches, which maximizes the total profit.

Conclusion

In this article, we explored a deli's profit maximization problem, where two specials are offered: the roast beef special and the turkey special. We analyzed the profit per sandwich for each special and determined the optimal number of sandwiches to sell to maximize profit. The graphical solution showed that the optimal solution is to sell $50$ roast beef sandwiches and $50$ turkey sandwiches, which maximizes the total profit. The algebraic solution showed that the optimal solution is to sell $0$ roast beef sandwiches and $100$ turkey sandwiches, which maximizes the total profit.

References

• [1] "Profit Maximization Problem" by [Author]
• [2] "Linear Programming" by [Author]

Appendix

The following is a list of variables used in this article:

• $x$: number of roast beef sandwiches sold
• $y$: number of turkey sandwiches sold
• $P$: total profit
• $S$: total number of sandwiches sold
• $2.30$: profit per sandwich for roast beef special
• $3.10$: profit per sandwich for turkey special
• $100$: maximum number of sandwiches that can be sold
  A Deli's Profit Maximization Problem: Q&A =====================================

Introduction

In our previous article, we explored a deli's profit maximization problem, where two specials are offered: the roast beef special and the turkey special. We analyzed the profit per sandwich for each special and determined the optimal number of sandwiches to sell to maximize profit. In this article, we will answer some frequently asked questions (FAQs) related to the deli's profit maximization problem.

Q: What is the profit per sandwich for each special?

A: The profit per sandwich for the roast beef special is $\$2.30$, and the profit per sandwich for the turkey special is $\$3.10$.

Q: How many slices of bread are used for each special?

A: The roast beef special uses two slices of bread, and the turkey special uses three slices of bread.

Q: What is the total profit function?

A: The total profit function is given by:

$$P = 2.30x + 3.10y$$

where $x$ is the number of roast beef sandwiches sold and $y$ is the number of turkey sandwiches sold.

Q: What is the constraint on the total number of sandwiches sold?

A: The constraint on the total number of sandwiches sold is given by:

$$S = x + y \leq 100$$

where $S$ is the total number of sandwiches sold.

Q: How can we solve this problem graphically?

A: We can solve this problem graphically by plotting the profit function $P = 2.30x + 3.10y$ and the constraint $S = x + y \leq 100$. The profit function is a linear function, and the constraint is a linear inequality.

Q: How can we solve this problem algebraically?

A: We can solve this problem algebraically by using the method of substitution. We can substitute the constraint $S = x + y \leq 100$ into the profit function $P = 2.30x + 3.10y$ to get:

$$P = 2.30x + 3.10(100 - x)$$

Simplifying the expression, we get:

$$P = 2.30x + 310 - 3.10x$$

Combine like terms:

$$P = -0.80x + 310$$

Q: What is the optimal solution?

A: The optimal solution is to sell $50$ roast beef sandwiches and $50$ turkey sandwiches, which maximizes the total profit.

Q: Why is the optimal solution to sell $50$ roast beef sandwiches and $50$ turkey sandwiches?

A: The optimal solution is to sell $50$ roast beef sandwiches and $50$ turkey sandwiches because this combination of sandwiches maximizes the total profit. The profit per sandwich for the roast beef special is $\$2.30$, and the profit per sandwich for the turkey special is $\$3.10$. By selling $50$ roast beef sandwiches and $50$ turkey sandwiches, the deli can maximize the total profit.

Q: What is the maximum profit that can be achieved?

A: The maximum profit that can be achieved is $\$380$.

Q: How can the deli achieve the maximum profit?

A: The deli can achieve the maximum profit by selling $50$ roast beef sandwiches and $50$ turkey sandwiches.

Conclusion

In this article, we answered some frequently asked questions (FAQs) related to the deli's profit maximization problem. We provided the profit per sandwich for each special, the total profit function, the constraint on the total number of sandwiches sold, and the optimal solution. We also explained why the optimal solution is to sell $50$ roast beef sandwiches and $50$ turkey sandwiches.

References

• [1] "Profit Maximization Problem" by [Author]
• [2] "Linear Programming" by [Author]

Appendix

The following is a list of variables used in this article:

• $x$: number of roast beef sandwiches sold
• $y$: number of turkey sandwiches sold
• $P$: total profit
• $S$: total number of sandwiches sold
• $2.30$: profit per sandwich for roast beef special
• $3.10$: profit per sandwich for turkey special
• $100$: maximum number of sandwiches that can be sold