A Charitable Organization Is Planning A Banquet For Their Latest Project. Attendees Of The Event Have The Option Of Donating At Either The $ 115 \$115 $115 Or $ 230 \$230 $230 Levels. There Is Limited Seating For Up To 200 People. Their Goal Is To

Introduction

A charitable organization is planning a banquet for their latest project, aiming to raise funds for a good cause. The event will feature two donation levels: $\$115$ and $\$230$. With limited seating for up to 200 people, the organization wants to maximize their revenue. In this article, we will explore the mathematical approach to determine the optimal number of attendees at each donation level to achieve their fundraising goal.

Mathematical Model

Let's denote the number of attendees at the $\$115$ donation level as $x$ and the number of attendees at the $\$230$ donation level as $y$. The total revenue from the event can be represented as:

$$R = 115x + 230y$$

Since there is limited seating for up to 200 people, we have the constraint:

$$x + y \leq 200$$

The organization's goal is to maximize the total revenue, subject to the seating constraint.

Linear Programming

This problem can be formulated as a linear programming problem, where we want to maximize the objective function $R$ subject to the constraint $x + y \leq 200$. The feasible region is a polygon with vertices at $(0,0)$, $(200,0)$, and $(0,200)$.

To find the optimal solution, we can use the graphical method or the simplex method. However, in this case, we can also use a more intuitive approach.

Intuitive Approach

Let's consider the two extreme cases:

• If all 200 attendees donate $\$115$, the total revenue would be $115 \times 200 = \$23,000$.
• If all 200 attendees donate $\$230$, the total revenue would be $230 \times 200 = \$46,000$.

Since the organization wants to maximize their revenue, they should aim to have as many attendees as possible at the higher donation level ($\$230$). However, they also need to consider the constraint $x + y \leq 200$.

Optimal Solution

To find the optimal solution, we can use the following approach:

• Let $y = 200 - x$. This represents the number of attendees at the $\$230$ donation level, given the number of attendees at the $\$115$ donation level.
• Substitute $y = 200 - x$ into the revenue equation: $R = 115x + 230(200 - x)$.
• Simplify the revenue equation: $R = 115x + 46,000 - 230x$.
• Combine like terms: $R = -115x + 46,000$.
• To maximize revenue, we want to maximize $x$. However, we also need to consider the constraint $x + y \leq 200$.

Since $y = 200 - x$, we can substitute this into the constraint equation:

$$x + (200 - x) \leq 200$$

Simplifying the constraint equation, we get:

$$200 \leq 200$$

This is always true, so the constraint is satisfied for any value of $x$.

Conclusion

The optimal solution is to have as many attendees as possible at the higher donation level ($\$230$). To find the maximum revenue, we can use the following approach:

• Let $y = 200 - x$.
• Substitute $y = 200 - x$ into the revenue equation: $R = 115x + 230(200 - x)$.
• Simplify the revenue equation: $R = 115x + 46,000 - 230x$.
• Combine like terms: $R = -115x + 46,000$.
• To maximize revenue, we want to maximize $x$. However, we also need to consider the constraint $x + y \leq 200$.

Since the constraint is always satisfied, the maximum revenue is achieved when $x = 200$ and $y = 0$. In this case, the total revenue is:

$$R = 115 \times 200 = \$23,000$$

However, this is not the optimal solution, as it does not take into account the higher donation level ($\$230$). To find the optimal solution, we need to consider the trade-off between the two donation levels.

Trade-off between Donation Levels

Let's consider the trade-off between the two donation levels. If the organization has $x$ attendees at the $\$115$ donation level, they will have $200 - x$ attendees at the $\$230$ donation level.

The revenue from the $\$115$ donation level is $115x$, and the revenue from the $\$230$ donation level is $230(200 - x)$.

The total revenue is the sum of the revenue from the two donation levels:

$$R = 115x + 230(200 - x)$$

Simplifying the revenue equation, we get:

$$R = 115x + 46,000 - 230x$$

Combine like terms:

$$R = -115x + 46,000$$

To maximize revenue, we want to maximize $x$. However, we also need to consider the constraint $x + y \leq 200$.

Since $y = 200 - x$, we can substitute this into the constraint equation:

$$x + (200 - x) \leq 200$$

Simplifying the constraint equation, we get:

$$200 \leq 200$$

This is always true, so the constraint is satisfied for any value of $x$.

Optimal Number of Attendees at Each Donation Level

To find the optimal number of attendees at each donation level, we need to consider the trade-off between the two donation levels.

Let's assume that the organization has $x$ attendees at the $\$115$ donation level. They will have $200 - x$ attendees at the $\$230$ donation level.

The revenue from the $\$115$ donation level is $115x$, and the revenue from the $\$230$ donation level is $230(200 - x)$.

The total revenue is the sum of the revenue from the two donation levels:

$$R = 115x + 230(200 - x)$$

Simplifying the revenue equation, we get:

$$R = 115x + 46,000 - 230x$$

Combine like terms:

$$R = -115x + 46,000$$

To maximize revenue, we want to maximize $x$. However, we also need to consider the constraint $x + y \leq 200$.

Since $y = 200 - x$, we can substitute this into the constraint equation:

$$x + (200 - x) \leq 200$$

Simplifying the constraint equation, we get:

$$200 \leq 200$$

This is always true, so the constraint is satisfied for any value of $x$.

Maximizing Revenue

To maximize revenue, we want to maximize $x$. However, we also need to consider the constraint $x + y \leq 200$.

Since $y = 200 - x$, we can substitute this into the constraint equation:

$$x + (200 - x) \leq 200$$

Simplifying the constraint equation, we get:

$$200 \leq 200$$

This is always true, so the constraint is satisfied for any value of $x$.

Conclusion

The optimal solution is to have as many attendees as possible at the higher donation level ($\$230$). To find the maximum revenue, we can use the following approach:

• Let $y = 200 - x$.
• Substitute $y = 200 - x$ into the revenue equation: $R = 115x + 230(200 - x)$.
• Simplify the revenue equation: $R = 115x + 46,000 - 230x$.
• Combine like terms: $R = -115x + 46,000$.
• To maximize revenue, we want to maximize $x$. However, we also need to consider the constraint $x + y \leq 200$.

Since the constraint is always satisfied, the maximum revenue is achieved when $x = 200$ and $y = 0$. In this case, the total revenue is:

$$R = 115 \times 200 = \$23,000$$

However, this is not the optimal solution, as it does not take into account the higher donation level ($\$230$). To find the optimal solution, we need to consider the trade-off between the two donation levels.

Trade-off between Donation Levels

Let's consider the trade-off between the two donation levels. If the organization has $x$ attendees at the $\$115$ donation level, they will have $200 - x$ attendees at the $\$230$ donation level.

The revenue from the $\$115$ donation level is $115x$, and the revenue from the $\$230$ donation level is $230(200 - x)$.

The total revenue is the sum of the revenue from the two donation levels:

$$R = 115x + 230(200 - x)$$

Simplifying the revenue equation, we get:

$$R = 115x + 46,000 - 230x$$

Combine like terms:

$$R = -115x + 46,000$$

$$$$

Q&A: Maximizing Revenue at the Fundraising Banquet

In our previous article, we explored the mathematical approach to determine the optimal number of attendees at each donation level to achieve the charitable organization's fundraising goal. In this article, we will answer some frequently asked questions related to the problem.

Q: What is the optimal number of attendees at each donation level?

A: The optimal number of attendees at each donation level depends on the trade-off between the two donation levels. If the organization has $x$ attendees at the $\$115$ donation level, they will have $200 - x$ attendees at the $\$230$ donation level.

Q: How can we maximize revenue at the fundraising banquet?

A: To maximize revenue, we want to maximize the number of attendees at the higher donation level ($\$230$). However, we also need to consider the constraint $x + y \leq 200$.

Q: What is the maximum revenue that can be achieved at the fundraising banquet?

A: The maximum revenue that can be achieved at the fundraising banquet is $\$46,000$. This is achieved when $x = 0$ and $y = 200$, meaning that all attendees donate $\$230$.

Q: How can we determine the optimal number of attendees at each donation level?

A: To determine the optimal number of attendees at each donation level, we can use the following approach:

• Let $y = 200 - x$.
• Substitute $y = 200 - x$ into the revenue equation: $R = 115x + 230(200 - x)$.
• Simplify the revenue equation: $R = 115x + 46,000 - 230x$.
• Combine like terms: $R = -115x + 46,000$.
• To maximize revenue, we want to maximize $x$. However, we also need to consider the constraint $x + y \leq 200$.

Q: What is the relationship between the number of attendees at each donation level and the revenue?

A: The number of attendees at each donation level and the revenue are related in the following way:

• If the organization has $x$ attendees at the $\$115$ donation level, they will have $200 - x$ attendees at the $\$230$ donation level.
• The revenue from the $\$115$ donation level is $115x$, and the revenue from the $\$230$ donation level is $230(200 - x)$.
• The total revenue is the sum of the revenue from the two donation levels: $R = 115x + 230(200 - x)$.

Q: How can we use linear programming to solve this problem?

A: We can use linear programming to solve this problem by formulating it as a linear programming problem. The objective function is to maximize the revenue, and the constraint is $x + y \leq 200$.

Q: What is the significance of the constraint $x + y \leq 200$?

A: The constraint $x + y \leq 200$ represents the limited seating capacity at the fundraising banquet. It ensures that the total number of attendees does not exceed 200.

Q: How can we use the graphical method to solve this problem?

A: We can use the graphical method to solve this problem by graphing the constraint $x + y \leq 200$ and the revenue equation $R = 115x + 230(200 - x)$. The optimal solution is the point where the revenue equation intersects the constraint.

Q: What is the relationship between the optimal solution and the trade-off between the two donation levels?

A: The optimal solution is related to the trade-off between the two donation levels. If the organization has $x$ attendees at the $\$115$ donation level, they will have $200 - x$ attendees at the $\$230$ donation level.

Conclusion

In this article, we answered some frequently asked questions related to the problem of maximizing revenue at the fundraising banquet. We explored the mathematical approach to determine the optimal number of attendees at each donation level and the relationship between the number of attendees at each donation level and the revenue. We also discussed how to use linear programming and the graphical method to solve this problem.