Problem 45 - Algebra 2

Question

For Exercises \(43-45,\) use the following information.The girls' softball team is sponsoring a fund-raising trip to see a professional baseball game. They charter a \(60-\) passenger bus for \(\$ 525 .\) In order to make a profit, they will charge \(\$ 15\) per person if all seats on the bus are sold, but for each empty seat, they will increase the price by \(\$ 1.50\) per person. What is the maximum profit the team can make with this fund-raiser, and how many passengers will it take to achieve this maximum?

Step-by-Step Solution

Answer

The team can make a maximum profit of \(\$1312.5\) with 35 passengers.

Step 1: Define the Profit Function

The profit function is defined based on the number of passengers (let this be \( x \)). The cost for each passenger initially is \( \\(15 \). If there are \( 60 - x \) empty seats, the charge per person increases by \( \\)1.50 \times (60 - x) \). Thus, the price per ticket is \( 15 + 1.5(60 - x) \). The profit is then given by the formula: \( P(x) = x \times (15 + 1.5(60 - x)) - 525 \).

Step 2: Expand the Profit Function

Expand the function to simplify: \( P(x) = x (15 + 90 - 1.5x) - 525 \). Simplifying, we get \( P(x) = 105x - 1.5x^2 - 525 \).

Step 3: Find the Vertex of the Parabola

The profit function \( P(x) = -1.5x^2 + 105x - 525 \) is a quadratic equation. The maximum profit occurs at the vertex of the parabola. For a quadratic equation in the form \( ax^2 + bx + c \), the vertex x-coordinate is at \( x = -\frac{b}{2a} \).

Step 4: Calculate the Vertex X-Coordinate

Using the vertex formula where \( a = -1.5 \) and \( b = 105 \), calculate \[ x = -\frac{105}{2 \times (-1.5)} = \frac{105}{3} = 35 \].

Step 5: Evaluate the Maximum Profit

Substitute \( x = 35 \) back into the profit equation to find the maximum profit:\[ P(35) = -1.5(35)^2 + 105(35) - 525 \]. Calculate to find \[ P(35) = -1.5(1225) + 3675 - 525 = -1837.5 + 3675 - 525 = 1312.5 \]. Thus, the maximum profit is \(\$1312.5\).

Key Concepts

Profit Maximization

Maximizing profit means finding the condition under which profit reaches its highest possible value. In many real-world scenarios, similar to the softball team's fundraiser, we use mathematical functions to calculate this.

To determine maximum profit, a function is used to represent the expected earnings from an event, such as the number of passengers multiplied by the ticket price, minus costs. In this case, the profit was modeled by the equation:

• Given profit function: \[ P(x) = x imes (15 + 1.5(60 - x)) - 525 \]
• Simplified to: \[ P(x) = 105x - 1.5x^2 - 525 \]

Here, maximizing involves finding the peak of the parabola described by the profit function.

Understanding this concept involves recognizing how pricing adjustments (such as charging more for fewer passengers) impact overall profit.

Vertex of a Parabola

The vertex of a parabola is crucial in determining points of maximum or minimum value for quadratic equations, which visually depict a curve. For the girls' softball team profit function, finding the vertex determines the number of passengers that maximizes profit.

With any quadratic equation of the form \( ax^2 + bx + c \), you can find the vertex's x-coordinate using:

• \( x = -\frac{b}{2a} \)

In our example, the profit function is \( P(x) = -1.5x^2 + 105x - 525 \), giving the vertex x-coordinate as:

• \( x = -\frac{105}{2 imes (-1.5)} = 35 \)

This calculation points to the optimal number of passengers needed to maximize profit, which is the main reason finding the vertex of a parabola is useful in these problems.

The vertex not only informs decision-making but is a fundamental part of analyzing and understanding the behavior of quadratic functions.

Fund-Raising Problems

Organizations, like school teams, often hold fundraisers to meet budgetary goals, and understanding mathematical concepts is key to their success. In the softball team's example, they aim to make the most money by leveraging available resources, like the bus seats.

Fundraisers need careful consideration:

• Balancing ticket prices while considering demand
• Understanding customer needs

By using mathematics, the team deciphers the optimal number of participants for maximum profit, thus ensuring efficiency.

Solving fundraising problems often involves constructing mathematical models, akin to the profit function, to gauge outcomes and make informed decisions. Knowing these calculations gives fundraisers a powerful tool for strategizing.

Quadratic Equations

Quadratic equations are a type of equation frequently encountered in mathematics and various applications. These are polynomials of degree two, typically expressed in the form \( ax^2 + bx + c = 0 \).

They create a parabolic graph that can open upwards or downwards, a direction determined by the sign of the coefficient \( a \). Their practical applications can be seen in projects like the team’s fundraiser where they represent scenarios involving variables like costs, revenue, or population growth.

• Quadratic formula is used to solve them: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• Occasionally, solving by factoring or completing the square is effective

For the softball team, the quadratic equation didn't just plot potential outcomes but also solved to find practical values such as the number of passengers needed to reach maximum profitability.

Understanding the nuances of quadratic equations can unlock solutions to diverse, complex problems both in academic exercises and real-world situations.