Problem 40 - Beginning and Intermediate Algebra

Question

The senior class at Richmont High School is selling t-shirts to raise money for its prom. The equation \(R(p)=-25 p^{2}+600 p\) describes the revenue, \(R,\) in dollars, as a function of the price, \(p,\) in dollars, of a t-shirt. That is, the revenue is a function of price. a) Determine the revenue if the group sells each shirt for \(\$ 10\) b) Determine the revenue if the group sells each shirt for \(\$ 15\) c) If the senior class hopes to have a revenue of \(\$ 3600,\) how much should it charge for each t-shirt?

Step-by-Step Solution

Answer

In summary: a) The revenue when the price is $10 per t-shirt is $3500. b) The revenue when the price is $15 per t-shirt is $3375. c) To achieve a revenue of $3600, the senior class should charge $12 per t-shirt.

Step 1: a) Price = $10

To determine the revenue when the price is \(10, substitute \)p=10$ into the given equation: \[R(10) = -25(10)^2 + 600(10)\] \[R(10) = -25(100) + 6000\] \[R(10) = -2500 + 6000\] \[R(10) = 3500\] The revenue when the price is \(10 per t-shirt is \)3500.

Step 2: b) Price = $15

To determine the revenue when the price is \(15, substitute \)p=15$ into the given equation: \[R(15) = -25(15)^2 + 600(15)\] \[R(15) = -25(225) + 9000\] \[R(15) = -5625 + 9000\] \[R(15) = 3375\] The revenue when the price is \(15 per t-shirt is \)3375. For part c, set the given revenue equal to the equation and solve for the price.

Step 3: c) Revenue = $3600

In order to find the price that results in a revenue of \(3600, set the given revenue equal to the equation and solve for \)p$: \[3600 = -25p^2 + 600p\] Rearrange the equation to find the quadratic equation to solve for p: \[25p^2 - 600p + 3600 = 0\] Now, solve the quadratic equation using the quadratic formula: \[p = \frac{-(-600) \pm \sqrt{(-600)^2 - 4(25)(3600)}}{2(25)}\] Calculate the values of the discriminant and the two possible prices: \[p = \frac{600 \pm \sqrt{(360000 - 360000)}}{50}\] Since the discriminant is zero, there is only one solution: \[p = \frac{600}{50}\] \[p = 12\] To achieve a revenue of \(3600, the senior class should charge \)12 per t-shirt.

Key Concepts

Revenue Calculation

Revenue plays a significant role in financial math and can be calculated using various mathematical models. In this exercise, the revenue generated from selling t-shirts is given as a function of the t-shirt price, demonstrating a practical example of revenue calculation.

Revenue is simply the amount of money earned from sales, before any costs or expenses are deducted. In mathematical terms, revenue is typically formulated as:

• Revenue = Price × Quantity Sold

In this specific case, we use the quadratic equation to model revenue, as different pricing strategies yield different quantities sold. The equation used is:\[ R(p) = -25p^2 + 600p \]where:

• \( p \) is the price of one t-shirt
• \( R(p) \) is the revenue as a function of the price

By substituting various values for \( p \), you can determine how changing the price impacts the revenue, thus making strategic decisions based on the outcomes.

High School Algebra

High school algebra introduces students to the manipulation and understanding of mathematical equations, and it's essential for solving real-life problems. In this exercise, we use algebra to evaluate a quadratic function and solve for variables.

With the problem focusing on substitution and solving equations, it engages:

• Substituting specific numbers into equations to find outcomes
• Simplifying and rearranging equations to solve for unknown variables

The task of plugging in the t-shirt prices into the function \(R(p)\), such as \(p=10\) or \(p=15\), demonstrates the substitution process. It shows how inputs into functions influence outputs. Furthermore, solving for a specific revenue goal involves understanding how to manipulate and solve algebraic equations. These skill sets are foundational for higher-level math and practical problem-solving in everyday life.

Quadratic Equations

Quadratic equations are a staple of algebra that allow for calculating variables when faced with complex relationships. These equations often involve terms squared, and they are pivotal in modeling various scenarios like projectile motion or revenue calculations, as seen here.

The standard format of a quadratic equation is given by:\[ ax^2 + bx + c = 0 \]In our revenue problem, the equation \(-25p^2 + 600p = 3600\) needs to be rearranged to form \(25p^2 - 600p + 3600 = 0\). This is typical when solving for an unknown variable within quadratic functions.

To find the variable \(p\) that satisfies the equation, you can use the quadratic formula, which is:\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this scenario, the quadratic formula determines the optimal price to reach a desired revenue, showcasing how these equations relate algebra and real-world financial outcomes. Using this method allows for understanding which strategies maximize profit given constraints and desired revenue goals.