Solving Quadratic Functions For Profit Maximization In Business

In the world of business, understanding the relationship between price, demand, and profit is crucial for success. Mathematical models often play a vital role in this understanding, allowing businesses to predict outcomes and make informed decisions. Quadratic functions are one such tool, frequently used to model scenarios where the relationship between variables is not linear. This article delves into how a quadratic function can model a store's daily profit based on the selling price of a T-shirt, and how to determine the price points that would generate a specific profit target. We'll explore the equation needed to solve this problem and discuss various methods for finding the solution.

Modeling Profit with a Quadratic Function

Let's consider the quadratic function $y = -10x^2 + 160x - 430$, which models a store's daily profit (y) for selling a T-shirt priced at x dollars. Quadratic functions are characterized by their parabolic shape, which can either open upwards or downwards. In this case, the coefficient of the $x^2$ term is -10, which is negative. This indicates that the parabola opens downwards, meaning there is a maximum point (vertex) representing the price that yields the highest profit. Understanding this is crucial because it tells us that as the price increases or decreases beyond a certain point, the profit will decline.

The given function encapsulates several key factors that influence profit. The $-10x^2$ term suggests that as the price (x) increases, there's a diminishing return on profit. This could be due to decreased demand at higher prices. The $160x$ term indicates that there's a positive relationship between price and profit, meaning that initially, increasing the price increases profit. However, this positive effect is counteracted by the $-10x^2$ term as the price continues to rise. The constant term, -430, represents the fixed costs or other factors that contribute to a loss even when no T-shirts are sold. This could include rent, utilities, or other overhead expenses. Analyzing each component of the quadratic function provides valuable insights into the store's financial dynamics. For instance, the coefficient of the linear term (160) represents the marginal profit gained from each additional dollar increase in price, up to a certain point. However, the quadratic term (-10) signifies the rate at which this marginal profit decreases as the price goes up. This balance between the linear and quadratic terms determines the overall shape of the profit curve.

Setting a Profit Target: The Equation to Solve

Now, let's address the core question: What equation do we need to solve to find the selling price or prices that would generate $50 in daily profit? To determine this, we set the profit (y) equal to 50 and solve for x. This gives us the equation:

$50 = -10x^2 + 160x - 430$

This equation is a quadratic equation, and solving it will provide the selling price(s) at which the store would make a $50 profit. Rearranging the equation to standard form, we subtract 50 from both sides to set the equation equal to zero:

$0 = -10x^2 + 160x - 480$

This standard form equation ($ax^2 + bx + c = 0$) is essential for applying various methods to find the solutions for x. Understanding how to transform the original equation into this standard form is the first crucial step in solving for the price points that yield the desired profit. This transformation allows us to use standard algebraic techniques, such as factoring, completing the square, or applying the quadratic formula, to find the roots of the equation. Each of these methods provides a systematic approach to finding the values of x that satisfy the equation, representing the selling prices that result in a $50 daily profit. The standard form also makes it easier to analyze the coefficients and determine the nature of the solutions, such as whether they are real or complex, and whether there are one, two, or no real solutions. This preliminary analysis can help in choosing the most appropriate solution method and interpreting the results in a practical context.

Methods for Solving the Quadratic Equation

There are several methods to solve the quadratic equation $-10x^2 + 160x - 480 = 0$. Let's explore the most common ones:

1. Factoring

Factoring involves expressing the quadratic expression as a product of two binomials. To make factoring easier, we can first simplify the equation by dividing all terms by -10:

$0 = x^2 - 16x + 48$

Now, we look for two numbers that multiply to 48 and add up to -16. These numbers are -4 and -12. Therefore, we can factor the equation as follows:

$0 = (x - 4)(x - 12)$

Setting each factor equal to zero gives us the solutions:

$x - 4 = 0$ or $x - 12 = 0$

$x = 4$ or $x = 12$

So, the selling prices that would generate a $50 daily profit are $4 and $12.

Factoring is an efficient method when the quadratic equation can be easily factored, meaning that the roots are integers or simple fractions. The process involves identifying the coefficients and constants in the quadratic equation and finding two binomial expressions that, when multiplied, yield the original quadratic expression. This method relies on the ability to recognize patterns and relationships between the coefficients, such as the sum and product of the roots. When factoring is possible, it provides a direct and straightforward way to find the solutions, without the need for more complex formulas or calculations. However, not all quadratic equations can be factored easily, and in such cases, alternative methods like the quadratic formula or completing the square are more appropriate. The success of factoring depends largely on the nature of the roots and the ability to quickly identify the factors. If the roots are irrational or complex, factoring may not be a feasible method, and other techniques will be required to solve the equation.

2. Quadratic Formula

The quadratic formula is a universal method for solving any quadratic equation of the form $ax^2 + bx + c = 0$. The formula is:

$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

For our equation, $-10x^2 + 160x - 480 = 0$, we have $a = -10$, $b = 160$, and $c = -480$. Plugging these values into the quadratic formula, we get:

$x = \frac{-160 ± \sqrt{160^2 - 4(-10)(-480)}}{2(-10)}$

$x = \frac{-160 ± \sqrt{25600 - 19200}}{-20}$

$x = \frac{-160 ± \sqrt{6400}}{-20}$

$x = \frac{-160 ± 80}{-20}$

This gives us two solutions:

$x = \frac{-160 + 80}{-20} = \frac{-80}{-20} = 4$

$x = \frac{-160 - 80}{-20} = \frac{-240}{-20} = 12$

Again, we find that the selling prices that would generate a $50 daily profit are $4 and $12.

The quadratic formula is a powerful and versatile tool for solving any quadratic equation, regardless of the complexity or nature of its roots. Unlike factoring, which requires identifying specific factors, the quadratic formula provides a direct calculation of the solutions using the coefficients of the quadratic equation. This makes it particularly useful when factoring is difficult or impossible, such as when the roots are irrational or complex. The formula involves substituting the values of the coefficients a, b, and c into the expression and performing the arithmetic operations to find the solutions. The discriminant ($b^2 - 4ac$) within the formula provides valuable information about the nature of the roots. If the discriminant is positive, there are two distinct real roots; if it is zero, there is exactly one real root (a repeated root); and if it is negative, there are two complex roots. This makes the quadratic formula not only a method for finding solutions but also a tool for analyzing the characteristics of the quadratic equation itself. The application of the quadratic formula requires careful attention to detail in the substitution and simplification steps, but it offers a reliable and systematic approach to solving quadratic equations in a wide range of contexts.

3. Completing the Square

Completing the square is another method to solve quadratic equations. It involves transforming the equation into a perfect square trinomial. Starting with our simplified equation:

$x^2 - 16x + 48 = 0$

Move the constant term to the right side:

$x^2 - 16x = -48$

To complete the square, we take half of the coefficient of the x term (-16), square it ((-8)^2 = 64), and add it to both sides:

$x^2 - 16x + 64 = -48 + 64$

$(x - 8)^2 = 16$

Now, take the square root of both sides:

$x - 8 = ±4$

This gives us two solutions:

$x = 8 + 4 = 12$

$x = 8 - 4 = 4$

Again, we find the same selling prices: $4 and $12.

Completing the square is a method for solving quadratic equations that transforms the equation into a form where a perfect square trinomial is equal to a constant. This technique involves manipulating the equation algebraically to create a squared binomial, which can then be solved by taking the square root of both sides. The process begins by isolating the terms containing $x^2$ and $x$ on one side of the equation and moving the constant term to the other side. Then, a specific constant is added to both sides to complete the square, making the left side a perfect square trinomial. This constant is determined by taking half of the coefficient of the x term, squaring it, and adding the result to both sides. Once the square is completed, the equation can be written in the form $(x - h)^2 = k$, where h and k are constants. Taking the square root of both sides yields two possible solutions for x, which can be found by solving the resulting linear equations. Completing the square is a valuable technique because it provides a systematic approach to solving quadratic equations and also lays the foundation for understanding the quadratic formula, which is derived from the process of completing the square. While it may be more complex than factoring in some cases, completing the square is a reliable method for solving any quadratic equation, especially when factoring is not straightforward.

Interpreting the Solutions

The solutions x = 4 and x = 12 represent the two selling prices for the T-shirt that would result in a daily profit of $50. This is because a quadratic function forms a parabola, and a horizontal line (representing a specific profit target) can intersect the parabola at two points. These points correspond to the two different selling prices that yield the same profit.

Interpreting the solutions of a quadratic equation within the context of the problem is crucial for understanding their practical implications. In the case of the store's profit model, the two solutions for x represent the selling prices that would generate the desired profit level. However, it's important to consider whether both solutions are viable and meaningful in the real world. For example, if one of the solutions were negative, it would not make sense as a selling price. Similarly, if one of the prices were excessively high or low, it might not be practical due to market conditions or customer behavior. The quadratic function models the relationship between price and profit, and the solutions provide specific price points that meet the profit target. However, a comprehensive analysis should also consider other factors such as production costs, demand elasticity, and competitive pricing to determine the optimal selling price. The shape of the parabola, defined by the quadratic function, also provides insights into the profit behavior. The vertex of the parabola represents the maximum profit and the price at which it is achieved. Understanding the overall shape of the profit curve helps in making strategic decisions about pricing and production levels. Therefore, while the solutions of the quadratic equation are valuable, they should be interpreted in conjunction with other relevant business considerations.

Conclusion

In this article, we explored how a quadratic function can be used to model a store's daily profit based on the selling price of a T-shirt. We determined the equation needed to find the selling prices that would generate a $50 profit and discussed three methods for solving quadratic equations: factoring, the quadratic formula, and completing the square. Each method provides a way to find the solutions, and the choice of method often depends on the specific equation and personal preference. The solutions obtained, x = 4 and x = 12, represent the selling prices that would result in a $50 daily profit, demonstrating the practical application of quadratic functions in business decision-making.

Understanding the relationship between mathematical models and real-world scenarios is essential for effective problem-solving. Quadratic functions, with their parabolic nature, are particularly useful for modeling situations where there is a maximum or minimum value, such as profit maximization or cost minimization. The ability to set up and solve quadratic equations provides a powerful tool for analyzing and optimizing these scenarios. By exploring different solution methods and interpreting the results in context, businesses can make informed decisions that lead to better outcomes. The application of mathematical concepts like quadratic functions extends beyond the classroom and into practical domains, highlighting the importance of mathematical literacy in various fields. In conclusion, the use of quadratic functions in modeling profit and other business-related scenarios exemplifies the value of mathematics in real-world decision-making.