Maximizing Profit from Sales Through Algebraic Equations

Understanding how to determine the profit of a sale given algebraic equations is a crucial skill in both business and mathematics. In this article, we explore how algebraic equations can help us model and optimize profit. We will provide detailed steps on how to derive the profit function, find its maximum, and implement it in practical scenarios.

Introduction to the Problem

Consider a scenario where the cost and selling price of a product depend on the number of gadgets produced or sold. We are given the following functions:

C(gadgets) 1/6 x^2 5x 22

P(gadget) 30 - 1/3 x

Where:

G(X) is the cost of producing x gadgets. P(x) is the selling price of one gadget.

The gross income when selling x gadgets is x * P(x). The profit function P(x) is the difference between the gross income and the cost of production. Thus, we can write:

P(x) x * P(x) - C(x)

Substituting the given functions:

P(x) x(30 - 1/3 x) - (1/6 x^2 5x 22)

Let's simplify the equation:

P(x) 3 - (1/3)x^2 - (1/6)x^2 - 5x - 22

P(x) 3 - (2/6)x^2 - 5x - 22

P(x) -1/3x^2 25x - 22

Determining the Maximum Profit

To find the maximum profit, we need to find the vertex of the parabola defined by the profit function. The coefficient of x^2 is negative, indicating that the parabola opens downwards, and thus, the vertex will give us the maximum profit.

Using Calculus to Find the Vertex

To find the maximum, we take the derivative of P(x):

dP/dx - (2/6)x 25

Setting the derivative equal to zero:

- (2/6)x 25 0

- (1/3)x 25 0

x 25

The profit is maximized when x 25. To confirm, we can find the second derivative:

d^2P/dx^2 - (2/6) -1/3

Since the second derivative is negative, the function has a maximum at x 25.

Calculating the Maximum Profit

We substitute x 25 into the profit function to find the maximum profit:

P(25) - (1/3)(25^2) 25(25) - 22

P(25) - (1/3)(625) 625 - 22

P(25) - 625/3 625 - 22

P(25) 395

The maximum profit is $395 when 25 gadgets are sold.

Conclusion

Through algebraic equations, we can accurately model the cost and revenue of a product, and use derivatives to find the maximum profit. By optimizing the number of units sold, we can maximize the profit. The process involves:

Defining the cost and revenue functions. Formulating the profit function. Using calculus to find the vertex of the parabola.

Understanding these steps is essential for businesses looking to make informed decisions that lead to financial success.

Key Takeaways:

Profit maximization involves finding the vertex of a parabola defined by the profit function. Calculus derivatives help identify the maximum profit point. Algebraic equations can be used to model cost, revenue, and profit in business scenarios.