Problem 15 - Beginning and Intermediate Algebra

Question

A manufacturer's revenue, \(R(x)\) in dollars, from the sale of \(x\) calculators is given by \(R(x)=12 x .\) The company's cost, \(C(x)\) in dollars, to produce \(x\) calculators is \(C(x)=8 x+2000\). a) Find the profit function, \(P(x),\) that defines the manufacturer's profit from the sale of \(x\) calculators. b) What is the profit from the sale of 1500 calculators?

Step-by-Step Solution

Answer

a) The profit function is \(P(x) = 4x - 2000\). b) The profit from the sale of 1500 calculators is $4000.

Step 1: Determine the profit function

To find the profit function P(x), we need to subtract the cost function C(x) from the revenue function R(x). That is, P(x) = R(x) - C(x). We are given R(x) = 12x and C(x) = 8x + 2000. Thus, P(x) = R(x) - C(x) = (12x) - (8x + 2000).

Step 2: Simplify the profit function

Now we simplify by subtracting the terms and combining like terms: P(x) = 12x - 8x - 2000 = 4x - 2000. So, the profit function P(x) is P(x) = 4x - 2000. #b) What is the profit from the sale of 1500 calculators?#

Step 3: Substitute the value into the profit function

To find the profit from the sale of 1500 calculators, we substitute x = 1500 into the profit function P(x) = 4x - 2000: P(1500) = 4(1500) - 2000.

Step 4: Calculate the profit

Now we proceed with the calculation: P(1500) = 6000 - 2000 = 4000. Hence, the profit from the sale of 1500 calculators is $4000.

Key Concepts

Revenue Function

A revenue function is a crucial concept in business mathematics. It represents the income from selling a certain number of products, such as calculators in this case. The revenue function is typically denoted as \( R(x) \). For this exercise, the revenue function is given as \( R(x) = 12x \). This means for every calculator sold, the company earns $12.
Understanding the revenue function helps businesses forecast potential earnings based on different sales levels. By analyzing the revenue function, companies can make informed decisions on pricing and production to maximize profit.
It's important to note that revenue is not the same as profit. Expenses and costs need to be accounted for, and that's where the cost function and profit function come into play.

Cost Function

The cost function, \( C(x) \), represents the total cost of producing a specific number of items. For the calculator sale scenario, the cost function is given by \( C(x) = 8x + 2000 \). This expression means that producing each calculator costs \(8, and there are fixed costs of \)2000, regardless of the number of calculators produced.
The fixed cost is an essential aspect, as it includes expenses that do not vary with production level, such as rent or salaries. Variable costs, on the other hand, refer to costs that change with the number of items produced, like materials or labor per unit.
Understanding both fixed and variable costs allows companies to budget accurately and calculate the break-even point, which is the number where total revenue equals total costs.

Calculators

Calculators aren't just devices for arithmetic; in this context, they're the product being analyzed for profit. Each calculator adds $12 to revenue and incurs $8 in production costs plus a share of the fixed costs.
From a mathematical perspective, calculators serve as variables in the equation representing the business problem. Studying the production and sale of calculators helps illustrate how businesses make decisions about pricing and production amounts.
Analyzing "calculators" in terms of units sold helps businesses to figure out how many need to be sold to achieve profitability, considering all associated costs and the determined selling price.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is vital for solving many mathematical problems, including the calculation of profit functions. In this exercise, simplifying allows us to take complex functions like \( P(x) = 12x - (8x + 2000) \) and reduce them to a more manageable form, \( 4x - 2000 \).
Combining like terms, which means adding or subtracting coefficients of similar variables, is a key part of this process. For instance, \( 12x - 8x \) simplifies to \( 4x \).
Simplification helps not only in finding solutions but also in making the functions easier to interpret and use for further calculations, such as determining the profit for various sales volumes.