Problem 71 - Algebra 2

Question

Profit A craftsman makes and sells violins. The function \(C(x)=1000+700 x\) represents his cost in dollars to produce \(x\) violins. The function \(I(x)=5995 x\) represents the income in dollars from selling \(x\) violins. a. Write and simplify a function \(P(x)=I(x)-C(x) .\) a. Find \(P(30),\) the profit earned when he makes and sells 30 violins.

Step-by-Step Solution

Answer

The profit earned when the craftsman makes and sells 30 violins is $149,000.

Step 1: Write the Function for Profit

The function for profit is given by \(P(x) = I(x) - C(x)\). Here \(I(x)\) is the income from selling x violins, and \(C(x)\) is the cost of making x violins. Hence, the function for profit becomes \(P(x) = 5995x - (1000 + 700x)\).

Step 2: Simplify the Function for Profit

Simplify the function for profit by combining the like terms. It simplifies as: \(P(x) = 5995x - 1000 - 700x = 5000x - 1000 \).

Step 3: Calculate Profit for 30 violins

The profit for a given number of violins can be found by substituting the number of violins into the profit function. Let's find \(P(30)\) : \(P(30) = 5000 * 30 - 1000 = 149000\).

Key Concepts

Cost Function

The cost function, denoted as \(C(x)\), is a mathematical representation of the total cost incurred to produce a certain number of items—in this case, violins. In our exercise, the cost function is given as \(C(x) = 1000 + 700x\).

This function is composed of two parts:

• A fixed cost, which is \(1000\) dollars. This is a constant amount that doesn't change regardless of how many violins are made, perhaps covering overheads or initial setup expenses.
  
• The variable cost component, \(700x\), indicates that each additional violin made costs \(700\) dollars. As such, this term hinges on the number of violins, \(x\), produced.
  

Understanding the cost function helps in efficiently managing production costs, and it serves as one half of our profit calculation.

Revenue Function

The revenue function, \(I(x)\), represents the income generated from selling products—in our example, violins. Mathematically, it's expressed as \(I(x) = 5995x\).

Here’s how it’s broken down:

• The coefficient \(5995\) is the price at which each violin is sold.
  
• Multiplying this by \(x\), the number of violins sold, gives the total revenue.
  

This function is crucial because it shows the potential income from sales. By examining \(I(x)\), one can understand how income changes with different levels of production and sales. Revenue functions help in setting sales targets and understanding potential market scenarios.

Simplifying Expressions

When dealing with functions like our profit function \(P(x) = I(x) - C(x)\), simplifying expressions is a key mathematical skill. Initially, our profit function was \(P(x) = 5995x - (1000 + 700x)\). Simplification involves combining like terms and removing unnecessary brackets.

Here’s how we simplify:

• The terms \(5995x\) and \(-700x\) are combined since they are both multiplied by \(x\). This yields \(5995x - 700x = 5295x\).
  
• Next, the constant term \(-1000\) is simplified to find the overall profit expression, leading to \(P(x) = 5000x - 1000\).
  

By simplifying the profit function to \(P(x) = 5000x - 1000\), we make it easier to evaluate and assess under different sales scenarios, streamlining further calculations.

Substitution Method

The substitution method is used in solving equations by replacing variables with specific values. In our exercise, it's a straightforward way to find the profit for a given number of violins.

Once we have the simplified profit function \(P(x) = 5000x - 1000\), we can substitute \(x = 30\) to calculate the profit for selling 30 violins.

• This involves substituting 30 in place of \(x\): \(P(30) = 5000 \times 30 - 1000\).
  
• After doing the arithmetic, we get \(P(30) = 150000 - 1000 = 149000\).
  

This method is critical in applied mathematics as it allows for practical, numerical results from algebraic expressions, ensuring theoretical models have real-world application.