Problem 68 - Algebra for College Students

Question

Suppose that the profit function for selling \(n\) items is given by $$ P(n)=-n^{2}+500 n-61,500 $$ Evaluate \(P(200), P(230), P(250)\), and \(P(260)\). -1500; 600; 1000; 900

Step-by-Step Solution

Answer

-1500; 600; 1000; 900

Step 1: Evaluate P(200)

We need to substitute \(n = 200\) into the profit function \(P(n) = -n^2 + 500n - 61,500\). So, \(P(200) = -(200)^2 + 500 \times 200 - 61,500\). Calculate each part:\(- (200)^2 = -40,000\)\(500 \times 200 = 100,000\)Substituting these back, we get:\(P(200) = -40,000 + 100,000 - 61,500 = -1,500\).

Step 2: Evaluate P(230)

Substitute \(n = 230\) into the profit function:\(P(230) = -(230)^2 + 500 \times 230 - 61,500\).Calculate each part:\(- (230)^2 = -52,900\)\(500 \times 230 = 115,000\)Substituting these back, we get:\(P(230) = -52,900 + 115,000 - 61,500 = 600\).

Step 3: Evaluate P(250)

Substitute \(n = 250\) into the profit function:\(P(250) = -(250)^2 + 500 \times 250 - 61,500\).Calculate each part:\(- (250)^2 = -62,500\)\(500 \times 250 = 125,000\)Substituting these back, we get:\(P(250) = -62,500 + 125,000 - 61,500 = 1,000\).

Step 4: Evaluate P(260)

Substitute \(n = 260\) into the profit function:\(P(260) = -(260)^2 + 500 \times 260 - 61,500\).Calculate each part:\(- (260)^2 = -67,600\)\(500 \times 260 = 130,000\)Substituting these back, we get:\(P(260) = -67,600 + 130,000 - 61,500 = 900\).

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function that can be written in the standard form of \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). The graph of a quadratic function is a parabola.
In the context of profit functions, this parabola can open upwards or downwards. If \( a \) is positive, it opens upwards like a smile. If \( a \) is negative, it opens downwards like a frown.
In our example, the profit function \( P(n) = -n^2 + 500n - 61,500 \) is quadratic with \( a = -1 \), \( b = 500 \), and \( c = -61,500 \).
This function opens downwards, indicating there is a maximum point of profit which can be particularly important in maximizing business performance.

• The degree of the quadratic is 2 because of the highest power of \( n \).
• Quadratic functions can be used to model various real-world scenarios, including profit maximization and loss minimizations in business.

Algebraic Expressions

Algebraic expressions combine numbers and variables using arithmetic operations, like addition, subtraction, multiplication, and division. These are symbolic representations to help solve equations or model scenarios in everyday mathematics.
With the profit function \( P(n) = -n^2 + 500n - 61,500 \), we see a classic example of an algebraic expression in action.
Each part of the expression serves a specific role:

• The term \(-n^2\) represents a declining factor as the number of items \( n \) increases.
• The \( 500n \) term signifies a linear increase in profit per item.
• The constant term \(-61,500\) indicates the base expenses or setup cost the business must cover.

Understanding each piece can provide insights into how varying item sales impact the overall profit outcome.
Breaking down complex expressions into components helps in solving and evaluating the profit function efficiently.

Substitution Method

Substituting values into an algebraic expression is a fundamental tool when dealing with functions or equations. The goal is to replace the variable with a given number and perform the necessary calculations to evaluate the function.
For the profit function \( P(n) = -n^2 + 500n - 61,500 \), substitution allows calculating profit for specific sales numbers.
Here’s how:

• Identify the variable, which is \( n \) in our function.
• Substitute a specific number for \( n \).
• Perform arithmetic to simplify the expression.

For example, when evaluating \( P(200) \), we replace \( n \) with 200, using step-by-step calculations:

• Calculate \(- (200)^2 = -40,000\).
• Compute \( 500 \times 200 = 100,000\).
• Finally, simplify \(-40,000 + 100,000 - 61,500 = -1,500\).

Substitution makes it manageable to evaluate profits accurately for each given scenario.

Profit Calculation

Profit calculation is an essential aspect of financial analysis for businesses, revealing how much money is earned after covering expenses.
The profit function \( P(n) = -n^2 + 500n - 61,500 \) allows us to calculate the profit from selling\( n \) items.

• The term \(-n^2\) suggests that beyond a certain point, increasing the number of items sold might decrease total profit.
• The linear term \( 500n \) represents the income from each item sold.
• Constant term \(-61,500\) accounts for initial costs that are independent of the number of items sold.

Calculating profit also enables businesses to identify optimal sales quantities for maximizing returns.
Knowing how to evaluate profits at different sales levels provides strategic insights into managing resources effectively.