Problem 68 - Intermediate Algebra

Question

In Company \(\mathrm{S}\), the profit function for selling \(n\) items is given by \(P(n)=-n^{2}+500 n-61,500\). Compute \(P(200), P(230), P(250)\), and \(P(260)\).

Step-by-Step Solution

Answer

The profits are -1500, 600, 1000, and 900 for 200, 230, 250, and 260 items, respectively.

Step 1: Understand the Profit Function

The profit function given for Company \(S\) is \(P(n) = -n^2 + 500n - 61,500\). This is a quadratic equation where \(-n^2\) affects the shape of the parabola, \(500n\) is the linear term, and \(-61,500\) is the constant term. The variable \(n\) represents the number of items sold.

Step 2: Compute P(200)

Substitute \(n = 200\) into the profit function:\[ P(200) = -(200)^2 + 500 \cdot 200 - 61,500 \]Calculate each term:- \( -(200)^2 = -40,000 \)- \( 500 \times 200 = 100,000 \)Combine and simplify:\[ P(200) = -40,000 + 100,000 - 61,500 = -1,500 \]

Step 3: Compute P(230)

Substitute \(n = 230\) into the profit function:\[ P(230) = -(230)^2 + 500 \cdot 230 - 61,500 \]Calculate each term:- \( -(230)^2 = -52,900 \)- \( 500 \times 230 = 115,000 \)Combine and simplify:\[ P(230) = -52,900 + 115,000 - 61,500 = 600 \]

Step 4: Compute P(250)

Substitute \(n = 250\) into the profit function:\[ P(250) = -(250)^2 + 500 \cdot 250 - 61,500 \]Calculate each term:- \( -(250)^2 = -62,500 \)- \( 500 \times 250 = 125,000 \)Combine and simplify:\[ P(250) = -62,500 + 125,000 - 61,500 = 1,000 \]

Step 5: Compute P(260)

Substitute \(n = 260\) into the profit function:\[ P(260) = -(260)^2 + 500 \cdot 260 - 61,500 \]Calculate each term:- \( -(260)^2 = -67,600 \)- \( 500 \times 260 = 130,000 \)Combine and simplify:\[ P(260) = -67,600 + 130,000 - 61,500 = 900 \]

Key Concepts

Understanding the Profit Function

In the realm of business mathematics, a profit function is essential for understanding how the number of products sold affects a company's profitability. For Company \( S \), the given profit function is \( P(n) = -n^2 + 500n - 61,500 \). This function tells us the profit \( P \) that results from selling \( n \) items.

A few points to note:

• The term \( -n^2 \) suggests a quadratic relationship, which forms a "downward-opening" parabola. This implies that there is a peak profit point after which selling more items results in decreased profits.
• The \( 500n \) term means that initially, as sales increase, so does the profit due to the positive sign.
• The constant \( -61,500 \) represents fixed costs or other baseline expenses incurred by the company that need to be covered before making any profit.

By computing specific values like \( P(200), P(230), P(250), \) and \( P(260) \), you can find out exactly how many items must be sold to maximize profits without incurring losses due to overproduction.

Using the Substitution Method in Quadratic Equations

The substitution method is a powerful tool frequently used when working with equations, including those of polynomial nature like the profit function from above. In simple terms, this method involves replacing variable terms with specific values to find the result.

Here's how it works:

• You begin with the original equation, \( P(n) = -n^2 + 500n - 61,500 \).
• To find \( P(200) \), substitute \( n \) with \( 200 \) and calculate each term: \(-(200)^2 + 500 \times 200 - 61,500 \).

By repeating this process for \( P(230), P(250), \) and \( P(260) \), you can quickly determine the profit for each respective number of items sold. This method is straightforward and highly effective for evaluating polynomial functions at different points.

Exploring Polynomial Functions

Polynomial functions are mathematical expressions comprised of variables, coefficients, and exponents. In our context, the profit function \( P(n) = -n^2 + 500n - 61,500 \) is a quadratic polynomial, specifically of degree 2, because the highest exponent of \( n \) is 2.

Key features of polynomial functions include:

• The degree of a polynomial determines its general shape; degree 2 forms a parabola.
• Coefficients impact the steepness and direction of the graph's curve.
• Polynomial functions can model various real-life scenarios, such as profit dynamics, as seen here.

By understanding the structure of polynomial functions, students can better interpret how changes in variables affect outcomes. For example, adjusting the coefficients in the profit function could make the profit curve steeper or more shallow, altering the optimal number of items to be sold for maximum profit.