Problem 3 - Algebra 2

Question

BAKING The number of cookies produced in a factory each day can be estimated by \(C(w)=-w^{2}+16 w+1000\) , where \(w\) is the number of workers and \(C\) is the number of cookies produced. Divide to find the average number of cookies produced per worker.

Step-by-Step Solution

Answer

The average cookies produced per worker is \( -w + 16 + \frac{1000}{w} \).

Step 1: Understand the Function

The function given is \( C(w) = -w^2 + 16w + 1000 \). This represents the number of cookies produced based on the number of workers \( w \). Our goal is to find the average number of cookies produced per worker.

Step 2: Set up Average Calculation

To find the average number of cookies produced per worker, we divide the total number of cookies \( C(w) \) by the number of workers \( w \). Thus, the expression becomes \( \frac{C(w)}{w} \).

Step 3: Divide by Number of Workers

Substitute the expression for \( C(w) \) into our average formula: \( \frac{-w^2 + 16w + 1000}{w} \). Split the division for each term: \( \frac{-w^2}{w} + \frac{16w}{w} + \frac{1000}{w} \).

Step 4: Simplify Each Term

Simplify each part of the expression: \( \frac{-w^2}{w} = -w \), \( \frac{16w}{w} = 16 \), and \( \frac{1000}{w} \) remains as it is. Thus, the simplified average expression is \( -w + 16 + \frac{1000}{w} \).

Key Concepts

Quadratic Functions

A quadratic function is an equation that can be represented in the form \( f(x) = ax^2 + bx + c \). Such functions graph a parabola, which can either open upwards or downwards depending on the sign of the coefficient \( a \). Let's break this down a bit:

• **Coefficient \( a \)**: This determines the direction of the parabola. If \( a > 0 \), the parabola opens upward, and if \( a < 0 \), it opens downward.
• **Vertex**: The highest or lowest point of the parabola, which can be calculated using the vertex formula \( h = -\frac{b}{2a} \).
• **Axis of symmetry**: A vertical line that passes through the vertex, dividing the parabola into two mirror images.

In the given problem, the quadratic function \( C(w) = -w^2 + 16w + 1000 \) models the relationship between the number of workers \( w \) and the cookies produced. The parabola opens downward, indicating that after a certain number of workers, adding more workers actually results in fewer cookies produced. This happens because of diminishing returns, often a real-world scenario when too many workers start interfering with each other.

Algebraic Expressions

Algebraic expressions consist of variables, coefficients, and constants, and they can be manipulated using various algebraic techniques. Let's consider a few key components:

• **Variables**: Symbols that represent unknown numbers. In our exercise, \( w \) is the variable that signifies the number of workers.
• **Coefficients**: Numbers that multiply the variables, such as \( -1 \) and \( 16 \) in \( C(w) = -w^2 + 16w + 1000 \).
• **Constants**: Numbers without variables, like \( 1000 \) in the equation, which is the base level of cookies produced even when no workers are present.

When dealing with algebraic expressions, you often need to substitute values, simplify terms, or factor expressions. In the solution, simplifying \( \frac{-w^2}{w} + \frac{16w}{w} + \frac{1000}{w} \) involves dividing each term separately, which is an essential skill in evaluating real-world problems algebraically.

Rational Expressions

Rational expressions are ratios of two polynomials. They are expressed in the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \). Working with these requires:

• **Simplifying**: This involves dividing terms in the numerator and denominator by their greatest common factor.
• **Avoiding division by zero**: Since the denominator cannot be zero, it's crucial to identify any values that invalidate the expression, though in our exercise, \( w eq 0 \) must be maintained since we can't have zero workers.

In the example, dividing \( C(w) \) by \( w \) created the rational expression \( \frac{-w^2 + 16w + 1000}{w} \). Each term was divided by \( w \) separately to simplify it to \( -w + 16 + \frac{1000}{w} \), demonstrating how rational expressions can break down complex problems into simpler parts. Mastery of rational expressions is critical for solving a variety of algebraic problems effectively.