Problem 72 - Elementary Algebra

Question

The area of a certain rectangular pen is given by the formula \(A=14 w-w_{2},\) where \(w\) represents the width in feet. Determine the width that produces the maximum area.

Step-by-Step Solution

Answer

The width that produces the maximum area is 7 feet.

Step 1: Identify the Problem

We are given an area function \( A = 14w - w^2 \). The goal is to find the value of \( w \) that maximizes \( A \). This is a maximization problem.

Step 2: Understanding the Quadratic Function

The function \( A = 14w - w^2 \) is a quadratic function in the standard form \( A = aw^2 + bw + c \), where \( a = -1 \) and \( b = 14 \). It opens downwards because \( a < 0 \). Therefore, it has a maximum point.

Step 3: Use the Formula for Maximum of a Quadratic Function

The maximum value of a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). For our function, \( a = -1 \) and \( b = 14 \), so substitute these values into the formula: \[ w = -\frac{14}{2(-1)} = \frac{14}{2} = 7. \]

Step 4: Conclusion

The width \( w \) that produces the maximum area is 7 feet.

Key Concepts

Quadratic Functions

Quadratic functions are fundamental components in algebra, often appearing in the form \( f(x) = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable.

These functions graph as parabolas. The shape and orientation depend on the value of \( a \):

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), it opens downwards.

The highest or lowest point of a parabola is called the vertex, which plays a crucial role in optimization problems involving quadratics. In the given exercise, the function \( A = 14w - w^2 \) is quadratic, expressed in terms of width \( w \), showing the relationship between width and area. Here, \( a = -1 \) so the parabola opens downwards, indicating a maximum point that can be determined mathematically.

Quadratic functions are used to model various real-world scenarios, making them a critical concept for problem-solving.

Maximization Problems

Maximization problems are situations where we need to find the largest possible value for a particular function. These are prevalent in fields like economics, engineering, and in our case, geometry.

For quadratic functions that open downwards (like \( a < 0 \)), the maximum value occurs at the vertex of the parabola. This is precisely where the peak of the curve is located.

In our exercise, the area \( A = 14w - w^2 \) represents a parabola opening downwards because the coefficient of \( w^2 \) is negative. This suggests that the function has a maximum area that can be found by determining the vertex of the parabola.

• Identifying the vertex helps us solve such problems efficiently.
• The vertex provides the optimal width \( w \) for maximum area given the parameters.

Such problems enhance understanding of how mathematical principles apply directly to practical needs, like maximizing areas with specific conditions.

Vertex of a Parabola

The vertex of a parabola is an essential feature for understanding the behavior of quadratic functions. It is the point where the function achieves its maximum or minimum value.

There is a simple formula that helps find this crucial point: for a quadratic function \( ax^2 + bx + c \), the vertex occurs at \( x = -\frac{b}{2a} \).

• This formula derives from completing the square or using calculus to find critical points.
• It provides a quick way to determine the vertex without graphing.

In the provided problem, to find where \( A = 14w - w^2 \) reaches its maximum, we calculate the vertex using \( w = -\frac{14}{2(-1)} \).

Solving this, we get \( w = 7 \). Simplifying optimization becomes straightforward with this kind of tool, giving us insights into the optimal dimensions needed for the area maximization problem at hand.