Problem 51 A doorway has the shape of a par... [FREE SOLUTION]

Chapter 2: Problem 51

A doorway has the shape of a parabolic arch and is 9 feet high at the center and 6 feet wide at the base. If a rectangular box 8 feet high must fit through the doorway, what is the maximum width the box can have?

Short Answer

Expert verified

The maximum width of the box can be 2 feet.

Step by step solution

Understand the problem

We need to find out the maximum width of a box that is 8 feet tall, which can fit through a doorway shaped like a parabola. The parabola is 9 feet high at the center and 6 feet wide. This means we must find the maximum width such that the box can pass through the arch without exceeding 8 feet in height.

Construct the mathematical model

The parabolic arch can be modeled by a vertical parabola opening downwards. Assume the vertex is at (0, 9) with the parabola symmetric about the y-axis. Given that the arch is 6 feet wide, the points (-3, 0) and (3, 0) are on the parabola. We want to determine the width of a box at 8 feet high (height of y = 8 in the equation of the parabola).

Write the equation of the parabola

The standard form of a vertical parabola is given by \( y = a(x - h)^2 + k \), where (h, k) is the vertex. Here, the vertex is (0, 9), so \( h = 0 \) and \( k = 9 \), thus: \( y = a(x)^2 + 9 \). Since the parabola passes through point (3, 0), substituting in the equation gives: \( 0 = a(3)^2 + 9 \). Solve for \( a \): \( 9a = -9 \), giving \( a = -1 \). This results in the equation \( y = -x^2 + 9 \).

Solve for the maximum width of the box

We need to find the maximum \( x \) such that \( y = 8 \) since the box is 8 feet tall and the equation is \( y = -x^2 + 9 \). Set \( -x^2 + 9 = 8 \), which simplifies to \( -x^2 = -1 \) or \( x^2 = 1 \). Therefore, \( x = 1 \) or \( x = -1 \). This indicates that the box can be as wide as 2 feet \((2 \times x = 2)\) at 8 feet high when centered under the arch.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Functions

Quadratic functions play a pivotal role in modeling various real-world scenarios, including the unique shape of parabolic arches like the one described in the exercise. A quadratic function is typically represented by the equation \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. This equation forms a parabola, a symmetrical curve, on a graph.

Key characteristics of quadratic functions include:

The presence of a single turning point known as the "vertex," which is the highest or lowest point on the graph depending on whether the parabola opens upwards or downwards.
The parabola is symmetric around a vertical line called the "axis of symmetry" which passes through the vertex.
Zeros, or roots, which are the x-values where the parabola intersects the x-axis.

In the context of a parabolic arch, the quadratic function models the arch's curvature and provides vital information such as height and width at various points.

Vertex Form

The vertex form of a quadratic equation sheds light on the parabolic structure and makes it easy to identify the vertex at a glance. It is expressed as \( y = a(x - h)^2 + k \), where \( (h, k) \) represents the vertex of the parabola.

Advantages of using the vertex form include:

Easy identification of the parabola's vertex \((h, k)\), which is useful for graphing and solving practical problems.
Simplified calculations for transformations, like shifting or stretching the parabola.

In this exercise, the parabolic arch is described by the vertex form with \( h = 0 \) and \( k = 9 \). This indicates a vertex at \((0, 9)\), confirming that the arch reaches its maximum height of 9 feet right at its center.

Solving Quadratic Equations

Solving quadratic equations is essential to find specific points on a parabola, such as where a box can fit through our parabolic arch doorway. Quadratic equations can be solved using several methods:

**Factoring**: Writing the quadratic as a product of binomials and setting each equal to zero to solve for \( x \).
**Completing the Square**: Transforming the equation into a perfect square trinomial, often to easily derive the vertex form.
**Quadratic Formula**: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which provides solutions for any quadratic equation.

In our exercise, solving the equation \( y = -x^2 + 9 \) for \( y = 8 \) used basic algebra to find \( x = 1 \) and \( x = -1 \). These solutions show where the box can fit through, illustrating how understanding and solving quadratic equations help achieve practical outcomes.

91影视

A doorway has the shape of a parabolic arch and is 9 feet high at the center and 6 feet wide at the base. If a rectangular box 8 feet high must fit through the doorway, what is the maximum width the box can have?

Short Answer

Step by step solution

Understand the problem

Construct the mathematical model

Write the equation of the parabola

Solve for the maximum width of the box

Key Concepts

Quadratic Functions

Vertex Form

Solving Quadratic Equations

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