Problem 43 - Beginning and Intermediate Algebra

Question

Solve each problem. Polk Community College wants to construct a rectangular parking lot on land bordered on one side by a highway. It has \(280 \mathrm{ft}\) of fencing that is to be used to fence off the other three sides. What should be the dimensions of the lot if the enclosed area is to be a maximum? What is the maximum area?

Step-by-Step Solution

Answer

Dimensions: 140 feet by 70 feet. Maximum Area: 9800 square feet.

Step 1: Define Variables

Let the length of the lot parallel to the highway be denoted by \(L\), and the width of the lot perpendicular to the highway be denoted by \(W\).

Step 2: Set Up the Perimeter Equation

The total amount of fencing is used for the three sides, which gives us the equation: \[2W + L = 280\]

Step 3: Solve for L

Rearrange the perimeter equation to express \(L\) in terms of \(W\): \[L = 280 - 2W\]

Step 4: Set Up the Area Function

The area \(A\) of the rectangle is given by the product of its length and width, so: \[A = L \times W = (280 - 2W) \times W\]

Step 5: Express the Area Function as a Quadratic Equation

Expanding the expression for area, we get: \[A = 280W - 2W^2\]

Step 6: Find the Maximum Area

To find the maximum area, find the vertex of the quadratic function. Since the quadratic equation is in the form \(-2W^2 + 280W\), use the vertex formula \(W = -\frac{b}{2a}\) where \(a = -2\) and \(b = 280\). Substituting the values, we get: \[W = -\frac{280}{2(-2)} = 70\]

Step 7: Find the Corresponding Length

Using the value of \(W\), substitute back into the equation for \(L\): \[L = 280 - 2(70) = 140\]

Step 8: Calculate the Maximum Area

Using the values of \(L\) and \(W\), multiply them to get the maximum area: \[A = 140 \times 70 = 9800\, \text{square feet}\]

Key Concepts

Quadratic Equations

Understanding quadratic equations is crucial for solving problems involving maximization of areas and other scenarios. A quadratic equation is a second-degree polynomial of the form \[ ax^2 + bx + c = 0 \], where \(a\), \(b\), and \(c\) are constants and \(x\) is the variable. The graph of a quadratic equation is a parabola, which either opens upwards (if \(a > 0\)) or downwards (if \(a < 0\)).

In our parking lot problem, the area function derived from the perimeter equation forms a quadratic equation: \[A = 280W - 2W^2 \]. Understanding how quadratics work helps us find the maximum area by pinpointing the vertex of the parabola. This vertex represents the highest point on the graph when the parabola opens downwards (negative \(a\)).

Key properties of quadratic functions include:

• Roots or solutions where the graph intersects the x-axis (where \(y = 0\))
• The vertex, which gives us the maximum or minimum point
• The direction of the parabola, determined by the sign of \(a\)

A strong foundation in these properties enables solving more complex problems with ease.

Vertex Formula

For quadratic equations, the vertex formula is an essential tool to locate the maximum or minimum point of a parabola. The vertex form for a quadratic expression \[ax^2 + bx + c\] is given by the vertex formula \[x = -\frac{b}{2a}\]. This calculates the x-coordinate of the vertex, pinpointing the extreme point on the parabola.

In the parking lot problem, our area function is \[A = 280W - 2W^2 \] making \(a = -2\) and \(b = 280\). Substituting these into the formula yields: \[W = -\frac{280}{2(-2)} = 70\]. This tells us that the width (W) at the maximum area is 70 feet.

From there, we use the perimeter equation to determine the corresponding length: \[L = 280 - 2(70) = 140\]. Therefore, the dimensions that maximize the area are 70 feet for width and 140 feet for length. Using these dimensions, we calculate the maximum enclosed area: \[A = 140 \times 70 = 9800\, \text{square feet}\].

Understanding and applying the vertex formula simplifies our task of identifying the maximum point in quadratics, proving its utility in various real-world applications.

Perimeter and Area Relationships

Comprehending the relationship between perimeter and area is vital, especially for optimization problems like constructing a rectangular parking lot. The perimeter is the total length of all sides of a polygon, while the area is the enclosed space within the polygon. For a rectangle, the perimeter \(P\) and area \(A\) are connected as follows:

• Perimeter (three sides for our problem): \(2W + L = 280\)
• Area: \(A = L \times W\)

By manipulating these equations, we derived expressions that assisted in finding the desired dimensions. Rewriting the perimeter equation in terms of one variable (\(L = 280 - 2W\)), we substituted into the area formula to get a function only in terms of \(W\). This combination reveals how changes in width and length affect the overall area, guiding us towards maximizing the area.

This approach not only aids in solving specific problems but also reinforces fundamental geometric relationships. By observing how constraints (like a limited amount of fencing) influence dimensions and area, learners gain a deeper appreciation of geometry’s real-world applications.

Mastering these concepts ensures capability in tackling a wide array of mathematical and practical challenges efficiently.