Problem 41 - Precalculus Mathematics for Calculus

Question

A farmer has a rectangular garden plot surrounded by \(200 \mathrm{ft}\) of fense. Find the length and width of the garden if its area is \(2400 \mathrm{ft}^{2}\).

Step-by-Step Solution

Answer

The rectangle has dimensions 60 ft by 40 ft.

Step 1: Understand the Problem

We are given that the perimeter of the rectangle is 200 ft and the area is 2400 ft². We need to find both the length and the width of the rectangle. The formulas related to a rectangle are: the perimeter is given by \( P = 2l + 2w \) and the area is given by \( A = l \times w \).

Step 2: Set Up Equations

Using the perimeter formula, \( 2l + 2w = 200 \). Simplifying this gives \( l + w = 100 \). Using the area formula, \( l \times w = 2400 \). We now have two equations: \( l + w = 100 \) and \( l \times w = 2400 \).

Step 3: Solve One Equation for One Variable

From the equation \( l + w = 100 \), solve for \( l \) in terms of \( w \): \( l = 100 - w \).

Step 4: Substitute into the Area Equation

Substitute \( l = 100 - w \) into the area equation: \((100 - w) \times w = 2400\). Expand and rearrange this to form the quadratic equation \( w^2 - 100w + 2400 = 0 \).

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \( w^2 - 100w + 2400 = 0 \) using the quadratic formula \( w = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \), where \( a = 1 \), \( b = -100 \), \( c = 2400 \). Calculate \( w = \frac{{100 \pm \sqrt{{100^2 - 4 \times 1 \times 2400}}}}{2 \times 1} \).

Step 6: Calculate the Solution to the Quadratic

Calculate \( 100^2 - 4 \times 2400 = 10000 - 9600 = 400\). Thus, \( w = \frac{{100 \pm 20}}{2} \). The solutions are \( w = \frac{120}{2} = 60 \) and \( w = \frac{80}{2} = 40 \).

Step 7: Find Both Dimensions of the Rectangle

Using our solutions for \( w \), find \( l \) from the equation \( l = 100 - w \). When \( w = 60 \), \( l = 100 - 60 = 40 \). When \( w = 40 \), \( l = 100 - 40 = 60 \). Thus, the dimensions are \( l = 60 \) ft, \( w = 40 \) ft or vice versa.

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is the total distance around its outer edge. Imagine walking the entire boundary of a garden, and that distance would be the perimeter. It is calculated by adding together the lengths of all the sides. Since a rectangle has opposite sides of equal length, we have a formula:

• The formula for the perimeter is: \[ P = 2l + 2w \]where \( P \) is the perimeter, \( l \) is the length, and \( w \) is the width.
• In simpler terms, it's just doubling the sum of the length and the width.

In the context of the rectangular garden problem, the perimeter is given as \(200 \text{ ft}\). This valuable piece of information helps us form the equation: \[ 2l + 2w = 200 \]Simplifying that, we find \[ l + w = 100 \]. This equation becomes our first stepping stone to figuring out the garden's exact dimensions.

Area of a Rectangle

The area of a rectangle is the amount of space it covers. Visualize a floor you want to cover with tiles; that space represents the area. The formula for the area of a rectangle involves multiplying the length by the width.

• The formula is: \[ A = l \times w \]where \( A \) represents the area, \( l \) is the length, and \( w \) is the width.
• This formula tells you how many square units (e.g., square feet) will fit into the rectangle.

In the problem, we know the garden's area is \(2400 \text{ ft}^2\). Using this, we establish another key equation: \[ l \times w = 2400 \].When combined with the perimeter equation, this helps us navigate to the exact dimensions of the garden. Solving these equations together is crucial for uncovering the length and width.

Quadratic Equation

A quadratic equation is a type of polynomial equation that takes the form \[ ax^2 + bx + c = 0 \].These equations are characterized by their squared term (\(x^2\)). Solving them often involves finding two possible solutions using the quadratic formula. The whole process stems from algebraic concepts involving expansion and rearrangement.

• The quadratic formula used is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• This formula can handle any quadratic equation, providing the roots where the equation equals zero.

For our garden problem, a quadratic equation surfaces when we substitute \( l = 100 - w \) into the area equation \(2400 = l \times w\), resulting in: \[ w^2 - 100w + 2400 = 0 \]. Solving this quadratic equation gives us potential widths of the garden. By applying the quadratic formula, we find that the possible solutions for \( w \) are 60 and 40. These offer two potential dimensions for the garden: either \( l = 60 \text{ ft} \), \( w = 40 \text{ ft} \) or \( l = 40 \text{ ft} \), \( w = 60 \text{ ft} \). Solving quadratic equations is a key skill, especially in scenarios where outcomes aren't straightforward.