Problem 95 - Algebra and Trigonometry

Question

Fencing a Garden A determined gardener has 120 \(\mathrm{ft}\) of deer-resistant fence. She wants to enclose a rectangular vegetable garden in her backyard, and she wants the area that is enclosed to be at least \(800 \mathrm{ft}^{2} .\) What range of values is possible for the length of her garden?

Step-by-Step Solution

Answer

The length can be between 20 ft and 40 ft.

Step 1: Define Variables

Let \( L \) be the length and \( W \) be the width of the garden. We know the perimeter of the rectangle is \( 2L + 2W = 120 \).

Step 2: Solve for One Variable

Since \( 2L + 2W = 120 \), we can simplify to \( L + W = 60 \). Solve for \( W \) in terms of \( L \): \( W = 60 - L \).

Step 3: Express Area Inequality

The area \( A \) of the rectangle is \( A = L \times W \). Substituting the expression for \( W \), we have \( A = L(60 - L) \). We want \( A \geq 800 \).

Step 4: Set Up the Quadratic Inequality

Substitute into the inequality, \( L(60 - L) \geq 800 \), which simplifies to \( 60L - L^2 \geq 800 \) or rearranging gives us \( -L^2 + 60L - 800 \geq 0 \).

Step 5: Solve the Quadratic Inequality

The quadratic equation to solve is \( L^2 - 60L + 800 = 0 \). Use the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -60, c = 800 \). Substitute these values: \( L = \frac{60 \pm \sqrt{3600 - 3200}}{2} = \frac{60 \pm \sqrt{400}}{2} = \frac{60 \pm 20}{2} \).

Step 6: Calculate the Roots

The roots are \( L = \frac{60 + 20}{2} = 40 \) and \( L = \frac{60 - 20}{2} = 20 \). So the solutions to the equation are \( L = 40 \) and \( L = 20 \).

Step 7: Determine the Range for L

Since the inequality is \( -L^2 + 60L - 800 \geq 0 \), and it opens downward (as the coefficient of \( L^2 \) is negative), the solution is the interval between the roots: \( 20 \leq L \leq 40 \).

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is the total length of all its sides. For a rectangle, two sides are the length and the other two sides are the width. The formula for the perimeter is straightforward:

• Perimeter = 2(Length + Width), or simplified as
• Perimeter = 2L + 2W

Breaking this formula down, we see that you simply double the sum of the length and width. It helps us determine how much material is needed to surround a rectangular shape completely. For example, if you’re planning to fence a garden and know the total perimeter you want, you can use this formula to find possible length and width combinations. In the given problem, the gardener works with a perimeter of 120 ft, leading us to the equation:
\( 2L + 2W = 120 \). By dividing this equation by 2, we simplify it to \( L + W = 60 \). Understanding this setup is crucial for determining the individual dimensions when the perimeter is known.

Area of a Rectangle

The area of a rectangle tells us how much space is enclosed within. To find the area, you simply multiply the length by the width:

• Area = Length × Width, or simplified as
• Area = \( L \times W \)

This formula is helpful when you need to know the size of a surface, such as a garden or a room. In the context of the original exercise, the gardener wanted the area to be at least 800 square feet.
Using our simplified expression where width \( W \) is substituted by \( 60 - L \) from the perimeter equation, the area becomes \( L(60 - L) \). Setting up an inequality \( L(60 - L) \geq 800 \) helps to find suitable dimensions that meet the area requirement. It results in a quadratic inequality that needs solving to determine valid length values.

Quadratic Formula

When dealing with quadratic equations, especially those derived from expressions involving area or perimeter, the quadratic formula is a powerful tool. A quadratic equation typically looks like this:

• \( ax^2 + bx + c = 0 \)

The quadratic formula to find the roots of this equation is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula includes three components:

• \( b^2 - 4ac \) is called the discriminant; it determines the number and type of solutions.
• The square root component \( \sqrt{b^2 - 4ac} \) provides the variations needed to calculate both roots.

In the exercise, the gardener used it to solve \( L^2 - 60L + 800 = 0 \), a quadratic equation arising from the area inequality. After substituting the known values (\( a = 1, b = -60, c = 800 \)), the formula simplified to:
\[ L = \frac{60 \pm 20}{2} \]Yields roots of 40 and 20, determining that the length of the garden should be between these values to satisfy both the perimeter and area requirements.