Problem 39 - Precalculus Mathematics for Calculus

Question

A rectangular garden is \(10 \mathrm{ft}\) longer than it is wide. Its area is \(875 \mathrm{ft}^{2} .\) What are its dimensions?

Step-by-Step Solution

Answer

The garden's dimensions are 25 ft by 35 ft.

Step 1: Identify Variables

Let's denote the width of the rectangle by \( w \) (in feet). Since the garden is 10 feet longer than its width, its length is \( w + 10 \) feet.

Step 2: Set Up the Area Equation

The area of the rectangle is given by the formula \( ext{length} imes ext{width} = ext{area} \). Since the area is known to be 875 sq. feet, we can write:\[ w(w + 10) = 875\]

Step 3: Expand the Equation

By expanding the equation, we have\[ w^2 + 10w = 875\]

Step 4: Rearrange to Form a Quadratic Equation

Subtract 875 from both sides to set the equation to zero:\[ w^2 + 10w - 875 = 0\]

Step 5: Solve the Quadratic Equation

Use the quadratic formula where \( a = 1 \), \( b = 10 \), \( c = -875 \):\[w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Plug in the values:\[w = \frac{-10 \pm \sqrt{10^2 - 4(1)(-875)}}{2(1)}\] \[w = \frac{-10 \pm \sqrt{100 + 3500}}{2}\]\[w = \frac{-10 \pm \sqrt{3600}}{2}\]\[w = \frac{-10 \pm 60}{2}\]Calculating further gives two potential solutions, \( w = 25 \) or \( w = -35 \). Since a width can't be negative, \( w = 25 \).

Step 6: Determine the Length

The length is 10 feet more than the width. So, if \( w = 25 \), the length \( l = w + 10 = 25 + 10 = 35 \) feet.

Step 7: Verify the Solution

Calculate the area with these dimensions: \( 25 \, \text{ft} \times 35 \, \text{ft} = 875 \, \text{ft}^2 \). The calculations are correct.

Key Concepts

Rectangular Area Problem

A rectangular area problem involves finding the dimensions of a rectangle when given some of its properties, like the area or relationships between the length and width. In this exercise, we need to find the width and the length of a rectangular garden, given that it is 10 feet longer than it is wide, and its total area is 875 square feet. This is a common problem type where relationships between dimensions and area are explored and are usually solved using quadratic equations.

To approach such problems, remember these steps:

• Identify the variable for the width and express the length in terms of this variable.
• Write an equation based on the given area, involving the length and width.
• Use algebraic techniques such as expansion and rearrangement to form a quadratic equation.

Through these steps, you can systematically solve for the unknown dimensions.

The crucial understanding here is translating the word problem into a solvable mathematical equation that denotes the area of the rectangle. Recognizing the relationships and constraints provided in the problem helps in setting up the equation correctly.

Width and Length Problems

Width and length problems often involve finding the exact measurements of a rectangle when given specific relationships between the width and length. In the example, the garden's length is 10 feet longer than its width. By denoting the width as \( w \) and using the relationship, you define the length as \( w + 10 \).

Such statements are crucial as they allow you to express all dimensions in terms of a single variable, simplifying the problem.
Translating geometric properties into mathematical expressions is a key skill here.

• Start with clear definitions using variables for unknown dimensions.
• Express dependent dimensions using these variables, such as writing the length in terms of the width.
• This method makes it easier to substitute into equations that represent other properties, like the area.

Remember, understanding how to relate different dimensions through constraints is what allows you to move forward with the calculations.
Mastery in forming these relationships will make solving these types of problems much more intuitive.

Solving by Quadratic Formula

In problems involving the quadratic formula, such as the width and length of a rectangular garden, the key is to bring the equation to a standard quadratic form: \(ax^2 + bx + c = 0\). This form is essential for applying the quadratic formula to find the value of the unknown variable.
For this problem, we start by setting up the equation \(w(w + 10) = 875\), which simplifies to \(w^2 + 10w = 875\). Subtracting 875 from both sides, it becomes \(w^2 + 10w - 875 = 0\).

The quadratic formula is \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Applying this to our equation, with \(a = 1\), \(b = 10\), and \(c = -875\), allows us to find possible values for \(w\).

• Calculate the discriminant \((b^2 - 4ac)\).
• Find roots using the formula. Choose the one that makes sense in the context, as widths cannot be negative.

Using the quadratic formula efficiently solves for \(w = 25\), giving realistic dimensions.
The ability to employ the quadratic formula is essential in solving area problems like these. It delivers precise solutions when calculating dimensions from given equations.