Problem 48 - Basic Technical Mathematics with Calculus

Question

Solve the given problems. All numbers are accurate to at least two significant digits. A homeowner wants to build a rectangular patio with an area of \(20.0 \mathrm{m}^{2},\) such that the length is \(2.0 \mathrm{m}\) more than the width. What should the dimensions be?

Step-by-Step Solution

Answer

The dimensions of the patio are approximately 3.59 m by 5.59 m.

Step 1: Define Variables

Let the width of the patio be \( w \) meters. Since the length is \( 2.0 \;\mathrm{m} \) more than the width, we can express the length as \( w + 2.0 \;\mathrm{m} \).

Step 2: Set Up the Area Equation

The area of a rectangle is given by multiplying its length by its width. Therefore, the equation for the area is \( (w)(w + 2.0) = 20.0 \;\mathrm{m}^{2} \).

Step 3: Expand and Rearrange the Quadratic Equation

Expand the equation: \( w^2 + 2.0w = 20.0 \;\mathrm{m}^2 \). Rearrange it to form a standard quadratic equation: \( w^2 + 2.0w - 20.0 = 0 \).

Step 4: Use the Quadratic Formula

The quadratic formula is \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For our equation, \( a = 1 \), \( b = 2.0 \), and \( c = -20.0 \). Substitute these values into the formula.

Step 5: Calculate the Discriminant

Compute the discriminant \( b^2 - 4ac \): \( (2.0)^2 - 4(1)(-20.0) = 4 + 80 = 84 \).

Step 6: Solve for Width \(w\)

Substitute the discriminant into the quadratic formula: \( w = \frac{-2.0 \pm \sqrt{84}}{2} \). Calculate \( \sqrt{84} \approx 9.17 \). Thus, \( w = \frac{-2.0 \pm 9.17}{2} \).

Step 7: Determine Possible Solutions and Choose

Calculate the possible solutions: \( w = \frac{7.17}{2} \approx 3.585 \;\mathrm{m} \) and \( w = \frac{-11.17}{2} \) (not plausible as width can't be negative).

Step 8: Find the Length from Width

Using the width \( w = 3.585 \;\mathrm{m} \), calculate the length: \( w + 2.0 = 3.585 + 2.0 = 5.585 \;\mathrm{m} \).

Step 9: Verify the Calculation

Verify the area: \( 3.585 \times 5.585 \approx 20.0 \;\mathrm{m}^2 \), confirming our solution is correct.

Key Concepts

Area of a Rectangle

To understand the problem, we need to start with the basics of calculating the area of a rectangle. The area is the space inside the rectangle. It is calculated by multiplying the length by the width. So, the area formula is:

• Area = length × width

In our problem, the homeowner wants a patio with an area of 20.0 square meters. We know that one side (the length) is 2.0 meters longer than the other side (the width). Letting the width be represented by the variable \( w \), makes it easier to find the dimensions by setting up an equation:

• Area = \( w \times (w + 2.0) \) = 20.0

Recognizing how each dimension contributes is key to understanding any adjustments or changes you might make to either value when designing or constructing your own projects.

Solving Quadratics

After setting up the equation \( w(w + 2.0) = 20.0 \), it becomes crucial to know how to solve quadratic equations. A quadratic equation typically takes the standard form:

• \( ax^2 + bx + c = 0 \)

Our equation is rearranged to match this form as \( w^2 + 2.0w - 20.0 = 0 \). Solving quadratics often involves finding two solutions for the variable, known as roots. These can be real and positive, real and negative, or even complex numbers. However, when we talk about practical problems, like here with lengths of a patio, negative roots don't make sense. To solve a quadratic:

• Combine like terms to simplify it as much as possible.
• Consider factoring as a method if it is an easy expression.
• If typical factoring does not work, use methods like completing the square or the quadratic formula.

Observing how these methods are applied can make tackling future similar equations less daunting!

Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of any quadratic equation. This formula is written as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a \), \( b \), and \( c \) are constants from your equation \( ax^2 + bx + c = 0 \). In our problem, \( a = 1 \), \( b = 2.0 \), and \( c = -20.0 \).Using the formula involves calculating the discriminant \( b^2 - 4ac \), which determines the nature of the roots:

• If positive, two distinct real solutions exist.
• If zero, exactly one real solution exists.
• If negative, real solutions do not exist (roots are complex).

In the exercise, the discriminant is 84, a positive number, leading to two real solutions. This tells us that a positive width is feasible. By substituting the discriminant into the formula, the two potential widths were calculated, but only the positive width made sense here. This clear step-by-step clarifies how lengths are determined in a practical scenario using math.