Problem 36 - Basic Technical Mathematics with Calculus

Question

Use completing the square to solve the given problems. A rectangular storage area is \(8.0 \mathrm{m}\) longer than it is wide. If the area is \(28 \mathrm{m}^{2},\) what are its dimensions?

Step-by-Step Solution

Answer

Width: 2.63 m; Length: 10.63 m.

Step 1: Set up equations

Let the width of the rectangle be \( w \) meters. Then, the length of the rectangle is \( w + 8 \) meters. The area of a rectangle is given by the formula \( \text{Area} = \text{length} \times \text{width} \). So, we have the equation \( w(w + 8) = 28 \). Expand this to get \( w^2 + 8w = 28 \).

Step 2: Rearrange equation

Rearrange the equation to prepare for completing the square by moving 28 to the other side: \( w^2 + 8w - 28 = 0 \).

Step 3: Form the perfect square trinomial

To complete the square, take the coefficient of \( w \), which is 8, divide it by 2 to get 4, and then square it to get 16. Add and subtract this 16 in the equation: \( w^2 + 8w + 16 - 16 = 28 \). It becomes \( (w+4)^2 - 16 = 28 \).

Step 4: Simplify the equation

Simplify the equation by adding 16 to 28. This gives \( (w+4)^2 = 44 \).

Step 5: Solve for width using square roots

Take the square root of both sides to solve for \( w + 4 \): \( w + 4 = \pm \sqrt{44} \). Simplifying, \( \sqrt{44} \approx 6.63 \). So, \( w + 4 = 6.63 \) or \( w + 4 = -6.63 \). Since width cannot be negative, we only consider \( w + 4 = 6.63 \).

Step 6: Find the value of w

Subtract 4 from 6.63 to find \( w \): \( w = 6.63 - 4 \). Thus, \( w \approx 2.63 \) meters.

Step 7: Calculate the length

The length is \( w + 8 \). So, calculate it as \( 2.63 + 8 \approx 10.63 \).

Step 8: Conclusion

The dimensions of the rectangle are approximately 2.63 meters for the width and 10.63 meters for the length.

Key Concepts

Understanding Quadratic Equations

Quadratic equations are polynomial equations of degree 2, typically in the form of \( ax^2 + bx + c = 0 \). To fully grasp these equations, it's important to know:

• The variable, often denoted as \( x \), which you solve for.
• The coefficient \( a \), which multiplies \( x^2 \), should not be zero for a quadratic equation.
• The equation can yield two solutions due to its squared term.

In our rectangular area problem, the quadratic equation is \( w^2 + 8w - 28 = 0 \). Here, the width \( w \) takes the place of \( x \) and represents our unknown variable. The terms are arranged such that \( a = 1 \), \( b = 8 \), and \( c = -28 \). Solving this equation allows us to find the possible widths for the rectangle, from which further calculations, such as completing the square, can derive specific dimensions.

Applying Quadratic Equations to Rectangular Area Problems

Rectangular area problems often present themselves where you have to determine missing dimensions based on given area information. The formula for the area of a rectangle is:

• \( \ ext{Area} = \ ext{Length} \times \ ext{Width} \)

In this context, if either the width or length is expressed relative to the other, you can set up a quadratic equation. This is because substituting one dimension in terms of the other introduces a squared term when calculating the area.
For the given problem, the width \( w \) is unknown, while the length is given as \( w + 8 \). This gives the area equation \( w(w + 8) = 28 \), which leads to the quadratic equation \( w^2 + 8w - 28 = 0 \).
In rectangular problems, setting up the right quadratic equations helps provide clear steps towards solving for dimensions that fully satisfy the provided area.

Solving Equations Step by Step: Completing the Square Method

Completing the square is a method used to solve quadratic equations when they are difficult to factor directly or when a more precise solution is needed:

• To complete the square, rearrange the equation to have all terms on one side and zero on the other: \( w^2 + 8w - 28 = 0 \).
• Look at the coefficient of \( w \), which is 8. Divide it by 2 and square the result, \( \left( \frac{8}{2} \right)^2 = 16 \).

Add and subtract this squared number in your equation to maintain balance:
\( w^2 + 8w + 16 - 16 - 28 = 0 \).
Resulting in \( (w + 4)^2 = 44 \).
Next, solve for \( w + 4 \):

• Take the square root of both sides: \( w + 4 = \pm \sqrt{44} \).
• Only consider positive square roots if the context requires non-negative solutions (like positive widths).
• Subtract 4 from the positive root to find \( w \). Hence, \( w = \sqrt{44} - 4 \).

This method allows precise solutions and makes it easier to apply the results to real-world problems.