Problem 112 - Algebra and Trigonometry

Question

Find the dimensions of a rectangle whose perimeter is 22 feet and whose area is 24 square feet. (Section \(8.4,\) Example 5 )

Step-by-Step Solution

Answer

The dimensions of the rectangle are 3 feet by 8 feet.

Step 1: Set up equations

Using the given information, set up the system of equations. We get the following two equations: \(2*x + 2*y = 22\) and \(x*y = 24\). The first equation comes from the formula for the perimeter of a rectangle and the second equation comes from the formula for the area of a rectangle.

Step 2: Simplify Equations

Simplify the equations to make them easier to solve. The first equation \(2*x + 2*y = 22\) can be divided by 2 to simplify to \(x + y = 11\). Leave the second equation \(x*y = 24\) as is.

Step 3: Solve for One Variable

We can solve the equation \(x + y = 11\) for one variable to make it easier to substitute into the second equation. Let's solve for x to get \(x = 11 - y\).

Step 4: Substitute

Substitute \(x = 11 - y\) into the equation \(x*y = 24\) to solve for y. Substituting gives us \((11-y)*y = 24\).

Step 5: Solve for y

Solve for y in the equation \((11-y)*y = 24\). This is a quadratic equation \(y^2 - 11*y + 24 = 0\). The solutions for y are 8 and 3.

Step 6: Solve for x

We can substitute y = 8 and y = 3 back into the equation \(x = 11 - y\) to find the corresponding x values. When y = 8, x = 3 and when y = 3, x = 8.

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is the total distance around the outside of the rectangle. To calculate the perimeter, you add up the lengths of all four sides. For a rectangle with length x and width y, the formula is:
\[ Perimeter = 2x + 2y \]
In our example, with a perimeter of 22 feet, this mathematical relationship helps us establish the first equation of the system. When a student is tasked with finding a rectangle's dimensions, understanding the perimeter's importance allows for framing one part of the problem and further working towards a solution.

Area of a Rectangle

The area of a rectangle measures the space enclosed within its four sides. For a rectangle, the area can be found by multiplying the length and width. Mathematically, the formula looks like this:
\[ Area = x \times y \]
In our problem, the given area is 24 square feet which, when combined with the formula, yields the second equation of our system. Discussing rectangle areas in real-life context, such as flooring or painting walls, can aid students in visualizing the practical applications of this concept.

Substitution Method

The substitution method is a technique for solving systems of equations wherein one variable is expressed in terms of the other and then substituted into the other equation. By transforming either the equation for perimeter or area, we can substitute to find a solvable equation for the remaining variable.
For example, if we have the equation \(x + y = 11\) and we solve for \(x\), we can express \(x\) as \(x = 11 - y\). This expression can then replace every \(x\) in the area equation, leading us to a quadratic equation that we can solve. Teaching students to recognize instances where substitution is the best approach is vital for problem-solving efficiently.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in a single variable \(x\) with the standard form:
\[ ax^2 + bx + c = 0 \]
In this exercise, after substituting for \(x\), we get \( y^2 - 11y + 24 = 0 \) which is a quadratic equation in terms of \(y\). Solving quadratic equations is fundamental in algebra and can be tackled through factoring, completing the square, or using the quadratic formula. Here, factoring is the suitable method, leading to the values of \(y\) that, when substituted back into the rectangle's dimension equations, ultimately help us find the solution to the original problem.