Problem 82 - Algebra A Combined Function

Question

The length of a rectangle is \(\frac{3}{y-5}\) feet, while its width is \(\frac{2}{y}\) feet. Find its perimeter and then find its area.

Step-by-Step Solution

Answer

Perimeter: \( \frac{10(y-2)}{y(y-5)} \) feet, Area: \( \frac{6}{y(y-5)} \) square feet.

Step 1: Understand the Problem

We are given a rectangle where the length is \(\frac{3}{y-5}\) feet and the width is \(\frac{2}{y}\) feet. We need to calculate the perimeter and the area of this rectangle.

Step 2: Formula for Perimeter

The formula for the perimeter \(P\) of a rectangle is: \[ P = 2 \times (\text{Length} + \text{Width}) \].

Step 3: Substitute Length and Width

Substitute the given length and width into the perimeter formula: \[ P = 2 \times \left( \frac{3}{y-5} + \frac{2}{y} \right) \].

Step 4: Simplify the Perimeter Expression

To simplify \( \frac{3}{y-5} + \frac{2}{y} \), find a common denominator, which is \((y-5) \times y\): \[ \frac{3y}{y(y-5)} + \frac{2(y-5)}{y(y-5)} = \frac{3y + 2y - 10}{y(y-5)} = \frac{5y - 10}{y(y-5)} \].

Step 5: Substitute back into Perimeter Equation

Now place the simplified fraction back into the perimeter formula: \[ P = 2 \times \frac{5y - 10}{y(y-5)} = \frac{10y - 20}{y(y-5)} \].

Step 6: Perimeter Result

Thus, the perimeter of the rectangle is \( \frac{10(y-2)}{y(y-5)} \) feet.

Step 7: Formula for Area

The formula for the area \(A\) of a rectangle is: \[ A = \text{Length} \times \text{Width} \].

Step 8: Substitute Length and Width for Area

Substitute the given length and width into the area formula: \[ A = \frac{3}{y-5} \times \frac{2}{y} \].

Step 9: Simplify the Area Expression

Multiply the expressions: \[ A = \frac{3 \times 2}{(y-5) \times y} = \frac{6}{y(y-5)} \].

Step 10: Area Result

So, the area of the rectangle is \( \frac{6}{y(y-5)} \) square feet.

Key Concepts

rectangle perimeter

To find the perimeter of a rectangle, you need to understand the basic concept of what a perimeter actually is. The perimeter is the total distance around the outside of a shape. For a rectangle, you can calculate it by adding together all the sides.
The formula for finding the perimeter of a rectangle is:

• \( P = 2 \times ( ext{Length} + ext{Width}) \)

This formula is based on the fact that opposite sides of a rectangle are equal in length.
In our exercise, the length is \( \frac{3}{y-5} \) and the width is \( \frac{2}{y} \). By substituting these values into the formula, you can calculate the perimeter. Remember, you often need a common denominator to add fractions effectively. In this case, the common denominator is \( (y-5) \times y \). Once you have the simplified expression, you multiply by 2 to get the full perimeter.
There you have it! The perimeter helps us understand the boundary of the rectangle.

rectangle area

Finding the area of a rectangle involves understanding the number of square units that can cover its surface. The area is found by multiplying the length and the width together.
The basic formula you use for calculating the area of a rectangle is:

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

You substitute the given length and width into this formula. For our rectangle with length \( \frac{3}{y-5} \) and width \( \frac{2}{y} \), the multiplication of fractions proceeds as follows: \( \frac{3}{y-5} \times \frac{2}{y} \), resulting in \( \frac{6}{y(y-5)} \).
The result, \( \frac{6}{y(y-5)} \), tells us how much space is contained within the rectangle. Always express the area in square units, reflective of the two-dimensional space being measured.

rational expressions

Rational expressions are like fractions, but instead of numbers in the numerator and denominator, you have variables and expressions. Understanding rational expressions is key to solving problems involving geometric shapes where dimensions are given in terms of variables.
When dealing with the dimensions of a rectangle in our problem, you notice the expressions \( \frac{3}{y-5} \) and \( \frac{2}{y} \) representing the length and width. Each of these is a rational expression.
The critical step in working with rational expressions is finding a common denominator when you need to add or combine them, like when calculating the perimeter. Additionally, multiplying rational expressions, such as when finding the area, requires multiplying the numerators together and the denominators together.

• Simplify to the least complex form for ease of understanding and further calculations.

Mastering rational expressions involves practice, especially with operations such as addition, subtraction, multiplication, and division.

geometry in algebra

The intersection of geometry and algebra allows us to solve real-world problems, such as finding dimensions and attributes of shapes using algebraic expressions. This concept is often referred to as 'geometry in algebra.'
In the problem we're discussing, geometry in algebra enables us to use algebraic formulas to calculate the perimeter and area of a rectangle given as rational expressions. This application highlights algebra's power in providing exact measurements based on variable expressions.
Understanding this intersection helps bridge procedural knowledge and conceptual understanding—how changes in algebraic expressions affect geometric figures.

• For instance, changing the value of the variable \( y \) alters the measures of the rectangle due to the expressions \( \frac{3}{y-5} \) and \( \frac{2}{y} \).
• This illustrates how algebra can give precise geometrical insights.

Such skills are invaluable in math, engineering, and sciences, where precise calculations are necessary for design and analysis.