Problem 64 - College Algebra and Calculus: An Applied Approach

Question

A rectangular room with a perimeter of 50 feet is to have an area of at least 120 square feet. Within what bounds must the length be?

Step-by-Step Solution

Answer

The length of the rectangle must be between 10 and 15 feet.

Step 1: Establish the relationship between length and width

The formula for the perimeter of a rectangle is \(P = 2l + 2w\), with l representing the length and w the width of the rectangle. In this task, the perimeter is given as 50 feet. This can be solved to find the width in terms of length: \(w = (P - 2l)/2 = 25 - l\).

Step 2: Write the equation for the area

The area of a rectangle is given by \(A = lw\). Substitute the expression for W in this formula to get the equation for the area as a function of L: \(A(l) = l(25 - l)\).

Step 3: Determine the bounds for the length

The area must be at least 120 sq ft. Setting \(A(l)\geq120\) and solving the inequality for \(l\), we find that \(0 < l < 15\) and \(10 < l < 25\). However, we need to find the intersection of these ranges to ensure both the perimeter and area conditions are met. The intersection is \(10 < l < 15\).

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle measures the total distance around its boundary. To calculate the perimeter, you simply add up the lengths of all four sides. Since a rectangle has two lengths and two widths, you can use the formula:

• \( P = 2l + 2w \)

In this formula, \( l \) represents the length while \( w \) represents the width. For instance, if a rectangular room’s perimeter is given as 50 feet, you know that:

• \( 2l + 2w = 50 \)

This equation can be solved to express one variable in terms of the other. Typically, if you know the perimeter, you can rearrange the formula to express the width as a function of the length, like so:

• \( w = 25 - l \)

By understanding this relationship, you can explore different combinations of length and width that satisfy the given perimeter of the rectangle. This is particularly useful in optimization problems where maintaining certain constraints is essential.

Area of a Rectangle

The area of a rectangle indicates the amount of space enclosed within its borders. You calculate this by multiplying its length and width. The formula is represented as:

• \( A = lw \)

From the formula, you can substitute the width \( w \) with an expression you derive from the perimeter equation, as we did earlier \( w = 25 - l \). Thus, the area formula can be rewritten in terms of length:

• \( A(l) = l(25 - l) \)

This equation helps to determine how variations in length affect the area. In the given problem, we need the area to be at least 120 square feet. By substituting the expression for width, we derive a new focus: solving the inequality \( l(25 - l) \geq 120 \). Solving this inequality assists in restricting length to feasible values that satisfy both perimeter and area requirements.

Inequalities in Algebra

Inequalities in algebra allow us to explore ranges of values that satisfy certain conditions rather than pinpointing exact numbers. They are fundamental when finding bounds, like in our optimization problem with a rectangular room. In this context, we have:

• \( l(25 - l) \geq 120 \)

An inequality shows a broad spectrum of solutions rather than a single answer. To solve these inequalities, such as \( l(25 - l) \geq 120 \), we rearrange and simplify the expression:

• Rewriting gives: \( 25l - l^2 \geq 120 \)
• Rearranging: \( l^2 - 25l + 120 \leq 0 \)

Factoring or using a quadratic formula can solve such problems, finding critical points that divide our number line into regions that we test. Solving inequality provides meaningful bounds like \( 10 < l < 15 \), ensuring both the area requirement and perimeter conditions are met. In many optimization problems, accurately working through these inequalities is crucial to identify feasible solutions that adhere to multiple constraints.