Question

Sai is ordering new shelving units for his store. Each unit is 7 feet in length and extends from floor to ceiling. The total length of the walls in Sai's store is 119 feet, which includes a length of 21 feet of windows along the walls. If the shelving units cannot be placed in front of the windows, which of the following inequalities includes all possible values of \(r\), the number of shelving units that Sai could use? A) \(r \leq \frac{119-21}{7}\) B) \(r \leq \frac{119+21}{7}\) C) \(r \leq 119-21+7 r\) D) \(r \geq 119+21-7 r\)

Short answer

The correct inequality, which includes all possible values of \(r\), is \(r \leq \frac{119-21}{7}\).

Step-by-step solution

Calculate the available wall length for shelving units

(Subtract the length of the windows from the total wall length) The available wall length for the shelving units will be the total wall length minus the length of windows, which is: \(119 - 21 = 98\) feet.

Determine the maximum number of shelving units that can fit on the available wall length

(Divide the available wall length by the length of each shelving unit) We need to determine the maximum number of shelving units that can fit in the available wall length. Each shelving unit has a length of 7 feet. To find the maximum number of units, divide the available wall length by the length of each shelving unit: \(\frac{98}{7} = 14\)

Find the inequality that includes all possible values of \(r\)

(Compare each given inequality to Step 2 result) Now we need to find which inequality includes all possible values of \(r\). We found the maximum number of shelving units as 14 in Step 2, so the inequality should express that \(r\) cannot exceed this number. Let's analyze each option: A) \(r \leq \frac{119-21}{7}\) \\ After simplification, we get: \(r \leq \frac{98}{7}\) \\ \(r \leq 14\) B) \(r \leq \frac{119+21}{7}\) \\ After simplification, we get: \(r \leq \frac{140}{7}\) \\ \(r \leq 20\) C) \(r \leq 119-21+7 r\) \\ After simplification, we get: \(r \leq 98+7 r\) \\ This inequality is not able to represent the constraint since it will always give values of \(r\) greater than the maximum of 14. D) \(r \geq 119+21-7 r\) \\ After simplification, we get: \(r \geq 140-7 r\) \\ This inequality also does not represent the constraint on the shelving units' number. Out of the given options, Option A represents the inequality that contains all possible values of \(r\) and indicates that the number of shelving units should not exceed the maximum amount of 14. So, the correct answer is: \(r \leq \frac{119-21}{7}\)

Key concepts

Problem Solving

When tackling a problem like Sai's shelving unit dilemma, the first thing to do is to break the problem down into manageable parts. You start by identifying all the given information and restrictions. In this case, we know the total wall length and the length that is unavailable due to windows.

By breaking it down, you allow yourself to see the key pieces of information that directly contribute to the solution. Problem-solving often means thinking logically through each step. Consider each option carefully and apply logical reasoning to determine which choice best fits the problem statement. This approach helps not only in mathematics but in any problem-solving scenario.

Arithmetics

Arithmetics is fundamental in solving inequalities like the one Sai faces. The problem involves simple arithmetic operations such as subtraction and division, used in calculating available spaces and unit lengths. Begin by subtracting the lengths, allowing you to determine how much wall space remains after accounting for windows.

Once subtraction is done, division indicates how many shelving units can fit this space. The key arithmetic operations here help us establish the conditions for the number of shelving units. Arithmetics simplifies relationships and forms the basis of finding the inequalities that define feasible solutions.

Algebra

Algebra plays a significant role in turning the problem conditions into inequalities. Once we know the total available space for shelving, it's crucial to use algebra to derive the inequality that limits the number of shelves. Here, we would translate the maximum allowable shelves into an algebraic inequality:

• Subtract the window length from the total wall length.
• Divide by the shelf length to find the maximum number of shelves.
• Formulate the inequality to represent the relationship \( r \leq \frac{98}{7} \).

Using algebra, we turn word problems into mathematical expressions, making them easier to work with and understand. It allows us to find exact limits like the maximum number of shelving units possible.

Mathematics Education

In mathematics education, concepts such as inequalities offer a rich ground for building skills associated with logical thinking and practical application. This kind of exercise demonstrates the blending of arithmetic operations with algebra to solve real-world problems. It's an opportunity for students to understand not just the how, but the why behind each step.

Educators should encourage students to visualize problems, use diagrams, and relate abstract mathematical concepts to realistic scenarios. By focusing on foundational skills and encouraging critical thinking through problem-solving exercises, students learn to approach complex mathematical challenges with confidence.