Problem 35 - Elementary Algebra

Question

At a store, the daily number of sales \(S\) is approximately related to the number of employees \(E\) by \(S=100+15 \sqrt{E+6}\) (a) Determine the approximate number of sales if there are 19 employees. (b) Determine the number of employees the store would need to produce 310 sales.

Step-by-Step Solution

Answer

(b) How many employees are required to produce 310 sales? Answer: (a) The approximate number of sales when there are 19 employees is 175. (b) The store needs 190 employees to produce 310 sales.

Step 1: (a) Calculate the sales with 19 employees

Given that E = 19, we will substitute this value in the equation: \(S=100+15\sqrt{E+6}\). \(S = 100 + 15\sqrt{19 + 6}\) \(S = 100 + 15\sqrt{25}\) Now, calculate the square root of 25 and multiply by 15: \(S = 100 + 15 \times 5\) \(S = 100 + 75\) Thus, the approximate number of sales when there are 19 employees is: \(S = 175\)

Step 2: (b) Calculate employees needed for 310 sales

To find the number of employees required to produce 310 sales, we will set S = 310 and solve for E in the equation: \(S=100+15\sqrt{E+6}\) \(310 = 100 + 15\sqrt{E + 6}\) Now, subtract 100 from both sides of the equation: \(210 = 15\sqrt{E + 6}\) Divide both sides by 15: \(14 = \sqrt{E + 6}\) To remove the square root, square both sides of the equation: \(196 = E + 6\) Now, subtract 6 from both sides to solve for E: \(E = 196 - 6\) Therefore, the number of employees needed to produce 310 sales is: \(E = 190\)

Key Concepts

Problem Solving

When tackling a problem in algebra, such as finding the relationship between variables like sales and employees, we use methods that help us isolate and solve for unknowns. The key is breaking down the problem into manageable steps.

• Begin by understanding the question: What are you solving for?
• Identify the known quantities or given data.
• Use mathematical operations to manipulate the equation and isolate the unknown variable.

Start by substituting given or known values into the equation, as seen in part (a) of our exercise. This allows you to compute the desired results directly from the formula.
For part (b), reverse the process. Given a result, work backwards to find the initial condition. Problem solving in algebra often involves such back-and-forth steps using both forward substitution and reverse engineering through the equation.

Square Roots

Square roots are a fundamental concept in algebra, especially when dealing with equations involving squared numbers or variables under square root signs. A square root essentially asks what number, when multiplied by itself, gives the original number.

• For example, the square root of 25 is 5, because 5 × 5 equals 25.
• To calculate square roots, either use arithmetic knowledge of perfect squares, or a calculator for non-perfect squares.

In our exercise, the daily sales equation "\(S = 100 + 15 \sqrt{E + 6}\)" involved simplifying a square root. We needed to find the square root of \(E + 6\). This simplification is pivotal, as it affects the whole equation. When solving for unknowns, especially in equations with square roots, accurately calculate the square roots to avoid errors in later steps.

Algebraic Equations

Algebraic equations are the backbone of analyzing relationships between quantities. They consist of variables, constants, and usually an '=' sign indicating equality of two expressions.

• Start by identifying the variables and constants in the equation.
• Understand that the goal is often to solve for one variable in terms of others.
• Manipulate the equation using algebraic operations: addition, subtraction, multiplication, division, and naturally, usage of roots and powers.

In our context, the equation "\(S = 100 + 15 \sqrt{E+6}\)" describes how sales \(S\) are related to employees \(E\). To solve part (b), we set \(S\) to a desired number and used algebraic manipulation to isolate \(E\). This involved getting rid of the constant and dividing by coefficients before squaring both sides to eliminate the square root. Understanding these operations is crucial for solving more complex algebraic equations efficiently.