Problem 55 - College Algebra and Calculus: An Applied Approach

Question

Candles A candle company sells three types of candles for $$\$ 15$$, $\$ 10$$, and $$\$ 5$$ per unit. In one year, the total revenue for the three products was $$\$ 550,000$$, which corresponded to the sale of 50,000 units. The company sold half as many units of the $$\$ 15$$ candles as units of the $$\$ 10$$ candles. How many units of each type of candle were sold?

Step-by-Step Solution

Answer

The company sold 15,625 units of $15 candles, 31,250 units of $10 candles, and 3,125 units of $5 candles.

Step 1: Define Variables

Let \(x\) be the amount of $15 candles, \(y\) be the amount of $10 candles, and \(z\) be the amount of $5 candles.

Step 2: Create Equations

We know that in one year, the total revenue for the three products was $550,000, which corresponds to: \(15x + 10y + 5z = 550,000\). And there were 50,000 units sold in total, this corresponds to: \(x+y+z=50,000\). Lastly, it is stated that the number of units for the $15 candle is half that of the $10 candle: \(x = 0.5y\).

Step 3: Solve System of Equations with Substitution

Now, take the equation \(x=0.5y\) and substitute \(x\) into the other two equations. The first equation becomes \(7.5y + 10y + 5z = 550,000\) which simplifies to \(17.5y + 5z = 550,000\). The second becomes \(0.5y + y + z = 50,000\), which simplifies to \(1.5y+z=50,000\). Now, subtract the second simplified equation from the first one to find \(y\). So, we get \(16y=500,000\), which gives \(y=31,250\). Substituting \(y\) back into the equation we have \(x=0.5(31,250)=15,625\) and \(z=50,000-(31,250+15,625) = 3,125\).

Step 4: Check your Solution

Replace your calculated \(x\), \(y\), and \(z\) in your original equations to verify your answer.

Key Concepts

Algebraic Equations

When we talk about algebraic equations, we're referring to mathematical statements that assert the equality of two algebraic expressions. These expressions can include numbers, variables (such as x, y, z), and arithmetic operations (like addition and multiplication). In the context of our candle company problem, each equation represents a certain relationship or condition that must be satisfied.

For example, the equation derived from the total revenue, 15x + 10y + 5z = 550,000, indicates that the number of \(15, \)10, and \(5 candles, when multiplied by their respective prices and summed up, should result in the total annual revenue of \)550,000. Understanding how to construct these equations is crucial for translating a word problem into a form that can be solved mathematically.

An algebraic equation effectively serves as a scale balance—both sides need to equal out. So when you solve an equation, you're essentially finding the value(s) for the variable(s) that make the equation true. In many real-life applications, such as the candle company scenario, algebraic equations allow us to solve for unknown quantities that we are interested to find, such as the number of individual candle units sold.

Substitution Method

Substitution Method is a strategy used to solve systems of equations where one equation is manipulated to express one variable in terms of another, which is then substituted into another equation. This is a common tactic when confronted with multiple algebraic equations within a system.

In our candle company example, the substitution method is used to deduce the number of each type of candle sold. Once the equations are set up, we solve one equation for a variable—in this case, x—and then replace that variable with its equivalent expression in the other equations, as shown by x = 0.5y. This technique progressively reduces the number of variables in the system, making it more manageable to solve.

By continuously substituting and simplifying, eventually, you reach a point where one variable can be isolated and solved for directly, as seen in our solution for y. This value can then be substituted back into previous equations to find the remaining unknowns. The method is particularly effective when working with linear relationships because it translates these interconnected conditions into a series of steps that sequentially unlock each variable's value.

Linear Systems

A linear system is a collection of two or more linear equations involving the same set of variables. Our exercise contains a linear system because it involves equations that, when graphed, would each produce a straight line. It is 'linear' because no variable is raised to a power higher than one.

In such systems, we're often interested in finding the point(s) where all the equations intersect, as this represents the solution set—values for variables that satisfy all equations simultaneously. Depending on the system, there could be one solution, no solution, or infinitely many solutions.

For the candle problem, the linear system consists of the equations derived from the contextual information given—total revenue and units sold, plus the relationship between types of candles sold. The goal is to find how many of each type of candle (variables x, y, z) were sold that satisfies all the given conditions at once. This demonstrates the practical utility of algebra in solving complex real-world problems by breaking them down into simpler, solvable units, showcasing how multiple pieces of information can be coordinated through mathematical relationships.