Problem 63 - College Algebra

Question

Describe a number of business ventures. For each exercise, a. Write the cost function, \(C\). b. Write the revenue function, \(R\). c. Determine the break-even point. Describe what this means. You invest in a new play. The cost includes an overhead of \(\$ 30,000,\) plus production costs of \(\$ 2500\) per performance. A sold-out performance brings in \(\$ 3125 .\) ( In solving this exercise, let \(x\) represent the number of sold-out performances.)

Step-by-Step Solution

Answer

cost function: \(C(x) = \$30,000 + \$2,500x\), revenue function: \(R(x) = \$3,125x\), break-even point: 48 performances.

Step 1: Determining the cost function C

The cost includes a fixed overhead of \(\$30,000\) plus a variable cost of \(\$2,500\) per performance. Therefore, the cost function is defined as \(C(x) = \$30,000 + \$2,500x\), where \(x\) represents the number of sold-out performances.

Step 2: Determining the revenue function R

The revenue is solely based on the number of sold-out performances, where each performance brings in \(\$3,125\). Therefore, the revenue function is defined as \(R(x) = \$3,125x\).

Step 3: Determining the break-even point

The break-even point occurs when the cost equals the revenue, or \(C(x) = R(x)\). So, we set \(C(x) = R(x)\) and we solve for \(x\): \[ \$30,000 + \$2,500x = \$3,125x \] \[ \$625x = \$30,000 \] \[ x = \$30,000 / \$625 \] \[ x = 48 \] Thus, the break-even point is at 48 sold-out performances.

Key Concepts

Cost Function

In algebraic terms, the cost function symbolically represents the total cost of production as a function of the number of goods produced or services provided. It typically consists of fixed costs, such as rent and salaries, which do not change with the production level, and variable costs, which fluctuate with the quantity produced.

For the given business venture of investing in a new play, the cost function can be formulated as follows:
\[\begin{equation}C(x) = Fixed\underline{\phantom{xxx}}Costs + (Variable\underline{\phantom{xxx}}Cost\underline{\phantom{xxx}}per\underline{\phantom{xxx}}Performance \times Number\underline{\phantom{xxx}}of\underline{\phantom{xxx}}Performances)\underline{\phantom{xxx}}= 30000 + 2500x\underline{\phantom{xxx}}\end{equation}\]
This equation makes it clear that for each additional performance, the cost increases linearly by the amount of the variable cost.

Revenue Function

The revenue function is a critical concept in business and economics, indicating the total revenue generated by the sale of goods or services. It depends on the number of units sold and the price per unit.

In the context of the new play, where revenue is generated per sold-out performance, the revenue function is articulated as:
\[\begin{equation}R(x) = Revenue\underline{\phantom{xxx}}Per\underline{\phantom{xxx}}Performance \times Number\underline{\phantom{xxx}}of\underline{\phantom{xxx}}Performances = 3125x\underline{\phantom{xxx}}\end{equation}\]
This conveys that the revenue has a direct relationship with the number of performances, and, therefore, it's also a linear function of the variable 'x'.

Algebraic Problem-Solving

Identifying the Break-Even Point

One of the most common applications of algebraic problem-solving in business is finding the break-even point. This is the exact point where total costs and total revenue are equal, meaning there is no net loss or gain in the venture.

To determine this pivotal moment for the play, we equate the cost and revenue functions:
\[\begin{equation}C(x) = R(x)\end{equation}\]After setting up this equation and solving for 'x', we find the break-even point, which informs the business owner of the minimum number of sold-out performances needed to cover all costs.

Linear Functions

Linear functions are the simplest type of algebraic function, graphically represented by a straight line. They have constant rates of change, which makes them predictable and easy to analyze. Both cost and revenue functions, in the presented scenario, exemplify linear functions because they increase by a fixed amount with each additional unit of 'x'.

Understanding the properties of linear functions is fundamental when interpreting business models, as they allow for clear visualization of how costs and revenues change with production or sales levels.