Problem 36 - A Graphical Approach to Precalculus with Limits

Question

do each of the following. (a) Express the cost \(C\) as a function of \(x,\) where \(x\) represents the number of items as described. (b) Express the revenue \(R\) as a function of \(x .\) (c) Determine analytically the value of \(x\) for which revenue equals cost. (d) Graph \(y_{1}=C(x)\) and \(y_{2}=R(x)\) on the same \(x y\) -axes and interpret the graphs. Baking and Selling Cakes A baker makes cakes and sells them at county fairs. Her initial cost for the Pointe Coupee parish fair was \(\$ 40.00 .\) She figures that each cake costs \(\$ 2.50\) to make, and she charges \(\$ 6.50\) per cake. Let \(x\) represent the number of cakes sold. (Assume that there were no cakes left over.)

Step-by-Step Solution

Answer

Break-even point is at 10 cakes sold. Revenue equals cost at this point.

Step 1: Express Cost Function

The total cost to make the cakes includes a fixed initial cost and a variable cost that depends on the number of cakes made. Here, the initial cost is \(40, and the cost to make each cake is \)2.50. Thus, the cost function \( C(x) \) can be expressed as: \[C(x) = 40 + 2.5x\] where \( x \) is the number of cakes sold.

Step 2: Express Revenue Function

The revenue earned from selling cakes is determined by the price per cake and the number of cakes sold. The baker charges $6.50 per cake. Therefore, the revenue function \( R(x) \) is: \[R(x) = 6.5x\] where \( x \) is the number of cakes sold.

Step 3: Determine Break-Even Point

To find when revenue equals cost, set the cost function equal to the revenue function and solve for \( x \):\[C(x) = R(x) \Rightarrow 40 + 2.5x = 6.5x\]Simplifying gives:\[40 = 4x\]Dividing both sides by 4 gives:\[x = 10\]Therefore, revenue equals cost when 10 cakes are sold.

Step 4: Graph Functions and Interpret

Graph \( y_1 = C(x) = 40 + 2.5x \) and \( y_2 = R(x) = 6.5x \) on the same axes. The line \( y_1 \) begins at 40 on the \( y \)-axis and rises with a slope of 2.5. The line \( y_2 \) passes through the origin with a slope of 6.5. The intersection point of these lines represents the break-even point, where revenues equal costs, which occurs at \( x = 10 \). Beyond this point, revenue exceeds costs, indicating profit.

Key Concepts

Cost Function

The cost function is a crucial concept in break-even analysis. It represents the total expenses incurred in producing a product or service. In our example, the baker's cost function encompasses both fixed and variable costs. The fixed cost, in this situation, is \(40, which includes initial expenses that do not change with the number of cakes produced. Meanwhile, the variable cost is associated with each unit produced — in this case, it's \)2.50 per cake.
This gives us the cost function formula:

• The fixed cost: \(40
• The cost per unit (cake): \)2.50

Thus, the complete cost function, represented by C(x), is expressed as: \[ C(x) = 40 + 2.5x \] where \( x \) represents the number of cakes sold. This function allows us to calculate the total cost for any number of cakes, making it integral to understanding the baker's expenses.

Revenue Function

Revenue function is equally vital in break-even analysis. It calculates the income generated from selling a particular product. In the scenario provided, the baker sells each cake for \(6.50. The revenue function therefore depends on the number of cakes sold, which we denote by \( x \).
Here is how we construct the revenue function:

• Selling price per cake: \)6.50

Hence, the revenue function, \( R(x) \), is given by: \[ R(x) = 6.5x \] This revenue function helps in determining how much money is earned based on the number of cakes sold. Understanding this supports predicting earnings and plays a key role when seeking to determine profitability.

Graphing Linear Functions

Graphing linear functions visually demonstrates how variables interact with each other. In our case, the cake cost and revenue are linear functions of \( x \), the number of cakes sold.

• The cost function \( y_1 = C(x) = 40 + 2.5x \) starts at 40 on the \( y \)-axis and ascends with a slope of 2.5.
• The revenue function \( y_2 = R(x) = 6.5x \) starts from the origin and has a steeper slope of 6.5.

Graphing these functions on the same axes illustrates their relationship. The point where the two lines intersect is particularly critical as it marks the break-even point — where total revenue equals total cost. Beyond this intersection, the revenue line sits above the cost line, indicating profit generation. Visual graphs like these powerfully depict how changes in \( x \) influence both cost and revenue.

Analytical Solutions

Analytical solutions refer to solving equations or problems logically through mathematical principles, without relying on numerical methods exclusively. To determine the break-even point analytically, we set the cost function equal to the revenue function and solve for \( x \).
The solution begins with the equation: \[ C(x) = R(x) \] Substituting the given functions, we have: \[ 40 + 2.5x = 6.5x \] By isolating \( x \), we simplify to: \[ 40 = 4x \] Then, divide both sides by 4 to find: \[ x = 10 \] Thus, the break-even point occurs when 10 cakes are sold. Analytical solutions like this demystify challenges, paving a path to grasp the relationships between different economic variables. Understanding this helps in strategic planning and resource allocation based on mathematical reasoning.