Problem 81 - Contemporary Precalculus

Question

At the Factory in Example \(14,\) the cost of producing \(x\) can openers is given by \(y=2.75 x+26,000\) (a) Write an equation that gives the average cost per can opener when \(x\) can openers are produced. (b) How many can openers should be made to have an average cost of \(\$ 3\) per can opener?

Step-by-Step Solution

Answer

Answer: The equation for the average cost per can opener is Average_cost = (2.75x + 26000) / x. To have an average cost of $3 per can opener, 104,000 can openers need to be produced.

Step 1: a) Finding the Equation for Average Cost Per Can Opener

To find the average cost per can opener, we need to divide the total cost (y) by the number of can openers produced (x). This can be represented as: Average_cost = Total_cost / Number_of_can_openers We are given the equation for the total cost: y = 2.75x + 26000 So, the equation for the average cost per can opener will be: Average_cost = (2.75x + 26000) / x

Step 2: b) Finding the Number of Can Openers for an Average Cost of $3 Per Can Opener

We need to find the value of x when the average cost per can opener is $3. We will use the equation we found in part (a): 3 = (2.75x + 26000) / x Now, we can solve this equation for x step by step: 1. Multiply both sides by x: 3x = 2.75x + 26000 2. Subtract 2.75x from both sides: 0.25x = 26000 3. Divide by 0.25 to solve for x: x = 26000 / 0.25 x = 104000 So, 104000 can openers should be made to have an average cost of $3 per can opener.

Key Concepts

Average Cost

The concept of average cost is crucial in many fields, especially in economics and business, where it helps in determining pricing strategies. Simply put, average cost is the total cost of production divided by the quantity of goods produced.

In mathematical terms, when you have a cost function like the one given in the original exercise, the average cost (\(AC\)) can be calculated by:

• Average Cost = Total Cost / Number of Units Produced

For instance, if we have a total cost function such as \(y = 2.75x + 26000\), then the average cost for producing \(x\) can openers becomes:

• \(AC = \frac{2.75x + 26000}{x}\)

By understanding this, you can determine the cost-effectiveness of producing items, helping you make better production or investment decisions. Calculating the average cost allows you to understand at what production level you can be most profitable.

Linear Equations

Linear equations are fundamental in algebra and precalculus. They represent relationships where variables change at a constant rate, forming a straight line on a graph.

In the context of a cost function, a linear equation like \(y = 2.75x + 26000\) represents the total cost \(y\) to produce \(x\) items. Each component of this equation has a specific meaning:

• \(2.75x\): Represents the variable cost per item.
• \(26000\): This is the fixed cost, which is independent of the number of items produced.

By plotting this equation, you can visualize how costs increase with the production of more units. Understanding how to manipulate and analyze linear equations is a key skill in solving real-world problems that involve rates of change or cost analysis.

Problem-solving in Precalculus

Problem-solving in precalculus often involves breaking down complex scenarios into simpler mathematical models, such as equations and functions. This process helps in understanding and solving practical and theoretical issues.

In the original problem, we applied several classic problem-solving steps:

• Identifying what needs to be calculated—here, the average cost and the necessary production level to achieve a specific average cost.
• Formulating the problem using algebraic expressions—setting up the average cost equation from the total cost expression.
• Manipulating the equation to find the desired variable such as the production level for a certain cost.

For example, solving \(3 = \frac{2.75x + 26000}{x}\) required algebraic manipulation to find \(x = 104000\). Such techniques are essential in precalculus, preparing students for more advanced problem-solving in calculus and related disciplines.