Problem 17 - Applied Calculus

Question

A photocopying company has two different price lists. The first price list is $$ 100\( plus 3 cents per copy; the second price list is $$ 200\) plus 2 cents per copy. (a) For each price list, find the total cost as a function of the number of copies needed. (b) Determine which price list is cheaper for 5000 copies. (c) For what number of copies do both price lists charge the same amount?

Step-by-Step Solution

Answer

(a) Cost functions: \( C_1(x) = 100 + 0.03x \), \( C_2(x) = 200 + 0.02x \). (b) The first price list is cheaper for 5000 copies. (c) Both charge the same for 10,000 copies.

Step 1: Define cost functions

To find the total cost as a function of the number of copies, we'll start by defining cost functions for each price list.For the first price list:\[ C_1(x) = 100 + 0.03x \]For the second price list:\[ C_2(x) = 200 + 0.02x \]where \( x \) is the number of copies.

Step 2: Evaluate costs for 5000 copies

Substitute \( x = 5000 \) into each cost function to find the cost.For the first price list, \[ C_1(5000) = 100 + 0.03 \times 5000 = 100 + 150 = 250 \]For the second price list, \[ C_2(5000) = 200 + 0.02 \times 5000 = 200 + 100 = 300 \]

Step 3: Compare costs for 5000 copies

The cost using the first price list is $250, and the cost using the second price list is $300. Since $250 < $300, the first price list is cheaper for 5000 copies.

Step 4: Find number of copies where costs are equal

To find the number of copies for which both price lists charge the same amount, set the two cost functions equal:\[ 100 + 0.03x = 200 + 0.02x \]Solve this equation for \( x \):\[ 0.03x - 0.02x = 200 - 100 \]\[ 0.01x = 100 \]\[ x = \frac{100}{0.01} = 10000 \]Thus, both price lists charge the same amount for 10,000 copies.

Key Concepts

Price Comparison

When faced with multiple pricing options, knowing how to compare them efficiently can be crucial. In our example of the photocopy company, we have two different pricing strategies to evaluate.

• The first price list includes a base price of $100 plus 3 cents per copy.
• The second price list starts at a higher base of $200 but charges only 2 cents per copy.

To make a decision on which pricing strategy is more affordable, we need to compare total costs for different numbers of copies. By substituting the number of copies into each respective cost function, we can determine which plan incurs less cost.
For instance, when evaluating costs for 5,000 copies, the first price plan results in $250, while the second plan leads to $300. In this scenario, the first pricing plan is clearly cheaper.
Understanding how base cost and variable costs impact total pricing helps in making informed decisions about which option is economically sensible.

Cost Analysis

Cost analysis involves breaking down different pricing components to understand their impact on total expenses. This allows businesses and consumers to see where their costs are coming from and thus identify the most cost-effective solution.

In the example of our photocopying company:

• The first pricing scheme is influenced by a $100 starting fee coupled with a small incremental cost of 0.03 per copy, meaning that while the base cost is lower, each additional copy is slightly more costly.
• The second pricing plan, on the other hand, has a higher base cost of $200, but only adds 0.02 for every extra copy.

This illustration shows the trade-offs between base costs and per-unit costs. If you're planning to make fewer copies, the first price list might be preferable due to the lower base cost. However, as the copy count increases, the initially cheaper looking plan might become more expensive, unless the per-unit cost is lower.
Performing a thorough cost analysis for both small and large volumes can clarify which pricing model saves money and when the savings kick in.

Linear Equations

Linear equations are crucial in determining the relationships between different variables. In price analysis, they help model costs effectively.

Here, the cost functions of photocopying companies are modeled as linear equations:

• The first plan uses the equation: \[ C_1(x) = 100 + 0.03x \]
  where 100 is the fixed component and 0.03x represents the variable component per copy.
• The second plan employs the equation: \[ C_2(x) = 200 + 0.02x \]
  where 200 is the initial fee with a lower variable cost per copy.

These equations allow us to predict total costs for any given number of copies by simply substituting \( x \) with the desired count. Additionally, linear equations help compare costs by setting them equal to solve for the number of copies that make both costs identical, as shown in the calculation where both plans are found to be equal at 10,000 copies.
Understanding how to set up and solve linear equations is essential for accurate cost predictions and making informed financial choices.