Problem 63 - Applied Calculus

Question

A factory installs new machinery that saves \(S(x)=1200-20 x\) dollars per year, where \(x\) is the number of years since installation. However, the cost of maintaining the new machinery is \(C(x)=100 x\) dollars per year. a. Find the year \(x\) at which the maintenance \(\operatorname{cost} C(x)\) will equal the savings \(S(x)\). (At this time, the new machinery should be replaced.) b. Find the accumulated net savings [savings \(S(x)\) minus \(\operatorname{cost} C(x)]\) during the period from \(t=0\) to the replacement time found in part (a).

Step-by-Step Solution

Answer

The machinery should be replaced after 10 years, and the accumulated net savings over this period is $6000.

Step 1: Set the two equations equal

To find the year when the maintenance cost equals the savings, set the cost equation equal to the savings equation: \[ C(x) = S(x) \] Substitute the provided equations: \[ 100x = 1200 - 20x \]

Step 2: Solve the equation for x

Combine like terms and solve for \(x\): \[ 100x + 20x = 1200 \] \[ 120x = 1200 \] Divide both sides by 120: \[ x = \frac{1200}{120} = 10 \] Thus, the maintenance cost will equal the savings after 10 years.

Step 3: Define net savings function

Net savings is calculated by subtracting the cost from the savings for any given year \(x\): \[ N(x) = S(x) - C(x) = (1200 - 20x) - (100x) \] Simplify the expression: \[ N(x) = 1200 - 20x - 100x = 1200 - 120x \]

Step 4: Find accumulated net savings

Calculate the accumulated net savings from \(t=0\) to \(t=10\) by integrating the net savings function \(N(x)\): \[ \int_0^{10} (1200 - 120x)\, dx \] This becomes: \[ \left[ 1200x - 60x^2 \right]_0^{10} \] Calculate this evaluation: \[ \left(1200(10) - 60(10)^2\right) - \left(1200(0) - 60(0)^2\right) \] \[ = 12000 - 6000 = 6000 \]

Step 5: Interpret results

The solution shows that the net savings accumulated over the 10-year period, before the maintenance costs equal the savings, is $6000.

Key Concepts

Function Optimization

In the context of factory machinery, function optimization involves determining the optimal point at which an operation should be adjusted or changed to maximize efficiency. Here, we consider the cost and savings equations related to the machinery's operation.

• Cost Equation: Maintenance costs increase linearly with time, given by \(C(x) = 100x\).
• Savings Equation: Yearly savings from machinery are represented as \(S(x) = 1200 - 20x\), showing a linear decrease over time.

This problem requires us to find the year \(x\) when these two linear functions meet, indicating that maintenance costs equal savings. By equating \(C(x) = S(x)\), we solve for \(x\) to find the turning point—10 years in this case. This marks the time when it is no longer worthwhile to maintain the machinery, as costs match savings. Function optimization, therefore, plays a crucial role in decision-making for cost-effective operations.

Integration

Integration is a fundamental concept in calculus used to find quantities such as areas under curves, accumulated values, and more. In this exercise, integration helps find the total net savings over the time the machinery operates.
We start with the net savings function \(N(x) = 1200 - 120x\), derived from subtracting the cost function from the savings function. To find the total accumulated net savings from 0 to 10 years, we use definite integration. The integral is \[\int_0^{10} (1200 - 120x) \, dx\]This represents the area between the net savings function and the x-axis from \(x=0\) to \(x=10\). Evaluating this definite integral gives us \[\left[ 1200x - 60x^2 \right]_0^{10}\]Both integration and evaluating the integral lead to the conclusion that the accumulated net savings over the first 10 years amount to $6000. Integration thus provides a powerful tool for calculating total values over a time period.

Cost Analysis

Cost analysis involves comparing different financial values to assess profitability and make strategic decisions. It is especially useful in manufacturing, where machinery and maintenance costs must be continually evaluated.
In this problem, cost analysis is first used to determine when the cost of maintaining machinery equals its savings. By setting \(C(x) = S(x)\), we identify at year 10 that the machinery costs as much as it saves. This is a key indicator for potential replacement.
Next, a further layer of cost analysis involves calculating net savings, \(N(x) = S(x) - C(x)\), which provides insight into profitability over time. The integration of this net savings function from year 0 to year 10 encapsulates the total excess savings before the cost becomes equal to the savings.

• Short-term savings may seem high, initially reducing maintenance costs.
• Over time, these savings dwindle as maintenance costs rise.
• Finally, the analysis reveals a threshold point where costs outgrow savings.

Thorough cost analysis here aids in strategic planning, ensuring timely machinery upgrades to maintain cost-efficiency for the factory.