Problem 86 - Calculus: Early Transcendentals

Question

A high-tech company purchases a new computing system whose initial value is \( V \). The system will depreciate at the rate \( f = f(t) \) and will accumulate maintenance costs at the rate \( g = g(t) \), where \( t \) is the time measured in months. The company wants to determine the optimal time to replace the system. (a) Let $$ C(t) = \frac{1}{t} \int^t_0 [f(s) + g(s)] \, ds $$ Show that the critical numbers of \( C \) occur at the numbers \( t \) where \( C(t) = f(t) + g(t) \). (b) Suppose that \( f(t) = \left\\{ \begin{array}{ll} \frac{V}{15} - \frac{V}{450}t & \mbox{if \) 0 < t \le 30 \(}\\\ 0 & \mbox{if \) t > 30 \(} \end{array} \right.\) and \( g(t) = \frac{Vt^2}{12,900} \) \( t > 0 \) Determine the length of time \( T \) for the total depreciation \( \displaystyle D(t) = \int^t_0 f(s) \, ds \) to equal the initial value \( V \). (c) Determine the absolute minimum of \( C \) on \( (0, T] \). (d) Sketch the graphs of \( C \) and \( f + g \) in the same coordinate system, and verify the result in part (a) in this case.

Step-by-Step Solution

Answer

The critical numbers occur where \( C(t) = f(t) + g(t) \). \( T = 45 \). The absolute minimum is at the critical point. Graphs support \( C(t) = f(t) + g(t) \).

Step 1: Setup the equation for critical numbers

We begin by considering the average cost function defined as \( C(t) = \frac{1}{t} \int^t_0 [f(s) + g(s)] \, ds \). In order for \( C(t) \) to have a critical number at \( t \), the derivative \( C'(t) \) must be zero or undefined. By differentiating \( C(t) \) using the quotient rule, we get \( C'(t) = \frac{t(f(t) + g(t)) - \int^t_0 [f(s) + g(s)] \, ds}{t^2} \). Setting the numerator to zero gives \( C(t) = f(t) + g(t) \), which is the condition for critical points.

Step 2: Solve for the depreciation time

The problem gives \( f(t) = \frac{V}{15} - \frac{V}{450}t \) for \( 0 < t \le 30 \). To calculate the length of time \( T \) such that \( D(t) = V \), solve \( D(t) = \int^t_0 f(s) \, ds \) by integrating. We find \( \int_0^t \left( \frac{V}{15} - \frac{V}{450}s \right) \, ds = V \). Solving this gives \( T = 45 \), meaning when the depreciation amount equals the initial value.

Step 3: Determine absolute minimum of C(t)

To find the absolute minimum, evaluate the function \( C(t) = \frac{1}{t} \int^t_0 [f(s) + g(s)] \, ds \). First, solve for the function values at critical points found in Step 1 and at the endpoints \( t = 0^+ \) and \( t = 45 \). The smallest value determines the absolute minimum. By comparison, the critical point yields the minimum cost configuration.

Step 4: Graph C and f+g, check Part (a) result

To graph \( C(t) \) and \( f(t) + g(t) \), first sketch these curves over the interval \( (0,45] \). The graph of \( f(t) + g(t) \) as a function of \( t \) will intersect \( C(t) \) at the critical point. Verify that where these graphs intersect, it supports the result of \( C(t) = f(t) + g(t) \) as the condition for critical numbers.

Key Concepts

Depreciation Analysis

In depreciation analysis, it is important to understand how the value of an asset decreases over time. Depreciation reflects the wear and tear or usage of the asset, leading to a decline in its useful value. For businesses, accurately calculating depreciation can significantly impact economic decisions, like when to replace equipment or upgrade systems.
Depreciation can be calculated using various methods:

• Straight-Line Depreciation: A fixed percentage of the asset's original value is deducted each period.
• Declining Balance Method: Depreciation occurs at a faster rate in the initial periods of the asset's life.
• Units of Production Method: Depreciation is based on usage or output rather than time.

These methods can adjust for maintenance needs and more accurately reflect an asset's real-world performance.

Definite Integrals

Definite integrals play a crucial role in various calculus applications, including optimization problems. A definite integral is used to calculate the area under a curve of a function over an interval \( [a, b] \). This area corresponds to the accumulation of quantities, such as total depreciation or total cost, over that interval.
To compute a definite integral, you would evaluate the integral of a function \( f(x) \) from \( a \) to \( b \), noted as \( \int_a^b f(x) \, dx \). The result gives a numerical value, indicating the net accumulation over the interval.
For example, in depreciation analysis, the definite integral helps compute the total depreciation amount over a given timeframe, providing insights to determine optimal asset replacement schedules.

Critical Points Mathematics

Understanding critical points is essential in mathematics, particularly in calculus optimization problems. Critical points of a function occur where the derivative is zero or undefined. At these points, the function might have local maxima, local minima, or a saddle point. These points help in identifying the behavior of cost functions or other related calculations.

• Finding Critical Points: Set the derivative of the function to zero (\( C'(t)=0 \)) and solve for \( t \).
• Interpreting Critical Points: Determine whether these points correspond to maxima, minima, or changing trends in data or costs.
• Verification: Use the second derivative test or analyze changes around the critical point to confirm its nature.

In the given exercise, setting \( C(t) = f(t) + g(t) \) identifies such critical situations, aiding in optimizing asset lifespan and cost efficiency.

Cost Function Analysis

In the realm of economics and operational analysis, cost functions are used to understand how the total cost of production varies with varying levels of output or time. In this context, analyzing cost functions helps businesses strategize budget allocations and operational efficiency.

• Average Cost Function: Defined as \( C(t) = \frac{1}{t} \int_0^t [f(s) + g(s)] \, ds \), it reflects the average cost of maintaining an asset over time \( t \).
• Total Cost: Includes all expenses involved in operating and maintaining equipment over time, encompassing depreciation and maintenance costs.
• Optimization: Lowering average costs involves analyzing their trend over time and adjusting strategies to find cost-effective operation windows.

By understanding these aspects, businesses can better manage funds, decide on investment durations, and synchronize maintenance and replacement schedules with cost minimization strategies.