Problem 103 - Calculus Volume 1

Question

A company purchases some computer equipment for $$\$ 20,500$$. At the end of a 3 -year period, the value of the equipment has decreased linearly to $$\$ 12,300$$. a. Find a function \(y=V(t)\) that determines the value \(V\) of the equipment at the end of \(t\) years. b. Find and interpret the meaning of the \(x\) - and \(y\) -intercepts for this situation. c. What is the value of the equipment at the end of 5 years? d. When will the value of the equipment be $$\$ 3000 ?$$

Step-by-Step Solution

Answer

a. \(V(t) = 20,500 - 2,733.33t\); b. x-intercept: 7.5 years, y-intercept: \$20,500; c. \$6,833.35; d. 6.4 years.

Step 1: Determine the Rate of Depreciation

We start by identifying the original value of the equipment \( V(0) = \\(20,500 \) and the value at the end of 3 years \( V(3) = \\)12,300 \). To find the rate of depreciation per year, we calculate the change in value over the duration:\[\text{Depreciation Rate} = \frac{V(3) - V(0)}{3} = \frac{12,300 - 20,500}{3} = \frac{-8,200}{3} = -2,733.33 \text{ (approximately)}\]This means the equipment loses approximately \( \$2,733.33 \) in value each year.

Step 2: Establish the Linear Function

We have a depreciation rate of \(-2,733.33\). The linear depreciation function can be expressed as:\[ V(t) = V(0) + \, \text{Depreciation Rate} \, \times \, t = 20,500 - 2,733.33 \, t \]

Step 3: Find the x-Intercept

The \(x\)-intercept is found by setting \(V(t) = 0\) in the depreciation function, because the value of the equipment will be zero.\[0 = 20,500 - 2,733.33 \, t\]Solving for \(t\) gives:\[2,733.33 \, t = 20,500\]\[t = \frac{20,500}{2,733.33} \approx 7.5 \, \text{years}\]The \(x\)-intercept is about 7.5 years, which is when the equipment will have no monetary value.

Step 4: Find the y-Intercept

The \(y\)-intercept is the initial value of the equipment when \(t = 0\):\[V(0) = 20,500\]This means the equipment's original value was \(\$20,500\).

Step 5: Calculate the Value at 5 Years

We use the linear function \(V(t) = 20,500 - 2,733.33 \, t\) to find the value at \(t = 5\):\[V(5) = 20,500 - 2,733.33 \times 5 = 20,500 - 13,666.65 = 6,833.35\]The value of the equipment after 5 years is approximately \(\$6,833.35\).

Step 6: Determine When the Value is $3000

To find when the value is \\(3000, set \(V(t) = 3,000\) in the equation:\[3,000 = 20,500 - 2,733.33 \, t\]Solving for \(t\):\[2,733.33 \, t = 20,500 - 3,000\]\[t = \frac{17,500}{2,733.33} \approx 6.4 \, \text{years}\]The equipment is worth \\)3000 after approximately 6.4 years.

Key Concepts

Rate of Depreciation

Understanding the rate of depreciation is essential when calculating the declining value of an asset over time. In this case, we have a linear depreciation model of computer equipment. The rate of depreciation is determined by the difference in value divided by the number of years. You start with an original equipment value of \(20,500. After 3 years, the value decreases to \)12,300. This gives us:

• Original value: \(20,500
• Value after 3 years: \)12,300

To find the rate of depreciation:\[\text{Depreciation Rate} = \frac{12,300 - 20,500}{3} = -2,733.33\]This calculation reveals that the equipment loses approximately $2,733.33 in value every year. The negative sign indicates a decrease over time. This rate provides the slope for the linear depreciation equation.

x-intercept

The x-intercept in a linear depreciation context represents the time when the value of the equipment will become zero. It is when the asset has fully depreciated and holds no salvage value. To find the x-intercept, set the value function to zero:\[0 = 20,500 - 2,733.33 \, t\]Solving for \(t\), the equation becomes:\[2,733.33 \, t = 20,500\] \[t = \frac{20,500}{2,733.33} \approx 7.5\]This tells us that approximately 7.5 years into owning the equipment, its value will drop to zero. It gives businesses an outlook on when they might need new investments or replacements.

y-intercept

In the context of a depreciation model, the y-intercept signifies the starting value of the equipment when no time has elapsed, or when \(t = 0\). It is where the depreciation line crosses the y-axis, showing the original purchase price of the asset.For the given function:

• At \(t = 0\), the starting value or y-intercept is clearly $20,500.

This intercept serves as a baseline in your depreciation formula, allowing you to calculate future values and understand the rate at which the equipment loses value over time.

Equipment Value Calculation

Calculating the value of the equipment after a given number of years involves using the linear depreciation function derived earlier. The function for this problem is:\[V(t) = 20,500 - 2,733.33 \, t\]To determine the equipment value at the end of 5 years:

• Insert \(t = 5\) into the function: \[V(5) = 20,500 - 2,733.33 \times 5 = 6,833.35\]

The equipment will be worth approximately $6,833.35 after 5 years. Understanding how to use the linear function like this is crucial for planning future budgets or investments related to asset management.