In 2012, National Textile installed a new textile machine in one of its factories at a cost of $200,000. The machine is depreciated linearly over 10 years with a scrap value of $10,000. (a) Find a function for the textile machine's book value in the t th year of use (0 ≤ t ≤ 10). V(t)

Let's work through the problem step by step.

(a) Find a function for the textile machine's book value in the t th year of use.

The book value of the machine is the original cost minus depreciation over time. We're told that:

• The initial cost is $200,000.
• The scrap value (the value of the machine after 10 years) is $10,000.
• The depreciation is linear, meaning the machine loses value by the same amount each year.

Depreciation per year:

The total depreciation over 10 years is the difference between the initial cost and the scrap value: $\text{Total depreciation} = 200,000 - 10,000 = 190,000$

Since the machine is depreciated linearly over 10 years, the annual depreciation is: $\text{Annual depreciation} = \frac{190,000}{10} = 19,000 \text{ per year}.$

Book value function:

Let $V(t)$ represent the book value of the machine at time $t$, where $t$ is the number of years since the machine was installed. The book value decreases by $19,000 each year, so: $V(t) = 200,000 - 19,000 \times t$ where $0 \le t \le 10$. This function gives the book value of the machine in year $t$.

(b) Sketch the graph of the function.

The function $V(t) = 200,000 - 19,000 \times t$ is a linear function, and its graph will be a line segment from $t = 0$ to $t = 10$.

• At $t = 0$ (the installation year), the book value is $200,000.
• At $t = 10$ (after 10 years), the book value is $10,000.

So the line will start at $(0, 200,000)$ and end at $(10, 10,000)$.

You can plot this line in a graphing tool, with the horizontal axis being the number of years ($t$) and the vertical axis being the book value ($V(t)$).

(c) Find the machine's book value in 2016.

To find the machine's book value in 2016, we need to determine the number of years that have passed since the installation of the machine in 2012. The difference between 2016 and 2012 is 4 years.

So, we plug $t = 4$ into the function $V(t)$:

$V(4) = 200,000 - 19,000 \times 4 = 200,000 - 76,000 = 124,000.$

Thus, the machine's book value in 2016 is $124,000.

(d) Find the rate at which the machine is being depreciated.

Since the depreciation is linear, the rate of depreciation is constant each year. From part (a), we already found that the machine loses $19,000 in value each year. Therefore, the rate of depreciation is:

$\text{Depreciation rate} = 19,000 \text{ per year}.$

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Recap:

• (a) The function for the book value is:
  $V(t) = 200,000 - 19,000 \times t \quad \text{(for } 0 \le t \le 10\text{)}.$
• (b) The graph is a line segment that starts at $(0, 200,000)$ and ends at $(10, 10,000)$.
• (c) The machine's book value in 2016 (after 4 years) is $124,000.
• (d) The rate of depreciation is $19,000 per year.

Would you like more details or further clarification on any part of this? Here are some related questions for you:

1. How would the depreciation change if the machine were depreciated using a different method, like exponential depreciation?
2. How do you calculate depreciation if a machine has an increasing rate of depreciation over time?
3. What happens to the book value if the machine is sold before the 10-year period?
4. How can you graph linear depreciation functions in a different coordinate system (e.g., logarithmic scale)?
5. If the scrap value were $5,000 instead of $10,000, how would that affect the depreciation rate?

Tip: When modeling depreciation, always double-check if it's linear or another method (like accelerated), as the function form changes based on that.