Question

For tax and accounting purposes, corporations depreciate the value of equipment each year. One method used is called "linear depreciation," where the value decreases over time in a linear manner. Suppose that two years after purchase, an industrial milling machine is worth $670,000, and five years after purchase, the machine is worth $130,000. Find a formula for the machine value V (in thousands of dollars) at time t ≥ 0 after purchase.

Answer 1

To calculate the linear depreciation formula, we used the provided value of the machine at two different times to calculate the slope and initial value. The formula for the machine value V (in thousands of dollars) at any time t after purchase is V = -$180,000 * t + $1,030,000.

Step-by-step explanation:

To find the formula for the machine value V (in thousands of dollars) at time t after purchase using linear depreciation, we need two data points which are given as follows:

• After 2 years (t=2), V = $670,000
• After 5 years (t=5), V = $130,000

First, let's find the slope (m) of the depreciation line, which is the rate at which the value is decreasing per year.

m = (Value at 5 years - Value at 2 years) / (5 - 2) = ($130,000 - $670,000) / (5 - 2) = -$540,000 / 3 years = -$180,000 per year

Now, we use one of the points to find the y-intercept (b), which represents the initial value of the machine. Using the point (2, $670,000), the linear equation takes the form:

V = m*t + b

Where V is value and t is time.

$670,000 = (-$180,000 * 2) + b

b = $670,000 + $360,000

b = $1,030,000

The linear depreciation formula for the value V of the machine at any time t is:

V = -$180,000 * t + $1,030,000

Note that V is in thousands of dollars, and t is the number of years after purchase.

Answer 2

Step-by-step explanation:

The method we have to use is linear depreciation.

We will first find the slope:

`Slope=(V_(2)-V_(1))/(Y_(2)-Y_(1))`

where:

V_2 is the machine value after 5 years.

V_1 is the machine value after 2 years.

Y_1 = 2 years

Y_2 = 5 years

`Slope=(130000-670000)/(5-2)`

`Slope=-180,000`

Now:

`Slope=(V_(2)-V)/(Y_(2)-Y)`

where:

V is the machine value at t ≥ 0.

Y is the years at t ≥ 0.

`-180,000=(130000-V)/(5-Y)`

Rearranging the above expression.

`V=130000+180000(5-Y)`

Formula for the machine value =
`V=130000+180000(5-Y)`