Problem 62 - Understanding Elementary Algebra with Geometry

Question

Round off to the nearest hundredth when necessary. A business buys a piece of machinery for 12,800 dollar Suppose that the value of the machinery depreciates linearly to a value of 2,400 dollar in 8 years. (a) Write an equation relating the value \(V\) of the machinery after \(n\) years. [ Hint: The value of the machinery in year 0 is 12,800 dollar (b) Find the value of the machine in year 5 (c) How long will it take for the machine to depreciate to a value of zero?

Step-by-Step Solution

Answer

a) V(n) = 12800 - 1300n. b) 6300 dollars. c) 9.85 years.

Step 1: Define Initial Values

The initial value of the machinery at year 0 is given by \[ V(0) = 12800 \] The value after 8 years is given by \[ V(8) = 2400 \]

Step 2: Determine Slope of Depreciation

The machinery's value depreciates linearly, which means we can find the slope (rate of depreciation) using the formula: \[ m = \frac{{V(8) - V(0)}}{8 - 0} = \frac{2400 - 12800}{8} = \frac{-10400}{8} = -1300 \]So, the rate of depreciation per year is -1300 dollars.

Step 3: Write the Linear Equation

Using the slope-intercept form of a linear equation \[ V(n) = V(0) + m \times n \] Substitute the known values: \[ V(n) = 12800 - 1300n \]This is the equation relating the value of the machinery after n years.

Step 4: Calculate Value in Year 5

To find the value of the machinery in year 5, substitute n = 5 into the equation: \[ V(5) = 12800 - 1300 \times 5 \]\[ V(5) = 12800 - 6500 = 6300 \] So, the value of the machinery in year 5 is 6300 dollars.

Step 5: Determine Time to Depreciate to Zero

To find how long it takes for the machinery to depreciate to zero, set V(n) = 0 and solve for n: \[ 0 = 12800 - 1300n \] \[ 1300n = 12800 \]\[ n = \frac{12800}{1300} \approx 9.85 \] Rounding to the nearest hundredth, it takes approximately 9.85 years for the value of the machinery to depreciate to zero.

Key Concepts

linear equation

Linear equations are fundamental in understanding various real-world problems, like the depreciation of machinery. A linear equation is an equation that makes a straight line when it is graphed. It has the general form:
\[y = mx + b\]
where:

• 'y' is the dependent variable (in our case, the value of the machinery, V)
• 'm' is the slope of the line (rate of depreciation)
• 'x' is the independent variable (in our case, the number of years, n)
• 'b' is the y-intercept (the initial value, in year 0)

For example, in the machinery depreciation problem, the linear equation is written as:
\[ V(n) = 12800 - 1300n \]
where V(n) represents the value of the machinery after 'n' years.

rate of depreciation

The rate of depreciation is a crucial concept that helps us understand how quickly an asset like machinery loses its value over time. In a linear depreciation model, this rate is constant. Consider our machinery depreciation problem:

• The machinery initially costs 12,800 dollars and
• The value drops to 2,400 dollars after 8 years.

The rate of depreciation can be calculated using the slope formula:
\[ m = \frac{V(8) - V(0)}{8 - 0} = \frac{2400 - 12800}{8} = \frac{-10400}{8} = -1300 \]
This tells us that the machinery loses 1,300 dollars in value each year. This rate is negative because the value decreases over time. Understanding this rate allows us to predict future values and understand how quickly the asset's worth will drop.

slope-intercept form

The slope-intercept form is a common way to express linear equations. It is written as:
\[ y = mx + b \]
where:

• 'm' is the slope of the line, showing how much y changes for a unit change in x (here, it represents the rate of depreciation)
• 'b' is the y-intercept, the value of y when x is 0 (here, it's the initial value of the machinery)

In the context of our example:
\[ V(n) = 12800 - 1300n \]
The slope (m) is -1300, indicating the machine loses 1,300 dollars each year. The y-intercept (b) is 12,800, meaning the machinery's value starts at 12,800 dollars. Using this form, you can easily identify both the initial value and the depreciation rate.