Chapter 0, Problem 62RE

a.

To determine

To write: an expression that gives value of y after x years.

Answer to Problem 62RE

  $y=100,000-10,000x,0\le x\le 10$

Explanation of Solution

Given information:

An 18-wheel truck is purchased for $100,000. Its value depreciates at the constant rate of $10,000 per year for 10 years.

Formula Used:

The slope-intercept equation of a line passing through the point $\left(0,b\right)$ and with slope m is:

  $y=mx+b$

It is given that the value of the truck depreciates at a constant rate. So, its equation must be linear.

The initial value of the truck is $100,000. So, the y -intercept of the equation representing the value of truck is $\left(0,100,000\right)$ .

Also, it is given that its value depreciates at the constant rate of $10,000 per year for 10 years.

So, $m=-10,000$

Thus, the required equation is:

$\begin{array}{c}y=mx+b\\ y=-10,000x+100,000\\ y=100,000-10,000x\end{array}$

The depreciation rate is provided for 10 years. So, the domain is $0\le x\le 10$ .

b.

To determine

To find: When will the value of truck be $55,000.

Answer to Problem 62RE

After 4.5 years

Explanation of Solution

Given information:

From part (a), the value y of truck after x years is given by the equation:

  $y=100,000-10,000x,0\le x\le 10$

The objective here is to find x for which $y=55,000$ .

Substitute $y=55,000$ in the given equation,

  $\begin{array}{c}y=100,000-10,000x\\ 55,000=100,000-10,000x\\ 10,000x=100,000-55,000\end{array}$

  $\begin{array}{c}10,000x=45,000\\ x=\frac{45,000}{10,000}\\ x=4.5\text{ years}\end{array}$

Hence, the value of truck will be $55,000 after 4.5 years.