Question

A charter fishing company buys a new boat for $224,000 and assumes that it will have a trade-in value of $115,200 after 16 years.
a. Find a linear model for the depreciated value V of the boat t years
after it was purchased.
b. What is the depreciated value of the boat after 10 years?
c. When will the depreciated value fall below $100,000?
d. Graph V for 0 ≤ t ≤ 30 and illustrate the answers from (b) and
(c) on the graph.

Answer 1

First, according to the information, we build a linear model, and then, the model is used to find the information asked.

a) The linear model is:

$V(t) = 224000 - 6800t$

b) The depreciated value of the boat after 10 years is of $156,000.

c) The depreciated value falls below $100,000 after 18.24 years.

d) The sketch is given at the end of the answer.

Item a:

• The initial value is of $224,000, thus, the y-intercept of the linear function is of $b = 224000$.
• The value decayed to $115,200 in 16 years, thus, the slope of the linear function is:

$m = \frac{115200 - 224000}{16} = -6800$

Then, the linear model is:

$V(t) = 224000 - 6800t$

Item b:

The value after 10 years is V(10), so:

$V(10) = 224000 - 6800(10) = 156000$

Then, the depreciated value of the boat after 10 years is of $156,000.

Item c:

This is t for which: V(t) = 100000. So

$V(t) = 224000 - 6800t$

$100000 = 224000 - 6800t$

$6800t = 224000 - 100000$

$6800t = 124000$

$t = \frac{124000}{6800}$

$t = 18.24$

The depreciated value falls below $100,000 after 18.24 years.

Item d:

The sketch is appended at the end of this answer.

A similar problem is given at brainly.com/question/24978772