Question

Question #b15b8

Answer 1

The value is about $233333 (or $230000 if you round to two significant digits).

Explanation:

We are assuming that `(dV)/dt=-k/(t+3)^2` for some constant `k>0`. Upon separation of variables, this can be written as `dV=-k/(t+3)^2dt`.

Integration of both sides leads to `V=k/(t+3)+C`.

We are given that `V(0)=500000` and `V(1)=400000` (the decrease in value is $100000 from time `t=0` to time `t=1`). Therefore, we get the following system of equations to solve for `k` and `C`:

`500000=k/3+C` and `400000=k/4+C`.

Solving the first equation for `C` and substituting into the second equation gives `400000=k/4+(500000-k/3)`, or
`100000=k/3-k/4=k/12`. Hence, `k=1200000` and
`C=500000-1200000/3=500000-400000=100000`.

Therefore, `V=V(t)=1200000/(t+3)+100000` and
`V(6)=1200000/9+100000 approx $233333`.

Answer 2

`V = 1200000 /(t+3) + 100000`

`V(6) = $233,333`

Explanation:

We must decode the given text to form an appropriate Differential Equation which we must then solve.

Depreciation reduces the value of an object. `(dV)/(dt)` represents the rate of change of the value of the machine. Thus the depreciation is `-(dV)/(dt)`.

We are told that the depreciation is inversely proportional to the square of `t+3`. Thus:

`-(dV)/(dt) prop 1/(t+3)^2 => -(dV)/(dt) =K/(t+3)^2`

Where `K` is the constant of proportionality. The above is a First order linear separable Ordinary Differential Equation. And we can "seperate the variables" to get:

`int \ dV = - K int \ 1/(t+3)^2 \ dt`

Which we can integrate to get:

`V = K/(t+3) + C`

Note that we have have two unknown constants and two conditions; thus:

"The initial value of the machine was $500,000":

`=> V=500000` when `t=0`

`:. 500000 = K/3 + C` ..... [A]

"its value decreased $100,000 in the first year":

`=> V=400000` when `t=1`

`:. 400000 = K/4 + C` ..... [B]

Eq[A] - Eq[B]:

`100000 = K/3 - K/4`
`:. K/12 = 100000 => K = 1200000`

Subs `K = 1200000` into Eq [A]:

`500000 = 1200000/3 + C`
`:. 500000 = 400000 + C`
`:. C = 100000`

Thus the solution to the DE is:

`V = 1200000 /(t+3) + 100000`

We seek the value when `t=6`:

`V = 1200000 /9 + 100000`
`\ \ = 400000/3 + 100000`
`\ \ = 400000/3 + 300000/3`
`\ \ = 700000/3`
`\ \ = 233333.33333 ...`
`\ \ = 233333 \ \` rounded to the nearest integer