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Question
an office machine is purchased for $7700. assume that its salvage value, v, in dollars, depreciates, according to a method called double - declining balance, by 40% each year and is given by v(t)=7700(0.60)^t, where t is the time, in years, after purchase.
a) find v(3) and explain its meaning.
b) find v(3) and explain its meaning.
c) when will the salvage value of the office machine be half of the purchase price?
a) v(3) = $1663.20
(round to the nearest cent as needed.)
the expression v(3) means that after 3 year(s), the salvage value
Step1: Calculate V(3)
Given $V(t)=7700(0.60)^t$, substitute $t = 3$.
$V(3)=7700\times(0.60)^3=7700\times0.216 = 1663.20$
The meaning of $V(3)$ is that after 3 years, the salvage - value of the office machine is $1663.20$ dollars.
Step2: Differentiate V(t)
First, if $V(t)=7700(0.60)^t$, using the formula for the derivative of an exponential function $y = a\cdot b^x$, where $y^\prime=a\cdot b^x\ln(b)$. Here $a = 7700$ and $b = 0.60$. So $V^\prime(t)=7700\times(0.60)^t\ln(0.60)$.
Then substitute $t = 3$:
$V^\prime(3)=7700\times(0.60)^3\times\ln(0.60)=1663.20\times\ln(0.60)\approx1663.20\times(- 0.5108)\approx - 850.03$
The meaning of $V^\prime(3)$ is that at $t = 3$ years, the rate of change of the salvage - value of the office machine is approximately $-850.03$ dollars per year. The negative sign indicates that the salvage - value is decreasing.
Step3: Solve for t when $V(t)=\frac{7700}{2}=3850$
Set $V(t)=3850$, so $7700(0.60)^t=3850$.
Divide both sides by 7700: $(0.60)^t=\frac{3850}{7700}=0.5$.
Take the natural logarithm of both sides: $t\ln(0.60)=\ln(0.5)$.
Solve for $t$: $t=\frac{\ln(0.5)}{\ln(0.60)}=\frac{-0.6931}{-0.5108}\approx1.36$ years.
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a) $V(3) = 1663.20$. The meaning is that after 3 years, the salvage - value of the office machine is $1663.20$ dollars.
b) $V^\prime(3)\approx - 850.03$. The meaning is that at $t = 3$ years, the rate of change of the salvage - value of the office machine is approximately $-850.03$ dollars per year (salvage - value is decreasing).
c) Approximately $1.36$ years.