Question

Let $I$ be the purchase value of an equipment and $V\left(t\right)$ be the value after it has been used for $t$ years. The value $V\left(t\right)$ depreciates at a rate given by differential equation $\frac{dV\left(t\right)}{dt}=-k(T-t),$ where $k>0$ is a constant and $T$ is the total life in years of the equipment. Then the scrap value $V\left(T\right)$ of the equipment is

• A. $I-\frac{k{T}^2}{2}$
• B. $I-\frac{k(T-10{)}^2}{2}$
• C. $e^{-kT}$
• D. $T^2-\frac{1}{k}$

Answer 1

The correct option is A $I-\frac{k{T}^2}{2}$

$\frac{dV\left(t\right)}{dt}=-k(T-t)\Rightarrow \int dV\left(t\right)=-k\int (T-t)dt$
$V\left(t\right)=\frac{k(T-t{)}^2}{2}+C$
$V\left(0\right)=I\Rightarrow I=\frac{k{T}^2}{2}+C\Rightarrow C=I-\frac{k{T}^2}{2}$
$\therefore$ The scrap value is $V\left(T\right)=0+C=I-\frac{k{T}^2}{2}$