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Q.

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

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a

$I - K T^{2}$

b

$T^{2} - \frac{1}{K}$

c

$I - \frac{K T^{2}}{2}$

d

$e^{- K T}$

answer is A.

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Detailed Solution

$\frac{d V (t)}{d t} = - K (T - t) \Rightarrow V (t) = \frac{K {(T - t)}^{2}}{2} + C$

At $t = 0, V = I \Rightarrow C = I - \frac{K T^{2}}{2}$

$∴ Scrap value V (T) \Rightarrow V (t = T) = C = I - \frac{K T^{2}}{2}$

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Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by differential equation dV(t)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(T) of the equipment is