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Let I be the purchase value of an equipment and $\mathrm{V}(\mathrm{t})$ be the value after it has been used for t years. The value $\mathrm{V}(\mathrm{t})$ depreciates at a rate given by differential equation $\frac{\mathrm{dV}(\mathrm{t})}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-\mathrm{t})$, where $\mathrm{k}>0$ is a constant and $\mathrm{T}$ is the total life in years of the equipment. Then the scrap value $\mathrm{V}(\mathrm{T})$ of the equipment is
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1655 Upvotes
Verified Answer
The correct answer is:
$\mathrm{I}-\frac{\mathrm{kT}}{2}$
$\mathrm{I}-\frac{\mathrm{kT}}{2}$
$$
\frac{d V}{d t}=-k(T-t) \Rightarrow d V=-k(T-t) d t
$$
Integrate
$$
V=\frac{-k(T-t)^2}{(-2)}+c \Rightarrow V=\frac{k(T-t)^2}{2}+c
$$
at $\mathrm{t}=\mathrm{0} \Rightarrow \mathrm{V}=\mathrm{I}$
$$
\mathrm{I}=\frac{\mathrm{kT}}{2}+\mathrm{c} \Rightarrow \mathrm{c}=\mathrm{I}-\frac{\mathrm{kT} \mathrm{T}^2}{2} \Rightarrow \mathrm{c}=\mathrm{V}(\mathrm{T})=\mathrm{I}-\frac{\mathrm{kT} \mathrm{T}^2}{2}
$$
\frac{d V}{d t}=-k(T-t) \Rightarrow d V=-k(T-t) d t
$$
Integrate
$$
V=\frac{-k(T-t)^2}{(-2)}+c \Rightarrow V=\frac{k(T-t)^2}{2}+c
$$
at $\mathrm{t}=\mathrm{0} \Rightarrow \mathrm{V}=\mathrm{I}$
$$
\mathrm{I}=\frac{\mathrm{kT}}{2}+\mathrm{c} \Rightarrow \mathrm{c}=\mathrm{I}-\frac{\mathrm{kT} \mathrm{T}^2}{2} \Rightarrow \mathrm{c}=\mathrm{V}(\mathrm{T})=\mathrm{I}-\frac{\mathrm{kT} \mathrm{T}^2}{2}
$$
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