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Problem: A person inherits a coin collection that has a value of $2000e^{\sqrt{t}}$ m.u. (monetary unit) after $t$ years. Assume a bank interest of 7% per year. At which time $t_0$ years (if there is one) is it best to sell the coin collection and deposit the money at the bank? (The life expectancy of the owner is disregarded and the collection is passed on to any surviving relatives, so $t$ is not limited by the remaining years of the current owner.)

Solution: The definition of ”best time” is unclear to me. I thought of it as follows: Let $V(t)=2000e^{\sqrt{t}}$ be the value of the coin collection at time $t$ years. The increased value of the collection per year should be less than the annual 7% interest of the collection’s value at $t$, when it is sold, i.e. $$ V(t+1)-V(t)<0.07V(t) $$ which has a ”break point” at $t=54$ years, but the answer is 57 years. Any thoughts what could be wrong?

mf67
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3 Answers3

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Consider the money you'd have at the time $T$, assuming you sold the coin collection at the time $t<T$ and immediately deposit the money in the bank. You'd have $$ W(T,t) = V(t)\cdot 1.07^{T-t} = 2000 e^{\sqrt{t} + (T-t)\ln (1.07)} $$ Considering $T$ to be constant, we want to find the best time of selling $t$. That means finding the maximum of function $$ f(t) = \sqrt{t} + (T-t)\ln(1.07)$$ We have $$ f'(t) = \frac{1}{2\sqrt{t}} - \ln(1.07)$$ $$ f'(t_0) = 0 \qquad \Leftrightarrow \qquad t_0= \frac{1}{4(\ln(1.07))^2} \approx 54.6127 $$ It does not depend on $T$, which means that no matter what $T\ge t_0$ you consider, if you want to have the most money in the and, it's best to sell the coin collection at time $t_0$.

The answer 57 years is wrong.

Adam Latosiński
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The "force of interest" is the rate of growth divided by the value of the asset, so if the value at time $t$ is $f(t)$, the force of interest is $${f'(t)\over f(t)},$$ the logarithmic derivative of $f$. Therefore we want to sell when $${1\over2\sqrt{t}}=\log{1.07}$$ which happens when $t=54.6127.$ If you have to sell after a whole number of years, you'd have to work out whether it's best to sell after $54$ or $55$ years, but in any event, $57$ seems incorrect.

saulspatz
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I would say your approach looks more or less correct. I would probably simplify it a bit in saying

$\frac {\frac {dV}{dt}}{V} = 0.07$

Well, it is simpler if you know calculus.

$\frac {dv}{dt} = \frac {2000e^{\sqrt t}}{2\sqrt t} = \frac {V}{2\sqrt t}\\ \frac{1}{2\sqrt t} = 0.07$

Using the approach in the OP, you should be able to get there with a few guesses in a guess and check approach.

user317176
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