41. Incorrect. The
answer is true, not false. The present
value of the value of wine is A_{t} = Ve^{-rt } = Kexp(t^{0.5} - rt). To determine the optimal time for selling
the wine optimize the present value function:
A_{t}' = (0.5t^{-0.5}-r)K exp(t^{0.5} - rt)
= (0.5t^{-0.5}-r)* A(t)=
0. Since A(t)¹0, (0.5t^{-0.5}-r)=
0 and t* = (4r^{2})^{-1}.
Check that A"(t) < 0 for a maximum.