41. Correct. The answer is true. 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)Kexp(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.