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* = (4r²)^-1. Check that A"(t) < 0 for a maximum.