Question

I have the following problem on a takehome assignment and am stuck.

A particular stock has value V(t) at time t where V(t)=Ke^(1/2) where K>0. Assume that alternative investments grow in value to e^(rt). Calculate the instantaneous rate of change in the value of the stock. Calculate the instantaneous rate of change in the value of alternative investments. What is the optimal time to sell the stock?

I believe the answer to this is to find the derivative of each and set them equal to each other.

for V'(t) I found = Ke^(1/2)*1/2t^(1/2)
and for the alternative I found = re^t

This is where I'm stuck, I don't see what the next move is that I'm supposed to make. Thank in advance for any help.

There's also a second half to this problem that I need help on, but I think this needs to be solved first.

Answer 1

V(t) at time t where V(t)=Ke^(1/2) where K>0.

You have a typo. there is no time in that function.

Answer 2

yes sorry,
it's Ke^t^(1/2)

Answer 3

so I will guess you mean
V = k e^(t/2)
the alternative is
A = a e^rt
D = value above alternate = V -A =ke^(t/2)- ae^rt

dD/dt = (k/2) e^(t/2) - a r e^rt

max or min when dD/dt = 0 or
(k/2)e^(t/2) = a r e^rt
k/(2ar) =e^rt /e^t/2 = e^(r t - t/2)
ln[ k/(2ar)] = t (r-1/2)
t = ln [ k/2ar)] / (r - 1/2)

Answer 4

I'm sorry that my post is confusing
what I meant was
Ke^(sqrt t)

It's a hand written assignment so it's hard for me to type it.

Answer 5

OH - ok
V = K e^t^(1/2)
then dV/dt = K e^t^(1/2) d/dt(t^1/2)
= K e^t^(1/2) [ (1/2) t^(-1/2)
= [K /2t^(1/2)] e^t^(1/2)
so
dD/dt = [K /2t^(1/2)] e^t^(1/2) -a r e^rt
at dD/dt = 0
[K /2t^(1/2)] e^t^(1/2) = a r e^rt
K/(2ar) = e^(r t -t^(1/2)) / t^(1/2)
ln[K/(2ar)] = (r t -t^(1/2)- ln t^(1/2)
ln[K/(2ar)] = (r t -t^(1/2)- (1/2)ln t
not much I can do with that. check my arithmetic

Answer 6

[K /2t^(1/2)] e^t^(1/2) = a r e^rt
K/(2ar) = e^(r t -t^(1/2)) * t^(1/2)
ln[K/(2ar)] = (r t -t^(1/2)+ ln t^(1/2)
ln[K/(2ar)] = (r t -t^(1/2)+ (1/2)ln t

There are no doubt more mistakes