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im kinda rusty with math, its been some time since i did this, so id like your help
i need to find the top of the following function:
Y(T)= (c*r^T - 10000)*( (r+0.02)^(50-T) )
c is starting cash, and r is interest, both are known
anyone know how to solve this one? really cant remember how its done...... thnx in advance
Differentiation? Sorry too lazy to do it myself
Differentiation of something to the power of x
http://mathworld.wolfram.com/d1img1366.gif
Product rule
http://mathworld.wolfram.com/d1img1386.gif
edit: then solve for derivative = 0, remmber to check is a maxima not minima
yikes!!!!
so i split it in (s=r+0.02)
f(T)=(c*r^T - 10000) and
g(T)=(r+0.02)^(50-T) = (s^50)*(s^-T)
f'(T)= c (ln r)r^T
g'(T)=(s^50)(ln s)(s^-T)
so f(T)*g'(T)+f'(T)*g(T)=
(c*r^T - 10000)(s^50)(ln s)(s^-T)+(c (ln r)r^T)(s^50-T)
try equaling THAT to 0 :'( makes me wish i took that graphic calculator back from my ex :P
well anyway thanks for your help
:huh:
*scratches head* :huh:
:blink: *no clue*
umm at first i thought it was compound interest, then i looked at ilw's post and now i'm just :blink:
It is possible and some bits do cancel out, I did it roughly and my final answer was nasty, but plausible i think.
I noticed i have an extra -ve sign in my g'(x) from taking the derivative of (50-T), if u don't get where this comes from, try following the stages in that 'something to the power of x' thing i posted before, with the actual function we have.
ie g'(T)=-(s^50)(ln s)(s^-T)
and i put the s to the power (50-T) back together ie
g'(T)=-ln(s) * s^(50-T)
setting the derivative to 0 and multiplying out the (c*r^T - 10000) I get
[c * ln(r) * r^T * s^(50-T)] - [c * ln(s) * s^(50-T) * r^T] + [10000* ln(s) * s^(50-T)] = 0
take the 10000 ln(s)... term across, cancel the s^(50-T) terms and do some jiggery pokery with brackets and logarithms and i got:
r^T = (10000 * ln(s)) / (c * ln(s/r) )
and this is equivalent to:
T = ln [(10000 * ln(s)) / (c * ln(s/r) )] / [ln(r)]
very messy answer, but an answer nevertheless
btw i've most likely made a mistake or 2 somewhere. I'm confident up to the g' stage though.
ok, tried to do the same as you ilw, and i think your right.........
(c*r^T-10000)(-(ln s)*s^(50-T))+c(ln r)r^T*s^(50-T)=0
multiply out
-c*r^T*(ln s)*s^(50-T)-10000(ln s)*s^(50-T)+c(ln r)r^T*s^(50-T)=0
divide by s^(50-T)
-c*r^T*(ln s)-10000(ln s)+c(ln r)r^T=0
sort
(c(ln r)-c(ln s0))r^T=10000(ln s)
c(ln (r/s))r^T=10000(ln s)
move
r^T=10000(ln s)/c(ln r/s)
take the ln of the function
T=(ln (10000(ln s)/c(ln r/s)))/(ln r)
THANKS FOR THE HELP MAN!!!!!!
it is compound interest, but the interest increases after you pay 10 000Quote:
Originally Posted by cpt_azad
wtf :blink:
no problem. Haven't had to do something like that in quite a while. :D
hehehe......... last time i had to do it was...... uhhmmm, july 2000
trouble is, im prolly gonna have to do alot of that stuf next year again at the university :P
hmmmm, just found out there is no solution........
since r/s will be a numer smaller than 0
ln (r/s) will be a negative numer
and you cant take the log (with a positive base) of a negative numer :(
so no real nummer is in the solution..........