need math help

[Storm]

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

[ilw]

Differentiation? Sorry too lazy to do it myself
Differentiation of something to the power of x

Product rule


edit: then solve for derivative = 0, remmber to check is a maxima not minima

[Storm]

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

well anyway thanks for your help

[tesco]

[Skiz]

*scratches head*

[S!X]

*no clue*

[cpt_azad]

umm at first i thought it was compound interest, then i looked at ilw's post and now i'm just

[ilw]

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.

[Storm]

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!!!!!!

umm at first i thought it was compound interest, then i looked at ilw's post and now i'm just

it is compound interest, but the interest increases after you pay 10 000

[DanB]

wtf