1. 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

2. 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

3. 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

5. *no clue*

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

7. 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)]

btw i've most likely made a mistake or 2 somewhere. I'm confident up to the g' stage though.

8. 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

9. wtf

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