+9 Two people start off with the same salary - 25$. Each month when they get their salary, if they have spent all of the previous money they will get a 10% raise. The goal both want is to save up to 10 milion dollars before the other one. So the question really is, how much money should you earn per month before you start saving up get to 10 milion dollars before the other one?
How the salary works:
You spend all of your salary after the first month(25$) - you get 27.5 the next month.
--------->
You spend all of your salary after the second month(27.5$) - you get 30.22 the next month.
How much salary should you have before you start saving it all to first get to 10 milion dollars?
And also, just because one gets lets say 10.3 milion and the other one 10.1 milion at the same month, they draw if that happens
+118654 I'm going to do this a bit at a time
Let it take y time periods to save 10,000,000
The idea is to minimize y
So i spend for x time periods then I save for y-x time periods
spending
end of first month after being paid T1= 25
end of second month after being paid T2= 25* 1.1^1
....
end of the xth month after being paid Tx= 25* 1.1^(x-1)
now I am going to start saving everything.
end of (x+1) month after being paid I'll have T(x+1)= 2*[25* 1.1^(x-1)]
end of (x+2) month after being paid I'll have T(x+2)= 3*[25* 1.1^(x-1)]
end of (x+(y-x)) month after being paid I'll have T(x+(y-x))= (y-x+1)*[25* 1.1^(x-1)]
we want
\((y-x+1)*[25* 1.1^{x-1}]=10^7\\ \text{I want y to be the subject}\\ (y-x+1)*[25* 1.1^{x-1}]=10^7\\ y[25* 1.1^{x-1}]+(-x+1)[25* 1.1^{x-1}]=10^7\\ y[25* 1.1^{x-1}]=10^7-(-x+1)[25* 1.1^{x-1}]\\ y[25* 1.1^{x-1}]=10^7+(x-1)[25* 1.1^{x-1}]\\ y=\frac{10^7+(x-1)[25* 1.1^{x-1}]}{[25* 1.1^{x-1}]}\\ \)
Now I want to find the stationary points for y dy/dx=0
\(y=\frac{10^7+(x-1)[25* 1.1^{x-1}]}{[25* 1.1^{x-1}]}\\ y=\frac{10^7}{[25* 1.1^{x-1}]}+\frac{(x-1)[25* 1.1^{x-1}]}{[25* 1.1^{x-1}]}\\ y=\frac{10^7}{25}*1.1^{1-x}+x-1\\\)
I'm too lazy to work this out with calculus so here is the graph.
\(y=\frac{10^7}{25}*1.1^{1-x}+x-1\\\)
So if I stop saving at around 111.7 months then it will take around 121.2 months to have $10,000,000
if x=111
y=10^7/25*1.1^(1-111)+111-1 = 121.1908899233864998 that is 122 months
if x=112
10^7/25*1.1^(1-112)+112-1 = 121.1735362939877271 that is 122 months too
if x=113
10^7/25*1.1^(1-113)+113-1 = 121.248669358170661 that is 122 months too
if x=114
10^7/25*1.1^(1-114)+114-1 = 121.4078812347006009 that is 122 months too
if x=115
10^7/25*1.1^(1-115)+115-1 = 121.6435283951823645 that is 122 months too
if x=116
10^7/25*1.1^(1-116)+116-1 = 121.9486621774385131 that is 122 months too
if x=117
10^7/25*1.1^(1-117)+117-1 = 122.3169656158531938 that is 123 months
if x=110
10^7/25*1.1^(1-110)+110-1 = 121.3099789157251498
if x=109
10^7/25*1.1^(1-109)+109-1 = 121.5409768072976648
if x=108
10^7/25*1.1^(1-108)+108-1 = 121.8950744880274312
if x=107
10^7/25*1.1^(1-107)+107-1 = 122.3845819368301744 that is 123 months
So if what I have done is correct
I can spend for 108 to 116 months and it will take still take 122 months to accumulate
$10,000,000
I cannot accumulate 10,000,000 in less than 122 months.
This needs to be carefully checked.
