Aresearcher has funds to buy enough computing power to number-crunch a problem in 5 years. computing power per dollar doubles every 23 months. when should he buy his computers to have finished the problem as soon as possible? give your answer as a decimal in months, accurate within 0.1 months suppose the problem would take c months on current computers. what is the largest value of c for which he should buy the computers immediately? give an answer as a decimal accurate to 0.1 months.

A) In t months, the number of months required to number-crunch the problem will be
  60*2^(-t/23)
By waiting t months, the researcher has made the total time f(t) to the solution of his problem be
  f(t) = t + 60*2^(-t/23)

The derivative of this is
  f'(t) = 1 + 60*ln(2)*(-1/23)*2^(-t/23)
We want to find the value of t that makes this be zero.
  0 = 1 - 60*ln(2)/23*2^(-t/23)
  2^(-t/23) = 23/(60*ln(2))
  (-t/23)*ln(2) = ln(23/(60*ln(2)))
  t = -23/ln(2)*ln(23/(60*ln(2))) ≈ 19.655

In order to finish his problem as soon as possible, the researcher should wait 19.7 months to buy his computers.

b) For this part of the problem, we want to find the value of "60" that makes t=0 be the solution. Taking the last expression and substituting t=0, 60=c, we get
  0 = -23/ln(2)*ln(23/(c*ln(2)))
  1 = 23/(c*ln(2)) . . . . . taking antilogs
  c = 23/ln(2) ≈ 33.2

The largest value of c for which he should buy the computers immediately is 33.2.

$400 per week is $1600 per month assuming theres 4 months