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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.
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.