Question

the number of times a new pop song has been downloaded t weeks after its initial release is given by
i(t)=\frac{4,000,000}{1 + 300e^{-0.45t}}
a) after how many weeks is the rate of change of the number of downloads maximized?
b) what is the rate of change of the number of downloads at the time found in part (a)?
c) how many times has the song been downloaded at the time found in part (a)?
a) the rate of change of the number of downloads is maximized after (square) weeks
(do not round until the final answer. then round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the derivative of a logistic function

For a function \( y=\frac{L}{1 + ae^{-kt}} \), the derivative \( y'=\frac{Lkae^{-kt}}{(1 + ae^{-kt})^2} \). Here \( L = 4000000\), \(a = 300\), \(k=0.45\). So \( l'(t)=\frac{4000000\times0.45\times300e^{-0.45t}}{(1 + 300e^{-0.45t})^2}=\frac{540000000e^{-0.45t}}{(1 + 300e^{-0.45t})^2}\).

Step2: Use the quotient rule to find the second - derivative

Let \(u = 540000000e^{-0.45t}\) and \(v=(1 + 300e^{-0.45t})^2\). Then \(u'=-243000000e^{-0.45t}\) and \(v'=2(1 + 300e^{-0.45t})(- 135e^{-0.45t})\).
By the quotient rule \(l''(t)=\frac{u'v - uv'}{v^2}\).
\(u'v=-243000000e^{-0.45t}(1 + 300e^{-0.45t})^2\)
\(uv'=-540000000e^{-0.45t}\times2(1 + 300e^{-0.45t})(-135e^{-0.45t})=145800000000e^{-0.9t}(1 + 300e^{-0.45t})\)
\(l''(t)=\frac{-243000000e^{-0.45t}(1 + 300e^{-0.45t})^2+145800000000e^{-0.9t}(1 + 300e^{-0.45t})}{(1 + 300e^{-0.45t})^4}\)
Set \(l''(t) = 0\). Factor out \( - 243000000e^{-0.45t}(1 + 300e^{-0.45t})\) from the numerator:
\(l''(t)=\frac{-243000000e^{-0.45t}(1 + 300e^{-0.45t})[(1 + 300e^{-0.45t})-600e^{-0.45t}]}{(1 + 300e^{-0.45t})^4}=0\)
\((1 + 300e^{-0.45t})-600e^{-0.45t}=0\)
\(1-300e^{-0.45t}=0\)
\(300e^{-0.45t}=1\)
\(e^{-0.45t}=\frac{1}{300}\)
Take the natural logarithm of both sides: \(-0.45t=\ln(\frac{1}{300})=-\ln(300)\)
\(t=\frac{\ln(300)}{0.45}\)
\(\ln(300)=\ln(3\times100)=\ln(3)+2\ln(10)\approx1.0986 + 4.6052=5.7038\)
\(t=\frac{5.7038}{0.45}\approx12.67\)

Answer:

\(12.67\)