Answer:
[tex]S(t) = -18t^\frac{5}{3}+2440[/tex]
Company will take approximately 14 months
Step-by-step explanation:
Given function that shows the change rate of computers,
[tex]S'(t)=-30 t^{\frac{2}{3}}[/tex]
Where,
t = number of months
On integrating,
[tex]\int S'(t) = -30\int t^{\frac{2}{3}} dt[/tex]
[tex]S(t) = -30(\frac{t^{\frac{2}{3}+1}}{\frac{2}{3}+1})+C[/tex]
[tex]S(t) = -30 (\frac{t^{\frac{2+3}{3}}}{\frac{2+3}{3}})+C[/tex]
[tex]S(t) = -30(\frac{t^\frac{5}{3}}{\frac{5}{3}})+C[/tex]
[tex]S(t) = -30(\frac{3t^\frac{5}{3}}{5})+C[/tex]
[tex]S(t) =-18t^\frac{5}{3}+C[/tex]
According to the question,
S(0) = 2,440,
[tex]\implies -18(0)^\frac{5}{3}+C=2440\implies C = 2440[/tex]
Hence, the required function,
[tex]S(t) = -18t^\frac{5}{3}+2440[/tex]
If S(t) = 1,000,
[tex]-18t^\frac{5}{3}+2440=1000[/tex]
[tex]-18t^\frac{5}{3}=-1440[/tex]
[tex]t^\frac{5}{3}=80[/tex]
[tex]t = (80)^\frac{3}{5}\approx 13.86[/tex]
