Question

Monthly sales of a particular personal computer are expected to decline at the following rate of S'(t) computers per​ month, where t is time in months and​ S(t) is the number of computers sold each month.
S'(t)= -30t^(2/3)

The company plans to stop manufacturing this computer when monthly sales reach 1,000 computers. If monthly sales now ​(t=​0) are 2,440 computers, find​ S(t). How long will the company continue to manufacture this​ computer?

Answer 1

Answer:

$S(t) = -18t^\frac{5}{3}+2440$

Company will take approximately 14 months

Step-by-step explanation:

Given function that shows the change rate of computers,

$S'(t)=-30 t^{\frac{2}{3}}$

Where,

t = number of months

On integrating,

$\int S'(t) = -30\int t^{\frac{2}{3}} dt$

$S(t) = -30(\frac{t^{\frac{2}{3}+1}}{\frac{2}{3}+1})+C$

$S(t) = -30 (\frac{t^{\frac{2+3}{3}}}{\frac{2+3}{3}})+C$

$S(t) = -30(\frac{t^\frac{5}{3}}{\frac{5}{3}})+C$

$S(t) = -30(\frac{3t^\frac{5}{3}}{5})+C$

$S(t) =-18t^\frac{5}{3}+C$

According to the question,

S(0) = 2,440,

$\implies -18(0)^\frac{5}{3}+C=2440\implies C = 2440$

Hence, the required function,

$S(t) = -18t^\frac{5}{3}+2440$

If S(t) = 1,000,

$-18t^\frac{5}{3}+2440=1000$

$-18t^\frac{5}{3}=-1440$

$t^\frac{5}{3}=80$

$t = (80)^\frac{3}{5}\approx 13.86$