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) = -20t^2/3
the company plans to stop manufacturing this computer when monthly sales reach 600 computers. if monthly sales now (t = 0) are 1,600 computers,
find s(t). how long will the company continue to manufacture this computer?
s(0) = □

Explanation:

Step1: Integrate the rate function

We know that if \(S^{\prime}(t)\) is the rate of change of \(S(t)\), then \(S(t)=\int S^{\prime}(t)dt\). Given \(S^{\prime}(t)=-\frac{200}{3}\), then \(S(t)=-\frac{200}{3}t + C\).

Step2: Find the constant of integration

When \(t = 0\), \(S(0)=1800\). Substitute into \(S(t)=-\frac{200}{3}t + C\), we get \(1800=-\frac{200}{3}(0)+C\), so \(C = 1800\). Then \(S(t)=-\frac{200}{3}t+1800\).

Step3: Solve for \(t\) when \(S(t)=600\)

Set \(S(t)=600\), so \(600=-\frac{200}{3}t + 1800\).
First, subtract \(1800\) from both sides: \(600 - 1800=-\frac{200}{3}t\).
\(- 1200=-\frac{200}{3}t\).
Then multiply both sides by \(-\frac{3}{200}\): \(t=\frac{-1200\times(-3)}{200}\).
\(t = 18\).

Answer:

\(18\)