Answer:
a). k = -0.1743
b). [tex]\lim_{t \to \infty}S=50[/tex] is be the saturation point.
c). 29000 units will be sold.
Step-by-step explanation:
The cumulative sales of a new product after t years is modeled by
S = [tex]50(1-e^{kt})[/tex]
a). During first year number of units sold were 8000.
That means for t = 1 and S = 8 (Since S is in thousands of units),
8 = [tex]50(1-e^{k})[/tex]
[tex]1-e^{k}=\frac{8}{50}[/tex]
[tex]1-e^{k}=0.16[/tex]
[tex]e^{k}=0.84[/tex]
By taking natural log on both the sides
[tex]ln(e^{k})=ln(0.84)[/tex]
k = -0.1743
b). To get the saturation point,
[tex]\lim_{t \to \infty}S= \lim_{t \to \infty}50(1-e^{kt})[/tex]
[tex]\lim_{t \to \infty}S= \lim_{t \to \infty}50(1-\frac{1}{e^{0.1743t}})[/tex]
Since [tex]\lim_{t \to \infty}\frac{1}{e^{0.1743t}}=0[/tex]
Therefore, [tex]\lim_{t \to \infty}S=50[/tex] will be the saturation point.
c). For t = 5, we have to find the number of units sold.
S = [tex]50(1-e^{-0.1743\times 5})[/tex]
= 50(1 - 0.4813)
= 29000 units will be sold after 5 years.
