Let's break this problem into parts and solve it step-by-step.
### Part (a): Finding [tex]\( S'(t) \)[/tex]
The formula for the total sales [tex]\( S(t) \)[/tex] is given by:
[tex]\[ S(t) = \frac{40 t^2}{t^2 + 150} \][/tex]
To find the derivative [tex]\( S'(t) \)[/tex], we will use the quotient rule. The quotient rule for a function [tex]\( \frac{u(t)}{v(t)} \)[/tex] states:
[tex]\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \][/tex]
Here, [tex]\( u(t) = 40 t^2 \)[/tex] and [tex]\( v(t) = t^2 + 150 \)[/tex]. Now, we need to compute the derivatives [tex]\( u'(t) \)[/tex] and [tex]\( v'(t) \)[/tex]:
[tex]\[ u'(t) = (40 t^2)' = 80 t \][/tex]
[tex]\[ v'(t) = (t^2 + 150)' = 2t \][/tex]
Now, applying the quotient rule:
[tex]\[ S'(t) = \frac{(80 t)(t^2 + 150) - (40 t^2)(2t)}{(t^2 + 150)^2} \][/tex]
Simplify the numerator:
[tex]\[ S'(t) = \frac{80 t (t^2 + 150) - 80 t^3}{(t^2 + 150)^2} \][/tex]
[tex]\[ S'(t) = \frac{80 t t^2 + 80 t \cdot 150 - 80 t^3}{(t^2 + 150)^2} \][/tex]
[tex]\[ S'(t) = \frac{80 t^3 + 12000 t - 80 t^3}{(t^2 + 150)^2} \][/tex]
[tex]\[ S'(t) = \frac{12000 t}{(t^2 + 150)^2} \][/tex]
### Simplified derivative:
[tex]\[ S'(t) = \frac{12000 t}{(t^2 + 150)^2} \][/tex]
### Part (b): Evaluating [tex]\( S(10) \)[/tex] and [tex]\( S'(10) \)[/tex]
First, let's evaluate [tex]\( S(10) \)[/tex]:
[tex]\[ S(10) = \frac{40 \cdot 10^2}{10^2 + 150} \][/tex]
[tex]\[ S(10) = \frac{40 \cdot 100}{100 + 150} \][/tex]
[tex]\[ S(10) = \frac{4000}{250} \][/tex]
[tex]\[ S(10) = 16 \][/tex]
So, the total sales after 10 months is 16,000 DVDs (since [tex]\( S \)[/tex] is in thousands).
Next, let's evaluate [tex]\( S'(10) \)[/tex]:
[tex]\[ S'(10) = \frac{12000 \cdot 10}{(10^2 + 150)^2} \][/tex]
[tex]\[ S'(10) = \frac{120000}{(100 + 150)^2} \][/tex]
[tex]\[ S'(10) = \frac{120000}{250^2} \][/tex]
[tex]\[ S'(10) = \frac{120000}{62500} \][/tex]
[tex]\[ S'(10) = 1.92 \][/tex]
So, the rate of change of sales after 10 months is 1920 DVDs/month (since [tex]\( S' \)[/tex] is in thousands per month).
From the above values [tex]\( S(10) = 16 \)[/tex] and [tex]\( S'(10) = 1.92 \)[/tex], the correct interpretation is:
### Answer for interpretation
D. After 10 months, the total sales are 16,000 DVDs and the sales are increasing at the rate of 1920 DVDs per month.
### Part (c): Estimating total sales after 11 months
To estimate the total sales after 11 months, we use the values from part (b):
[tex]\[ S(10) = 16 \][/tex]
[tex]\[ S'(10) = 1.92 \][/tex]
Estimating the sales after 11 months:
[tex]\[ S(11) \approx S(10) + S'(10) \][/tex]
[tex]\[ S(11) \approx 16 + 1.92 \][/tex]
[tex]\[ S(11) \approx 17.92 \][/tex]
Since [tex]\( S(t) \)[/tex] is in thousands of DVDs, after 11 months, the total sales will be approximately:
[tex]\[ 17,920 \text{ DVDs} \][/tex]
Therefore, the total sales after 11 months will be approximately [tex]\( \boxed{17920} \)[/tex] DVDs.
