Let's analyze the problem step-by-step and confirm the solution:
### Given:
The total sales, [tex]\( S(t) \)[/tex], of a certain movie in thousands of DVDs are given by the formula:
[tex]\[ S(t) = \frac{40 t^2}{t^2 + 150} \][/tex]
### (a) Find [tex]\( S'(t) \)[/tex].
The derivative, [tex]\( S'(t) \)[/tex], has already been provided:
[tex]\[ S'(t) = \frac{-80 t^3}{(t^2 + 150)^2} + \frac{80 t}{t^2 + 150} \][/tex]
### (b) Find [tex]\( S(10) \)[/tex] and [tex]\( S'(10) \)[/tex].
First, we need to find [tex]\( S(10) \)[/tex]:
[tex]\[ S(10) = \frac{40 \cdot 10^2}{10^2 + 150} \][/tex]
Evaluating inside the fraction:
[tex]\[ S(10) = \frac{40 \cdot 100}{100 + 150} = \frac{4000}{250} = 16.0 \][/tex]
So, [tex]\( S(10) = 16.0 \)[/tex] (already rounded to the nearest hundredth).
Next, we compute [tex]\( S'(10) \)[/tex]:
[tex]\[ S'(10) = \frac{-80 \cdot 10^3}{(10^2 + 150)^2} + \frac{80 \cdot 10}{10^2 + 150} \][/tex]
Evaluating each part separately:
1. [tex]\( \frac{-80 \cdot 10^3}{(10^2 + 150)^2} = \frac{-80 \cdot 1000}{(100 + 150)^2} = \frac{-80000}{250^2} = \frac{-80000}{62500} = -1.28 \)[/tex]
2. [tex]\( \frac{80 \cdot 10}{10^2 + 150} = \frac{800}{250} = 3.2 \)[/tex]
Now, adding the results of both parts:
[tex]\[ S'(10) = -1.28 + 3.2 = 1.92 \][/tex]
Thus, [tex]\( S'(10) = 1.92 \)[/tex] (already rounded to suitable precision).
### Final Answer:
The value of [tex]\( S(10) \)[/tex] rounded to the nearest hundredth is [tex]\( 16.0 \)[/tex].
The value of [tex]\( S'(10) \)[/tex] is [tex]\( 1.92 \)[/tex].
