Answer:
D. 40 < t < 60
Step-by-step explanation:
Given function,
[tex]N(t) = -3t^3 + 450t^2 - 21,600t + 1,100[/tex]
Differentiating with respect to x,
[tex]N(t) = -9t^2+ 900t - 21,600[/tex]
For increasing or decreasing,
f'(x) = 0,
[tex]-9t^2+ 900t - 21,600=0[/tex]
By the quadratic formula,
[tex]t=\frac{-900\pm \sqrt{900^2-4\times -9\times -21600}}{-18}[/tex]
[tex]t=\frac{-900\pm \sqrt{32400}}{-18}[/tex]
[tex]t=\frac{-900\pm 180}{-18}[/tex]
[tex]\implies t=\frac{-900+180}{-18}\text{ or }t=\frac{-900-180}{-18}[/tex]
[tex]\implies t=40\text{ or }t=60[/tex]
Since, in the interval -∞ < t < 40, f'(x) = negative,
In the interval 40 < t < 60, f'(t) = Positive,
While in the interval 60 < t < ∞, f'(t) = negative,
Hence, the values of t for which N'(t) increasing are,
40 < t < 60,
Option 'D' is correct.
