Question

1. Provide an appropriate response.

Answer 1

Answer:

D. 40 < t < 60

Step-by-step explanation:

Given function,

$N(t) = -3t^3 + 450t^2 - 21,600t + 1,100$

Differentiating with respect to x,

$N(t) = -9t^2+ 900t - 21,600$

For increasing or decreasing,

f'(x) = 0,

$-9t^2+ 900t - 21,600=0$

By the quadratic formula,

$t=\frac{-900\pm \sqrt{900^2-4\times -9\times -21600}}{-18}$

$t=\frac{-900\pm \sqrt{32400}}{-18}$

$t=\frac{-900\pm 180}{-18}$

$\implies t=\frac{-900+180}{-18}\text{ or }t=\frac{-900-180}{-18}$

$\implies t=40\text{ or }t=60$

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.