The productivity of a company during the day is given by  Q(t) = -t^3 + 9t^2 +12t  at time t minutes after 8 o'clock in the morning. At what time is the company most productive?

Mar 22, 2017

2:36 pm

Explanation:

The productivity is given as:

$Q \left(t\right) = - {t}^{3} + 9 {t}^{2} + 12 t$

To find the optimum productivity we seek a critical point of $Q \left(t\right)$, and would expect to find a maxima.

Differentiating wrt $t$ gives:

$\frac{\mathrm{dQ}}{\mathrm{dt}} = - 3 {t}^{2} + 18 t + 12$

At a critical point $\frac{\mathrm{dQ}}{\mathrm{dt}} = 0 \implies$

$- 3 {t}^{2} + 18 {t}^{+} 12 = 0$
$\therefore {t}^{2} - 6 {t}^{-} 4 = 0$
$\therefore t = 3 \pm \sqrt{13}$

We require $t > 0 \implies t = 3 + \sqrt{13}$

We can do a second derivative test to verify this is a maximum;

$\frac{{d}^{2} Q}{\mathrm{dt}} ^ 2 = - 6 t + 18$

When $t = 3 + \sqrt{13} \implies \frac{{d}^{2} Q}{\mathrm{dt}} ^ 2 < 0 \implies$ maximum

Thus the maximum productivity occurs when $t = 3 + \sqrt{13}$

ie, $t \approx 6.60555 \ldots$ which correspond to a duration of 6h36m

As $t = 0$ was t 8AM then the optimum production time would be 2:36 pm