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

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Answer 1 · 2017-03-22T12:05:36

2:36 pm

The productivity is given as:

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

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

Differentiating wrt $t$ $t$ gives:

$\frac{d Q}{d t} = - 3 t^{2} + 18 t + 12$ $(dQ)/dt = -3t^2 + 18t +12$

At a critical point $\frac{d Q}{d t} = 0 \Rightarrow$ $(dQ)/dt=0 =>$

$- 3 t^{2} + 18 t^{+} 12 = 0$ $-3t^2 + 18t^ +12 = 0$
$:. t^2 -6t^ -4 = 0$
$:. t=3+-sqrt(13)$

We require $t>0 => t=3+sqrt(13)$

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

$(d^2Q)/dt^2 = -6t + 18$

When $t=3+sqrt(13) => (d^2Q)/dt^2 <0 =>$ maximum

Thus the maximum productivity occurs when $t=3+sqrt(13)$

ie, $t ~~ 6.60555 ...$ which correspond to a duration of 6h36m

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

The productivity of a company during the day is given by Q(t) = -t^3 + 9t^2 +12t Q(t)=−t3+9t2+12t 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?