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?

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Answer 1 · 2017-03-22T12:05:36

2:36 pm

The productivity is given as:

# Q(t) = -t^3 + 9t^2 +12t #

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

Differentiating wrt #t# gives:

# (dQ)/dt = -3t^2 + 18t +12 #

At a critical point #(dQ)/dt=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