Application of the second derivative (efficiency study)

[cmnalo]

An efficiency study conducted for the packaging company showed that the # of cases packed by the avg. worker (t) hours after starting at 8am is given by:

-t^3 + 4t^2 + 12t ; (0) equal to or less than( t) equal to or less than (4)

At what time during the morning shift is the avg. worker at peak efficiency?

f(t) = -t^3 + 4t^2 + 12t

f'(t) = -3t^2 + 8t + 12

f"(t) = -6t +8

-6t+8 = 0

t= 4/3 Is this the critical point? I thought I needed to find where F"(t)=0. The answer is 9am so I know t=1. How do I get there?

 

[galactus]

I was incorrect. Soroban showed you.

 

[cmnalo]

The peak will be at inflection point when second derivative =0.
-6t+8=0
t=4/3
or 1.33
or 1h 20min.
Make the answer 9:20
Does this look correct?

 

[galactus]

You were right. I was missing something. It would've helped if I read the problem more carefully . Sorry. Not to worry. Soroban to the rescue.

 

[soroban]

Hello, cmnalo!

There must be a typo in the book . . .

A study showed that the # of cases packed by the avg. worker \(\displaystyle t\) hours
after starting at 8 am is given by: \(\displaystyle \;N(t) \:=\:-t^3\,+\,4t^2\,+\,12t,\;\;0\:\leq \:t\:\leq\:4\)

At what time during the morning shift is the avg. worker at peak efficiency?

Solving \(\displaystyle N'(t)\,=\,0\) might tell us when the maximum number of cases is packed.

My interpretation is: "Efficiency" = number of cases per hour.
. . Hence, \(\displaystyle N'(t)\) is the Efficiency which we must maximize.

We have: \(\displaystyle \,N(t)\:=\:-t^3\,+\,4t^2\,+\,12t\)

. .\(\displaystyle E(t)\:=\:N'(t)\:=\:-3t^2\,+\,8t\,+\,12\)

Then:. . . \(\displaystyle E'(t)\:=\:-6t\,+\,8\:=\:0\;\;\Rightarrow\;\;t\:=\:\frac{4}{3}\)


Therefore, peak efficiency occurs at 9:20 am.