Need help!

[gressi lika]

The number of units, N, produced each hour by a new employee after t hours on the
job satisfies
dN/dt=10e^0.1t
Assuming that she is unable to produce any goods at the beginning of her shift, calculate
how many she can produce per hour after eight hours. How many units per hour
will she be able to produce in the long run?

 

[Deleted member 4993]

The number of units, N, produced each hour by a new employee after t hours on the
job satisfies
dN/dt=10e^0.1t
Assuming that she is unable to produce any goods at the beginning of her shift, calculate
how many she can produce per hour after eight hours. How many units per hour
will she be able to produce in the long run?

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Please share your work/thoughts about this problem

 

[gressi lika]

Well, I think that the first step is to integrate N with respect t, it will give us 100e^01t+c
I'm not sure if I should do the first equation like 0=100e^0.1*0+c and it will give us c=-100
I really don't understand what should I do in this situation.

 

[HallsofIvy]

Very good! Yes, dN/dt=10e^(0.1t) leads to dN= 10e^(0.1t)dt and integrating both sides,N(t)= 100e^(0.1t)+c.

But, "Assuming that she is unable to produce any goods at the beginning of her shift" is, at best, ambiguous! I think they mean to say that there are no goods (from a previous shift perhaps) so N(0)= 0. Setting t=0 in N(t)= 100e^(0.1t)+ c, we have N(0)= 100+ c= 0 so c= -100. We have N(t)= 100e^(0.1t)- 100.

Setting t= 8, after 8 hours, she can produce N(8)= 100e^(0.8)- 100= 122.55 per hour.

"
How many units per hour will she be able to produce in the long run?"
Typically "in the long run" means "as t goes to infinity" but 100e^(0.1t)- 100 goes to infinity as t goes to infinity.