Calculus- application?

Question

The Camera Division of the Eastman Optical Company produces a 35mm single-lenses reflex camera. Eastman's training departments that after completing the basic tdepartmeraining program, a new, previously inexperienced employee will be able to assemble F(t)=50-30e^(-0.5t) model F cameras per day, t months after the employee starts to work on the assembly line.
A how many model F cameras can a new employee assemble per day after basic training?

I don't know what should I put for t...????

B. How manyy model F cameras can an employee with 1 month of experience assemble per day? Two months experience? six months experience?

for 1 month F(1)=50-30e^(-0.5(1))=31.804
for 2 months = 38.964
for 6 months= 48.506
Is this correct?

Answer 1

Well, I guess the answer is how many cameras the emplyee can p´roduce as as soon as he finishes his basic training, that is, t=0. So, F(0) = 50 - 30 = 20

Answer 2

Hi!

A) t=0 because t is the number of months that the employee has been there (after basic training). So we set it up like

F(0) = 50-30e^(-0.5[0])
F(0) = 50-30e^(0)
F(0) = 50-30
F(0) = 20


So he should be able to assemble 20 cameras per day.

B) Those are all correct!

Answer 3

t = 0 at the start.
That should give you 20 cameras.

Then as t increases, it slowly levels off at 50.

EVerything you did looks right except for the first part. Put t = 0.

Answer 4

If I am reading your formula correctly this F(t) could be a differential equation. Therefore you cannot solve it by just plugging in (t). You need first to separate the variables and integrate, consult separation of variables method in a diffeq book.

In the first part they are asking for the equation, like I previously mentioned they want the solved differential equation with the general solution and then in part b you can plug in the months for the particular solution.

Answer 5

if you take t=0 you get 20.

so PROBABLY that is after basic training

the rest is ok.