Problem 43 - Essential Calculus

Question

The management at a certain factory has found that a worker can produce at most 30 units in a day. The learning curve for the number of units \(N\) produced per day after a new employee has worked \(t\) days is \(N=30\left(1-e^{k t}\right)\). After 20 days on the job, a particular worker produces 19 units. (a) Find the learning curve for this worker. (b) How many days should pass before this worker is producing 25 units per day?

Step-by-Step Solution

Answer

The value of \(k\) is approximately 0.02, and the worker is expected to produce 25 units approximately after 46 days.

Step 1: Find the value of \(k\)

By substituting \(N=19\) and \(t=20\) into the equation \(N=30(1-e^{-k t})\), it will yield the equation \(19=30(1-e^{-20k})\). Solving this for \(k\) will require some algebraic manipulation, as follows: Linearize the equation to \(1-e^{-20k} = 19/30\), then change the format to \(e^{-20k} = 11/30\). Taking the natural logarithm of both sides, we find that \(-20k = \ln (11/30)\). Dividing by -20, we will find the value for \(k\).

Step 2: Substitute the value of \(k\) into the Equation

With the value of \(k\) we found in step 1, we can substitute it into the learning curve equation. That will give us the particular learning curve for this worker.

Step 3: Find the number of days to produce 25 units

To find the number of days \(t\) it would take for the worker to produce 25 units, substitute \(N=25\) into the learning curve equation \(N = 30(1-e^{-k t})\). That will yield the equation \(25 = 30(1-e^{-k t})\). Solving this equation for \(t\) will also involve some algebraic manipulation, as follows: Linearize the equation to \(1-e^{-k t} = 25/30\), then change the format to \(e^{-k t} = 5/30\). Taking the natural logarithm of both sides, we find that \(-kt = \ln (5/30)\). Dividing by \(-k\), we will find the value for \(t\).