Chapter 3.5, Problem 36E

(a)

To determine

To find: the learning curve for this employee.

Answer to Problem 36E

The learning curve of the employee is $N=30\left(1-e^{-0.0501t}\right)$ .

Explanation of Solution

Given information:

The management at a plastics factory has found that the maximum number of units a worker can produce in a day is 30. The learning curve for the number $N$ of units produced per day after a new employee has worked $t$ days is modeled by $N=30\left(1-e^{kt}\right)$ . After 20 days on the job, a new employee produces 19 units.

Calculation:

After 20 days on the job, a new employee produces 19 units.

Therefore substitute $t=20$ and $N=19$ in the model,

  $\begin{array}{c}N=30\left(1-e^{kt}\right)\\ 19=30\left(1-e^{k\left(20\right)}\right)\\ \frac{19}{30}=1-e^{20k}\\ e^{20k}=1-\frac{19}{30}\\ =\frac{11}{30}\\ 20k=\ln \left(\frac{11}{30}\right)\\ k=\frac{1}{20}\left(-1.0033\right)\\ =-0.0501\end{array}$

Hence, the learning curve of the employee is $N=30\left(1-e^{-0.0501t}\right)$ .

(b)

To determine

To find: how many days does the model predict will pass before this employee is producing 25 units per day.

Answer to Problem 36E

36 days should pass before this employee is producing 25 units per day.

Explanation of Solution

Given information:

Calculation:

Now employee produces 25 units per day.

Therefore substitute $N=25$ in $N=30\left(1-e^{-0.0501t}\right)$

  $\begin{array}{c}25=30\left(1-e^{-0.0501t}\right)\\ \frac{25}{30}=1-e^{-0.0501t}\\ e^{-0.0501t}=1-\frac{25}{30}\\ e^{-0.0501t}=\frac{5}{30}\\ -0.0501t=\ln \left(\frac{5}{30}\right)\\ t=\frac{1}{-0.0501}\left(-1.7917\right)\\ =35.76\\ \approx 36\end{array}$

Hence, 36 days should pass before this employee is producing 25 units per day.