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Sagot :
Let's solve the problem step by step.
### Part (a): Finding the learning curve (finding [tex]\( k \)[/tex] )
We are given the learning curve equation for the number of units produced per day:
[tex]\[ N = 30 \left(1 - e^{-k t}\right) \][/tex]
We know that after 20 days ([tex]\( t = 20 \)[/tex]), the new employee produces 19 units ([tex]\( N = 19 \)[/tex]). Therefore, we can write the equation:
[tex]\[ 19 = 30 \left(1 - e^{-20k}\right) \][/tex]
We'll solve this equation for [tex]\( k \)[/tex] by isolating it. Let's start by moving all terms involving [tex]\( k \)[/tex] to one side:
[tex]\[ \frac{19}{30} = 1 - e^{-20k} \][/tex]
Subtracting 1 from both sides:
[tex]\[ \frac{19}{30} - 1 = - e^{-20k} \][/tex]
Which simplifies to:
[tex]\[ \frac{19}{30} - \frac{30}{30} = - e^{-20k} \][/tex]
[tex]\[ \frac{-11}{30} = - e^{-20k} \][/tex]
Removing the negative signs from both sides:
[tex]\[ \frac{11}{30} = e^{-20k} \][/tex]
Next, we take the natural logarithm of both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ \ln \left(\frac{11}{30}\right) = -20k \][/tex]
Thus:
[tex]\[ k = -\frac{1}{20} \ln \left(\frac{11}{30}\right) \][/tex]
Therefore, the value of [tex]\( k \)[/tex] is:
[tex]\[ k = \frac{\log(11/30)}{20} \][/tex]
### Part (b): Calculating the number of days to produce 25 units per day
Now we know [tex]\( k \)[/tex]. We need to find out how many days ([tex]\( t \)[/tex]) it will take for the employee to produce 25 units per day ([tex]\( N = 25 \)[/tex]). We use the learning curve equation again with [tex]\( N = 25 \)[/tex]:
[tex]\[ 25 = 30 \left(1 - e^{-kt} \right) \][/tex]
Substituting [tex]\( k \)[/tex] from part (a):
[tex]\[ 25 = 30 \left(1 - e^{- \frac{\log(11/30)}{20} t} \right) \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ \frac{25}{30} = 1 - e^{- \frac{\log(11/30)}{20} t} \][/tex]
[tex]\[ \frac{5}{6} = 1 - e^{- \frac{\log(11/30)}{20} t} \][/tex]
Subtracting 1 from both sides:
[tex]\[ \frac{5}{6} - 1 = - e^{- \frac{\log(11/30)}{20} t} \][/tex]
Simplifying:
[tex]\[ -\frac{1}{6} = - e^{- \frac{\log(11/30)}{20} t} \][/tex]
Removing the negative signs:
[tex]\[ \frac{1}{6} = e^{- \frac{\log(11/30)}{20} t} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ \ln \left(\frac{1}{6}\right) = - \frac{\log(11/30)}{20} t \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{20 \ln \left(\frac{1}{6}\right)}{\log(11/30)} \][/tex]
Therefore, the number of days for the employee to produce 25 units per day is:
[tex]\[ t = \frac{20 \ln (6) + 360i \pi \ln (6)}{ (\log(11/30))^2 + 324\pi^2 } \][/tex]
In summary:
a) The learning curve for the employee is given by [tex]\( N = 30 \left(1 - e^{\frac{\log(11/30)}{20} t} \right) \)[/tex]
b) The employee will produce 25 units per day after approximately [tex]\( \frac{20 \ln (6) + 360i \pi \ln (6)}{ (\log(11/30))^2 + 324\pi^2 } \)[/tex] days.
### Part (a): Finding the learning curve (finding [tex]\( k \)[/tex] )
We are given the learning curve equation for the number of units produced per day:
[tex]\[ N = 30 \left(1 - e^{-k t}\right) \][/tex]
We know that after 20 days ([tex]\( t = 20 \)[/tex]), the new employee produces 19 units ([tex]\( N = 19 \)[/tex]). Therefore, we can write the equation:
[tex]\[ 19 = 30 \left(1 - e^{-20k}\right) \][/tex]
We'll solve this equation for [tex]\( k \)[/tex] by isolating it. Let's start by moving all terms involving [tex]\( k \)[/tex] to one side:
[tex]\[ \frac{19}{30} = 1 - e^{-20k} \][/tex]
Subtracting 1 from both sides:
[tex]\[ \frac{19}{30} - 1 = - e^{-20k} \][/tex]
Which simplifies to:
[tex]\[ \frac{19}{30} - \frac{30}{30} = - e^{-20k} \][/tex]
[tex]\[ \frac{-11}{30} = - e^{-20k} \][/tex]
Removing the negative signs from both sides:
[tex]\[ \frac{11}{30} = e^{-20k} \][/tex]
Next, we take the natural logarithm of both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ \ln \left(\frac{11}{30}\right) = -20k \][/tex]
Thus:
[tex]\[ k = -\frac{1}{20} \ln \left(\frac{11}{30}\right) \][/tex]
Therefore, the value of [tex]\( k \)[/tex] is:
[tex]\[ k = \frac{\log(11/30)}{20} \][/tex]
### Part (b): Calculating the number of days to produce 25 units per day
Now we know [tex]\( k \)[/tex]. We need to find out how many days ([tex]\( t \)[/tex]) it will take for the employee to produce 25 units per day ([tex]\( N = 25 \)[/tex]). We use the learning curve equation again with [tex]\( N = 25 \)[/tex]:
[tex]\[ 25 = 30 \left(1 - e^{-kt} \right) \][/tex]
Substituting [tex]\( k \)[/tex] from part (a):
[tex]\[ 25 = 30 \left(1 - e^{- \frac{\log(11/30)}{20} t} \right) \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ \frac{25}{30} = 1 - e^{- \frac{\log(11/30)}{20} t} \][/tex]
[tex]\[ \frac{5}{6} = 1 - e^{- \frac{\log(11/30)}{20} t} \][/tex]
Subtracting 1 from both sides:
[tex]\[ \frac{5}{6} - 1 = - e^{- \frac{\log(11/30)}{20} t} \][/tex]
Simplifying:
[tex]\[ -\frac{1}{6} = - e^{- \frac{\log(11/30)}{20} t} \][/tex]
Removing the negative signs:
[tex]\[ \frac{1}{6} = e^{- \frac{\log(11/30)}{20} t} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ \ln \left(\frac{1}{6}\right) = - \frac{\log(11/30)}{20} t \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{20 \ln \left(\frac{1}{6}\right)}{\log(11/30)} \][/tex]
Therefore, the number of days for the employee to produce 25 units per day is:
[tex]\[ t = \frac{20 \ln (6) + 360i \pi \ln (6)}{ (\log(11/30))^2 + 324\pi^2 } \][/tex]
In summary:
a) The learning curve for the employee is given by [tex]\( N = 30 \left(1 - e^{\frac{\log(11/30)}{20} t} \right) \)[/tex]
b) The employee will produce 25 units per day after approximately [tex]\( \frac{20 \ln (6) + 360i \pi \ln (6)}{ (\log(11/30))^2 + 324\pi^2 } \)[/tex] days.
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