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The camera division of Eastman Optical produces a certain model of digitalcamera. After completing the basic training program, the daily number ofcameras that an employee can assemble is given byQ(t) = 60 - 30e-0.51where Q represents the number of cameras that can be assembled, and t isthe number of months after completing the training.a.) How many cameras can be assembled by an employee who has worked for2 months? Type your answer (just the number) in the space below. Round tothe nearest whole number.b.) What is the maximum number of cameras that Eastman Optical can expectits employees to assemble? In other words, what is the limit as t approachesinfinity? Type your answer (just the number) in the space below.

Answer :

numberThe amount of cameras that can be assembled by the employee after training is given by;

[tex]Q=60-30e^{-0.5t}[/tex]

1. To find the number of cameras that can be assembled by an employee who has worked for two months, we insert t = 2 in the equation, we get;

[tex]\begin{gathered} Q=60-30e^{-0.5\times2} \\ Q=60-30e^{-1} \\ Q=48.96\approx49\text{ cameras} \end{gathered}[/tex]

Theref

2. The maximum number of cameras can be expected from an employee that has worked a long time ( close to "infinity" time), so as t approaches infinity, we get;

[tex]\begin{gathered} Q(\infty)=60-30e^{-0.5\times\infty} \\ -0.5\times\infty\text{ is simply }\infty,\text{ so} \\ Q=60-30e^{-\infty} \\ e^{-\infty}=0,\text{ so} \\ Q=60-30(0) \\ Q=60\text{ cameras} \end{gathered}[/tex]

Thus, the maximum number of cameras that can be expected is 60

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