(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPdt represents the rate at which performance improves. If M is the maximum level of performance of which the learner is capable, then a model for learning is given by the differential equation

dPdt=k(M−P(t))

where k is a positive constant.

Two new workers, Andy and John, were hired for an assembly line.
Andy could process 12 units per minute after one hour
and 15 units per minute after two hours. John could
process 10 units per minute after one hour
and 16 units per minute after two hours. Using the above
model and assuming that P(0)=0, estimate the maximum number of
units per minute that each worker is capable of
processing.

Andy:..............??

John: .......... ???

Note: Please show clearly the final answer

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givendP/dt=k(M−P(t))dP/(M−P(t)) =k dtdP/(M(1−P(t)/M)) =kdtdP/(1−P(t)/M) =Mkdtintegrate on both sides ∫dP/(1−P(t)/M) = ∫Mkdt-M(ln(1-P(t)/M))=Mkt +c(ln(1-P(t)/M))=-kt +c(1-P(t)/M)=e-kt +c(1-P(t)/M)=Ce-ktP(t)/M=1-Ce-ktgiven P(0)=00/M=1-Ce-k00=1-CC=1P(t)/M=1-e-ktP(t)=M(1-e-kt)Andy could process 12 units per minute after onehour 12=M(1-e-k)=>e-k=1-(12/M) ------------>(1)15 units per minute after two hours15=M(1-e-2k)from (1)15=M(1-(1-(12/M))2)15=M(1-(1-(24/M) +(144/M2))15=M((24/M) -(144/M2))15=((24) -(144/M))144/M=24-15144/M=9M=144/9M=16maximum number of units per minute that ANDY is capable ofprocessing. =16==================================P(t)=M(1-e-kt)John could process 10 units per minute after onehour 10=M(1-e-k)=>e-k=1-(10/M) ------------>(1)16 units per minute after two hours16=M(1-e-2k)from (1)16=M(1-(1-(10/M))2)16=M(1-(1-(20/M) +(100/M2))16=M((20/M) -(100/M2))16=((20) -(100/M))100/M=20-16100/M=4M=100/4M=25maximum number of units per minute that JOHN is capable ofprocessing. =25 ...