The amount A(t) of a certain item produced in a factory is given by A(t)= 4000+48(t-3)-4(t-3)^3 Where t is the number of hours of production since the beginning of the workday at 8 am. At what time is the rate of production increasing most rapidly?

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The amount A(t) of a certain item produced in a factory is given by A(t)= 4000+48(t-3)-4(t-3)^3 Where t is the number of hours of production since the beginning of the workday at 8 am. At what time is the rate of production increasing most rapidly?

Mathematics
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find double derivative and set equal to zero solve
and make sure the first derivative is positive
48-12(t-3)^2 is the first dirivative?

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looks right to me
when i got it equal to 0 i got t= 3
from there i dont know
you set the second derivative to zero -24(t - 3) = 0 t = 3
yes, then what?
it reaches the most rapid in 3 hours you start at 8 am plus 3 equal 11 am
yo man you just blew my mind! thank you sooo much! now i know but can you explain a bit? why the 2nd derivative?
sure the first derivative will give you the slope yes? if the slope is positive, then the rate at which it's being produced is increasing now here's the confusing part, so bare with me the second derivative will give you the slopes of the rate of production just like how when the first derivative hits zero, it's a max/min when the second derivative hits zero, it's when the rate of production is max/min so you solve for that make sense?
i have to go now but if this doesn't make sense, i'll try coming on again later good luck!
no it made perfecet sense! I just dont pay attention in class as much, ha! well ok thank you and have a good day!

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