amount is given by A(t) = 4000 + 48(t - 3) - 4(t - 3)^3

[Math wiz ya rite 09]

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 beggining of the workday at 8:00 am. At what time is the rate of production increasing most rapidly?

 

[skeeter]

find the time when A'(t) is a maximum.

 

[galactus]

Try finding f''(x). Set it to 0 and solve.

 

[Math wiz ya rite 09]

is the derivative -12(t^2 - 6t + 5)?

cause then its zeros are:

x=1,5

 

[skeeter]

A"(t) = 72 - 24t

72 - 24t = 0 at t = 3, which is the time when A'(t), the rate of production, is a maximum.

 

[Math wiz ya rite 09]

so 8:00 am + 3 hrs = 11:00 right

and by the way thanks

 

[Math wiz ya rite 09]

is that wat i do with the 3. or is there some other step?