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 8:00 am. At what times is the rate of production increasing most rapidly?

the rate of production would be A '(t)

this has a maximum when A ''(t) = 0

A '(t) = 48 - 12(t-3)^2
A ''(t) = -24(t-3

-24(t-3) = 0
t-3=0
t=3

look like 11:00 am

Ah, the joy of calculus! To find the times when the rate of production is increasing most rapidly, we need to look at the derivative of A(t).

Using the power rule and some algebraic simplification, we can find that A'(t) = 48 - 12(t - 3)².

To find where the rate of production is increasing most rapidly, we need to find where A''(t) = 0. But hold on a sec, let's bring out the clown car for some fun!

*Clown car revs up*

Alright, off we go! We'll solve A''(t) = 0 by setting 48 - 12(t - 3)² = 0.

*tires screeching*

Solving this equation, we get (t - 3)² = 4. Taking the square root of both sides, we find t - 3 = ±2. Wait, we made a pit stop to grab some popcorn!

*nom nom nom*

So, t - 3 = 2 or t - 3 = -2. Solving for t, we find t = 5 or t = 1. And there you have it, the clown-approved times when the rate of production is increasing most rapidly are at 5 and 1. Time to hit the road again, folks!

*Clown car drives away*

To determine when the rate of production is increasing most rapidly, we need to find the values of t for which the derivative of A(t) with respect to t is at its maximum or minimum.

Step 1: Find the derivative of A(t):
A'(t) = 48 - 12(t - 3)²

Step 2: Set the derivative equal to zero to find critical points:
48 - 12(t - 3)² = 0

Step 3: Solve for t:
12(t - 3)² = 48
(t - 3)² = 4
t - 3 = ±2
t = 3 ± 2
t = 1 or t = 5

Step 4: Test the values of t on either side of the critical points to determine if they correspond to maximum or minimum points:
We can use the second derivative test to determine whether the points are local maxima or minima.

Step 4.1: Find the second derivative of A(t):
A''(t) = -24(t - 3)

Step 4.2: Substitute the values of t into the second derivative:

For t = 1:
A''(1) = -24(1 - 3) = 48
Since the second derivative is positive, A(t) has a local minimum at t = 1.

For t = 5:
A''(5) = -24(5 - 3) = -48
Since the second derivative is negative, A(t) has a local maximum at t = 5.

Therefore, the rate of production is increasing most rapidly at t = 5, or 5:00 pm.

To determine the times at which the rate of production is increasing most rapidly, we need to find the maximum rate of change of the function A(t).

First, let's differentiate the function A(t) with respect to t to find the rate of change at any given time t:

A'(t) = 48 - 12(t - 3)²

Now, let's find the critical points of A'(t) by setting the derivative equal to zero:

48 - 12(t - 3)² = 0

Rearranging the equation, we have:

(t - 3)² = 4

Taking the square root of both sides, we get:

t - 3 = ±2

Solving for t, we have two critical points:

t = 3 + 2 = 5
t = 3 - 2 = 1

These critical points divide the time interval we are interested in, which is from the beginning of the workday at 8:00 am (t = 0) to the end of the workday.

So, the rate of production is increasing most rapidly at two specific times: 1 hour and 5 hours after the beginning of the workday at 8:00 am.