The current 𝐼 amperes in an electrical circuit after 𝑡 seconds can be modelled by 𝐼=𝑡3−6𝑡2+9𝑡, for 0≤𝑡≤3. What is the largest value of 𝐼 in the period 0≤𝑡≤3?

Answer 1

For I = t³ − 6t² + 9t:

dI/dt = 3t^(3-1) − (2×6)t^(2-1) + 9t^(1-1)

or 3t² − 12t + 9.

d²I/dt² = 6t − 12.

By the First Derivative Test, solve 3t² − 12t + 9 or (3t − 3)(t − 3) = 0

to obtain t=1 or t=3.

By the Second Derivative Test, 6t − 12 is -6 (less than 0) at t=1

which indicates a relative maximum in the interval 0 ≤ t ≤ 3.

By the Second Derivative Test, 6t − 12 is 6 (greater than 0) at t=3

which indicates a relative minimum in the interval 0 ≤ t ≤ 3.

The largest value of I = t³ − 6t² + 9t in the interval 0 ≤ t ≤ 3 is then

(1)³ − 6(1)² + 9(1) or 4 Ampères.