The voltage v(t) in a certain circuit is given by the equation: v(t)=4t^3-16t^2+19t-6 V, for 0<= t<=3 seconds. Determine the global maximum and the global minimum for v(t) on the time interval 0<=t<=3 seconds.

Hanna Webster

Hanna Webster

Answered question

2022-11-05

The voltage v(t) in a certain circuit is given by the equation:
v ( t ) = 4 t 3 16 t 2 + 19 t 6  for  0 t 3 seconds
Determine the global maximum and the global minimum for v(t) on the time interval 0 t 3 seconds.

Answer & Explanation

dobradisamgn

dobradisamgn

Beginner2022-11-06Added 17 answers

Given function :
v ( t ) = 4 t 3 16 t 2 + 19 t 6
We have to find the Global Maximum and Minimum in the interval [0,3]
First we calculate the derivative of the given function :
v ( t ) = 4.3 t 2 16.2 t + 19 v ( t ) = 12 t 2 32 t + 19
Then we equate the derivative to 0 and solve for t:
v ( t ) = 0 12 t 2 32 t + 19 = 0 t = ( 32 ) ± ( 32 ) 2 4 12 19 2 12 t = 32 ± 4 7 24 t = 8 ± 7 6 t 1.77 , 0.89
Both these points are in the given interval. So, they are both critical points.
When t=0,
v ( 0 ) = 4 ( 0 ) 3 16 ( 0 ) 2 + 19 ( 0 ) 6 v ( 0 ) = 6
When t=1.77
v ( 1.77 ) = 4 ( 1.77 ) 3 16 ( 1.77 ) 2 + 19 ( 1.77 ) 6 v ( 1.77 ) = 22.180932 50.1264 + 33.63 6 v ( 1.77 ) = 0.3155
When t=0.89
v ( 0.89 ) = 4 ( 0.89 ) 3 16 ( 0.89 ) 2 + 19 ( 0.89 ) 6 v ( 0.89 ) = 2.819876 12.6736 + 16.91 6 v ( 0.89 ) = 1.0563
When t=3
v ( 3 ) = 4 ( 3 ) 3 16 ( 3 ) 2 + 19 ( 3 ) 6 v ( 3 ) = 108 144 + 57 6 v ( 3 ) = 15
So,
Global Maximum is at t=3 where v(t)= 15
Global Minimum is at t=0 where v(t)= -6

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