s(t)= -3t^3+27t^2-36t+20; t≥0 a) Determine the time when velocity is at max b) Justify that the velocity at this time is max and not a min?

s(t) = -3t³ + 27t² - 36t + 20

a) v(t) => s '(t) = -9t² + 54t - 36

i.e. -9(t² - 6t + 4)

or, when -9(t - 3)² + 45

b) as the coefficient of t² is negative the function is  ∩ - shaped, hence a maximum turning point.

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s(t)= -3t^3+27t^2-36t+20; t≥0 . Take first derivative. That is velocity.

v= -9t^2+54t-36=0, t^2-6t+4=0. t=+(6+/-sqr(36-16))/2=3 +/- sqrt 5. There are two values. 3+sqrt 5 and 3-sqrt 5. Let us take the second derivative of v. dv/dt= -18 t+54.

dv/dt= -18(3+sqrt 5), dv/dt =-18(3-sqrt 5) Velocity is a maximum at  3+sqrt 5