Find the absolute minimum and absolute maximum values of f on the given interval.

Question

f(t) = 16 cos t + 8 sin 2t,    [0, 𝜋/2]

Answer 1

f(t) = 16cost + 8sin(2t)

f'(t) = -16sint + 16cos(2t) = -16sint + 16(1 - 2sin²t) = -16(2sin²t + sint - 1) = -16(2sint - 1)(sint + 1)

On the interval [0, π/2], f'(t) = 0 when t = π/6, The only critical number of f on the given interval is π/6.

To find the absolute extrema of f on the interval [0, π/2], evaluate f at each critical number and at the endpoints:

f(0) = 16

f(π/2) = 0 ⇐ absolute minimum value

f(π/6) = 12√3 ≈ 20.78 ⇐ absolute maximum value

Answer 2

f(t) = 16cost +8sin2t on the interval [0,pi/2]

find absolute maximum and minium

try the end points,

f(0) = 16(1) +8(0) = 16

f(pi/2)= 16(0) + 8(0) = 0

f'(t) = 16sint - 16cos2t = 0

sint -cos2t =0

sint= cos2t = 1- 2sin^2(t)

let x =sint

x = 1-2x^2

2x^2 +x -1 = 0

factor

(2x-1)(x+1) =0

set each factor =0, solve for x

x=-1, sint = 0, cost = sqr(1-sin^2(t)) = sqr1=1, t= 0

16cost +8sin2t = 16 + 8(0) = 16

t = 3pi/2 which is not in the interval [0,pi/2]

2x-1 = 0

2sint = 1

sint = 1/2

t = 30 degrees = pi/6 sin2t = sinpi/3 = sqr3/2 cost =cos30=sqr3/2

16cost +8sin2t = 8sqr3 +4sqr3 = 12sqr3 = about 12(1.732) = 20.785 = absolute maximum

give the absolute minimum f(pi/2) = 0

and absolute maximum = about 20.8