f'(t) = -16sin(t) + 16cos(t)
f'=0 when sin(t)=cos(t)
t = π/4
f(0) = 16
f(π/4) = 16/√2 + 8 = 8(1+√2) > 16
f(π/2) = 0
So, it looks like
max = 8(1+√2)
min = 0
Let's check the graph:
http://www.wolframalpha.com/input/?i=16+cos+t+%2B+8+sin+2t+for+0+%3C%3D+t+%3C%3D+pi%2F2
Find the absolute minimum and absolute maximum values of f on the given interval.
f(t) = 16 cos t + 8 sin 2t,
[0, π/2]
2 answers
nope max is actually 12sqrt(3)
1 answer