Posted by Leo on Monday, July 8, 2013 at 12:40am.
I'm asked to find max and min of: L(t)=12+2.8sin((2pi/365)(t80).
I find the derivative as:
L'(t)=(5.6*pi/365)*cos[(2pi/365)(t80)]
but I get lost afterwards.

Calculus 1  Steve, Monday, July 8, 2013 at 3:39am
You don't really need any derivatives for this. You know sin(z) has a max od 1 and a min of 1. So, when (2pi/365)(t80) is an odd multiple of pi/2, sin is 1 or 1, so the min/max of L is 12±2.8
If you want to find the values of t where these extrema occur, then recall that max/min occur when L'=0. So, when does
cos[(2pi/365)(t80)] = 0?
cos(z)=0 when z is an odd multiple of pi/2. So, for integer values of k, we need
(2pi/365)(t80) = (2k+1)*pi/2
t80 = (2k+1) pi/2 * 365/2pi
t80 = 365(2k+1)/4
t = 80 + 365(2k+1)/4
t = 80±365/4, 80±1095/4, ...
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