<s>[INST] Find maximum of a function I want to find the maximum of a function. 
$$ d = \frac{35}{3} + \frac{7}{3}\sin( \frac{2\pi}{365}t ) $$
I don't know if I applied the chain rule correctly.
$$ w = \frac{2\pi}{365}t $$
$$ w' = \frac{2\pi}{365} $$
$$ g = \sin(w) $$
$$ g' = \cos(w) $$
$$ f = \frac{35}{3} + \frac{7}{3}g $$
$$ f' = \frac{7}{3}g $$
$$ d' = \frac{7}{3}\sin( \frac{2\pi}{365}t ) \times \cos( \frac{2\pi}{365}t ) \times \frac{2\pi}{365} $$
I set it to $0$
$$ a\times b \times c = 0 $$
And it seems that I can't solve this because $x = 0$
$$ \arcsin(a) = \arcsin(0) \\
\frac{2\pi}{365}x = 0 \\
x = 0$$
 [/INST] Since $\sin(s)\in [-1,1]$ for any $s$, we have
$$d=\frac{35}{3} + \frac{7}{3}\sin\Big( \frac{2\pi}{365}t \Big)\leq \frac{35}{3} + \frac{7}{3}=\frac{42}{3}=14$$
for every $t$. Now, you have to find $t$ such that $\sin\Big( \frac{2\pi}{365}t \Big)=1$. By periodicity of $\sin$ this is true whenever $$\frac{2\pi}{365}t = \frac{\pi}{2}+2k\pi \iff t = \frac{365}{2\pi}\Big(\frac{\pi}{2}+2k\pi \Big)=\frac{365}{4}+k365$$ with $k\in \Bbb Z$.
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