Calculus rate of change questions The weekly sale of sandals $$ in thousands of pairs is given by $$S(t)=\frac{120}{t^2+100}$$ where $t$ is the number of weeks after the introduction of this style. After how many weeks do we have a maximum value?
I am confused as to what to do. Do we first need to find the derivative? Please help. 
 A: Hint:By AM-GM, $$ 
\frac1{t+\frac{100}t}\le \frac1{20}.
$$ (Equality when $t=\frac{100}t$.)
A: By the quotient rule we get $$S'(t)=120\frac{t^2+120-t\cdot 2t}{(t^2+100)^2}$$
The quotient rule $$(\frac{u}{v})'=\frac{u'v-uv'}{v^2}$$
A: This is a min/max problem. You basically need to find $t>0$ that makes the original expression give you the maximum value. You do that by first finding the derivative of the given function, setting the found derivative to zero and then solving it for $t$.
$$
S'(t)=\left(\frac{120}{t^2+100}\right)'=
\frac{(120t)'(t^2+100)-120t(t^2+100)'}{(t^2+100)^2}=\\
\frac{12,000-120t^2}{(t^2+100)^2}
$$
$$
\frac{12,000-120t^2}{(t^2+100)^2}=0\implies\\
12,000-120t^2=0\implies\\
t=\pm10
$$
Since we're only looking at $t$'s that are greater than zero, we choose the answer that's positive: $t=10$ weeks. That's the value that when plugged into the original function gives you a functional value that's maximal.
A: Hint:  Recall that you can find relative max and min by solving 
$$f'(t)=0$$
or by determining where $f'(t)$ doesn't exist. Then look at the behavior of the derivative to the left and right of your found critical values.
