# Math Help - Rate at maximum

1. ## Rate at maximum

A question asks at what time of day is the rate of power consumption the greatest. t>=0 with noon coresponding t=0

The rate of consumption I derived, which is correct is $450cos(\frac{pi*t}{12})+700$

I know I have to set the Rate equal to zero to get the maximum
$450cos(\frac{pi*t}{12})+700=0$
$450cos(\frac{pi*t}{12})=-700$
$cos(\frac{pi*t}{12})=\frac{-700}{450}$
$arccos(cos(\frac{pi*t}{12}))=arccos(\frac{-700}{450})$
$\frac{pi*t}{12}=arccos(\frac{-700}{450})$

I stopped right here because my calculator didnt like the Arccos(700/450). Can I get a point in the right direction?

2. ## Re: Rate at maximum

What was the original function?

3. ## Re: Rate at maximum

$E(t) = 700t + \frac{5400}{pi}sin(\frac{pi*t}{12})$