
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 $\displaystyle 450cos(\frac{pi*t}{12})+700$
I know I have to set the Rate equal to zero to get the maximum
$\displaystyle 450cos(\frac{pi*t}{12})+700=0$
$\displaystyle 450cos(\frac{pi*t}{12})=700$
$\displaystyle cos(\frac{pi*t}{12})=\frac{700}{450}$
$\displaystyle arccos(cos(\frac{pi*t}{12}))=arccos(\frac{700}{450})$
$\displaystyle \frac{pi*t}{12}=arccos(\frac{700}{450})$
I stopped right here because my calculator didnt like the Arccos(700/450). Can I get a point in the right direction?

Re: Rate at maximum
What was the original function?

Re: Rate at maximum
$\displaystyle E(t) = 700t + \frac{5400}{pi}sin(\frac{pi*t}{12})$