Question

PLEASE HELP!!!

Answer 1

By finding the derivative of that equation, we can find the least value. Derivative will give us
$v'=(20.8 x^{2} -458.3x+3500)'$
$41.6x-458.3$. Setting this to zero we find that x≈11.017

Alternatively, we can find the vertex coordinates of that equation. And again we'll get the same result. In the attached picture, you can see this result much more clearly. 

Finally, the year is going to be 1985