(1) 900a + 30b + c = 30
In the above equation substitute a = -1/90 and c = 0, and find the value of b.
Everything looks good up until the moment you introduce the derivative (which you take correctly - it's how you use it I'm not so sure of). Why not continue solving for b in the usual way? You've found c and a. Any original equation should give you b. I think you need to plug in different values into the derivative.