max R when R'=0
R' = -2x+400
R'=0 when x=200
In a way, you are on the right track. R is a parabola, whose vertex is midway between the roots. That is, at x=200. Using R' always works, though, even when R is not a nice easy function like a quadratic. This is a calculus class. All these max/min problems involve the derivative, so get used to that idea.
Find the maximum revenue using the following equations: R(x)=-x^2 +400x and C(x)=x^2+40x+100.
What I've done so far is use R(x) and solve for x, which got me 0 and 400. Is that how you're supposed to start it? What happens next? (I apologize; I'm really bad at teaching myself how to work problems without a teacher.)
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