(1) profit function p(x)=R(x)-C(x)=-2x2+250x-300, x ∈ [1, 100], x ∈ n;

The marginal profit function MP (x) = p (x+1)-p (x) = [-2 (x+1) 2+250 (x+1)-300]-(-2x2+250x-300)

(2) profit function p (x) =-2x2+250x-300 =-2 (x-62.5) 2+7512.5, x ∈ [1, 100], x ∈.

There is a maximum value p(x)max=75 12 (yuan);

Because Mp(x)=248-4x is a decreasing function, there is a maximum value of 244 when x= 1.