Answer
(a) $ -\frac{1}{10}x^2+1000x$
(b) $384,000\ dollars$.
(c) $ 5000\ units$, $ 2,500,000\ dollars$.
(d) $ 500\ dollars$.
Work Step by Step
(a) Based on the given conditions, we can find $R(x)=px=-\frac{1}{10}x^2+1000x$
(b) Let $x=400$, we have $R(400)=-\frac{1}{10}(400)^2+1000(400)=384,000\ dollars$.
(c) With $a=-\frac{1}{10}, b=1000$, to maximizes revenue, we need $x=-\frac{b}{2a}=-\frac{1000}{2(-1/10)}=5000\ units$, and $R(5000)=-\frac{1}{10}(5000)^2+1000(5000)=2,500,000\ dollars$.
(d) At $x=500$, we have $p(5000)=-\frac{1}{10}(5000)+1000=500\ dollars$.