Answer
(a) $-\frac{1}{10}x^2+150x$, $0\le x\le 1500$.
(b) $14,000$ dollars.
(c) $750$ units, $56,250$ dollars
(d) $75$ dollars.
Work Step by Step
Given $p=-\frac{1}{10}x+150, \ 0\le x\le 1500$, we have
(a) The revenue is $R(x)=px=-\frac{1}{10}x^2+150x$, $0\le x\le 1500$.
(b) Let $x=100$, we have $R(100)=-\frac{1}{10}(100)^2+150(100)=14,000$ dollars.
(c) A maximum happens when $x=-\frac{150}{2(-1/10)}=750$ units. And the maximum revenue is $R(750)=-\frac{1}{10}(750)^2+150(750)=56,250$ dollars
(d) The price at this point is $p=-\frac{1}{10}(750)+150=75$ dollars.
