Answer
a. $p=0$ or $p=3800$
b. $800
Work Step by Step
$R(p)=-\frac{1}{2}p^2+1900p,$
a. $R(p)=0,$
$-\frac{1}{2}p^2+1900p=0,$
$-\frac{1}{2}p(p-3800)=0,$
$-\frac{1}{2}p=0$ or $p-3800=0,$
$p=0$ or $p=3800$
b. $R(p)>1200000,$
$-\frac{1}{2}p^2+1900p>1200000,$
$-\frac{1}{2}p^2+1900p-1200000>0,$
Using the quadratic formula to solve the quadratic equation,
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a},$
$=\frac{-(1900) \pm \sqrt{(1900)^2-4(-\frac{1}{2})(-1200000)}}{2(-\frac{1}{2})},$
$=1900\pm \sqrt{1210000},$
$=1900\pm1100,$
$x=3000$ or $x=800$
$800
