Answer
a. $p=0$ or $p=1000$
b. From the price range of $x=276.4$ to $x=723.6$ will the revenue exceed $800000$
Work Step by Step
$R(p)=-4p^2+4000p,$
a. $R(p)=-4p^2+4000p=0,$
$-4p(p-1000)=0,$
$-4p=0, p=0$ or $p-1000=0, p=100$
b. $R(p)>800000,$
$-4p^2+4000p>800000,$
$-4p^2+4000p-800000>0,$
Solving the quadratic equation using the quadratic formula,
$x=\frac{-b \pm \sqrt {b^2-4ac}}{2a},$
$x=\frac{-400 \pm \sqrt {(400)^2-4(-4)(-800000)}}{2(-4)}=\frac{-4000\pm800\sqrt{5}}{-8}=500\pm100\sqrt5,$
$x=276.4$ or $x=723.6$
Thus, From the price range of $x=276.4$ to $x=723.6$ will the revenue exceed $800000$
