Answer
a. $R=-\frac{1}{6}x^2+100x,$
b. Domain of $R$ is $0\leq x\leq 600,$
c. $R(x)=13333.33,$
d. $R(x)=-\frac{1}{6}(x-300)^2+15000,$
e. $p=50$
Work Step by Step
$p=-\frac{1}{6}x+100,$
a. $R=xp,$
$R=x(-\frac{1}{6}x+100)=-\frac{1}{6}x^2+100x,$
b. Domain of $R$ is the quantity $x$ of certain product sold, which is $600\geq x\geq0,$
c. $R(x)=-\frac{1}{6}(200)^2+100(200)=13333.33,$
d. $R(x)=-\frac{1}{6}x^2+100x,$
$=-\frac{1}{6}(x^2-600x),$
$=-\frac{1}{6}(x^2-600x+90000)+15000,$
$=-\frac{1}{6}(x-300)^2+15000,$
Therefore, $x=300$ is going to maximize revenue and the maximum revenue is $15000,$
e. for $x=300,$ the price for the maximum revenue is,
$p=-\frac{1}{6}(300)+100=50$
