Answer
a.$R(x)=-\frac{1}{10}x^2+150x,$ $0\leq x\leq1500$
b.$R(100)=14000$
c. $R(x)=-\frac{1}{10}(x-750)^2+56250,$
Thus, $x=750$ maximizes the revenue and the maximum revenue is $56250$
Work Step by Step
$p=-\frac{1}{10}x+150,$
a. $R(x)=xp,$
$R(x)=x(-\frac{1}{10}x+150,$
$=-\frac{1}{10}x^2+150x,$ $0\leq x\leq1500$
b. $R(100)=-\frac{1}{10}(100)^2+150(100),$
$=-1000+15000,$
$=14000$
c. $R(x)=-\frac{1}{10}x^2+150x,$
$=-\frac{1}{10}(x^2-1500x+562500)+56250,$
$=-\frac{1}{10}(x-750)^2+56250,$
Thus, $x=750$ maximizes the revenue and the maximum revenue is $56250$
