Chapter 3 - Linear and Quadratic Functions - 3.3 Quadratic Functions and Their Properties - 3.3 Assess Your Understanding - Page 147: 86

$1900$ dollars. $1,805,000$ dollars.

Step 1. Given $R(p)=-\frac{1}{2}p^2+1900p$, let $a=-\frac{1}{2}, b=1900$, we know that the maximum of $R(p)$ happens when $p=-\frac{b}{2a}=-\frac{1900}{2(-1/2)}=1900$ dollars. Step 2. The maximum revenue is $R(1900)=-\frac{1}{2}(1900)^2+1900(1900)=1,805,000$ dollars.