Step 1

The demand function p(x) gives the price for x units sold, so in this problem x in the attendance, p(x) is the ticket price. From part (a):

\(\displaystyle{p}{\left({x}\right)}=-{\frac{{{1}}}{{{3000}}}}{x}+{19}\)

Multiply by x to get the revenue function

\(\displaystyle{R}{\left({x}\right)}={x}\cdot{p}{\left({x}\right)}=-{\frac{{{1}}}{{{3000}}}}{x}^{{2}}+{19}{x}\)

Step 2

\(\displaystyle\text{Find where }\ {R}'={0}\)

\(\displaystyle{R}'{\left({x}\right)}={0}=-{\frac{{{2}}}{{{3000}}}}{x}+{19}\)

\(\displaystyle{0}=-{\frac{{{1}}}{{{1500}}}}{x}+{19}\)

\(\displaystyle{\frac{{{1}}}{{{1500}}}}{x}={19}\)

\(\displaystyle{x}={28500}\)

\(\displaystyle\text{Since }\ {R}{''}=-{\frac{{{1}}}{{{1500}}}}{<}{0},{R}\ \text{ is concave donw here and this is a maximum}\)

Step 3

Plug the attendance x=28500 into p(x) to get the price

\(\displaystyle{p}{\left({28500}\right)}=-{\frac{{{1}}}{{{3000}}}}{\left({28500}\right)}+{19}=\${9.50}\)

Result

\(\displaystyle\${9.50}\)