Answer
The revenue is maximum (${\$} 72,000$) when $p={\$} 120$.
Work Step by Step
$R(p)=pq$
$=p(-5p+1200)$
$=-5p^{2}+1200p.$
The graph is a parabola.
The coefficients are
$a=-5$ (opens down),
$b=1200, $
$c=0$ (passes through the origin)
$\left[\begin{array}{llll}
\text{Vertex} & x_{V}=-b/(2a) & y_{V}=f(-b/(2a) & \\
& x_{V}=120 & y_{V}=72,000 & \\
& \bf V(120,72,000) & & \\\\
\text{Line of symmetry} & x=120\\ & & \\
\text{Zeros:} & -5p^{2}+1200=0 & & \\
& -5p(p-240)=0 & & \\
& p=0,p=240 & \bf\approx(0,0),(240,0) &
\end{array}\right]$
Sketch with the above data.
The revenue is maximum (${\$} 72,000$) when $p={\$} 120$
