Answer
The revenue is maximum (${\$} 20,000$) when $p={\$} 100$
Work Step by Step
$R(p)=pq$
$=p(-2p+400)$
$=-2p^{2}+400p.$
The graph is a parabola.
The coefficients are
$a=-2$ (opens down),
$b=400, $
$c=0$ (passes through the origin)
$\left[\begin{array}{llll}
\text{Vertex} & x_{V}=-b/(2a) & y_{V}=f(-b/(2a) & \\
& x_{V}=100 & y_{V}=20,000 & \\
& \bf V(100,20,000) & & \\\\
\text{Line of symmetry} & x=100\\ & & \\
\text{Zeros:} & -2p^{2}+400=0 & & \\
& -2p(p-200)=0 & & \\
& p=0,p=200 & \bf\approx(0,0),(200,0) &
\end{array}\right]$
Sketch with the above data.
The revenue is maximum (${\$} 20,000$) when $p={\$} 100$
