Answer
The revenue is maximum (${\$} 7500$) when $p={\$} 50$
Work Step by Step
$R(p)=pq$
$=p(-3p+300)$
$=-3p^{2}+300p.$
The graph is a parabola.
The coefficients are
$a=-3$ (opens down),
$b=300, $
$c=0$ (passes through the origin)
$\left[\begin{array}{llll}
\text{Vertex} & x_{V}=-b/(2a) & y_{V}=f(-b/(2a) & \\
& x_{V}=50 & y_{V}=7500 & \\
& \bf V(50,7500) & & \\\\
\text{Line of symmetry} & x=12.5\\ & & \\
\text{Zeros:} & -3p^{2}+300=0 & & \\
& -3p(p-100)=0 & & \\
& p=0,p=100 & \bf\approx(0,0),(100,0) &
\end{array}\right]$
Sketch with the above data.
The revenue is maximum (${\$} 7500$ when $p={\$} 50$
