Answer
The revenue is maximum (${\$} 625)$ when $p={\$} 12.50$
Work Step by Step
$R(p)=pq$
$=p(-4p+100)$
$=-4p^{2}+100\mathrm{p}.$
The graph is a parabola.
The coefficients are
$a=-4$ (opens down),
$b=100, $
$c=0$ (passes through the origin)
$\left[\begin{array}{llll}
\text{Vertex} & x_{V}=-b/(2a) & y_{V}=f(-b/(2a) & \\
& x_{V}=12.5 & y_{V}=625 & \\
& \bf V(12.5,625) & & \\\\
\text{Line of symmetry} & x=12.5\\ & & \\
\text{Zeros:} & -4p^{2}+100=0 & & \\
& -rp(p-25)=0 & & \\
& p=0,p=25 & \bf\approx(0,0),(25,0) &
\end{array}\right]$
Sketch with the above data.
The revenue is maximum (${\$} 625)$ when $p={\$} 12.50$
