Answer
a) $375$
b) $10$ thousand cameras or $20$ thousand cameras
c) $15$ thousand digital cameras
Work Step by Step
Let's note the variable by $x$ not to interfere with the constant $c$ in the standard form of a quadratic equation.
Given \begin{equation} R(x)=-3 x^2+90 x,\\
a= -3, b= 90, c= 0.
\end{equation} a) Set $x= 5$ to find the revenue of selling $5$ thousand cameras.
\begin{equation}
\begin{aligned}
R(5) & =-3(5^2)+90(5) \\
& =375.
\end{aligned}
\end{equation} b) Set $R(x)= 600$ to find the values of $x$, which is the number of cameras that must be sold to earn a revenue of $600$ thousand. .
\begin{equation}
\begin{aligned}
-3 x^2+90x & =600 \\
\frac{-3 x^2+90 x}{-3} & =\frac{600}{-3} \\
x^2-30 x & =-200\\
x^2-30 x + 200&= 0 \\\
\end{aligned}
\end{equation} Determine $x$: $$
\begin{aligned}
x& =\frac{-(-30) \pm \sqrt{(-30)^2-4 \cdot (200)}}{2} \\
& =\frac{30 \pm 10}{2}\\
& = 15\pm 5
\end{aligned}
$$ $$
\begin{aligned}
x & =15-5 \\
& =10 \\
x & =15+5 \\
& =20 \\
\end{aligned}
$$ c) The vertex of the revenue function will give us the maximum revenue that we are looking for. Use $a= -3$ and $b= 90$ into the following formula.
$$
\begin{aligned}
& x=\frac{-b}{2 a}=\frac{-(90)}{2(-3)}= 15 \\
& R_{max}=-3 \cdot 15^2+90\cdot 15=675.
\end{aligned}
$$ The vertex is $(15,675)$. This means that they must sell $15$ thousand digital cameras to maximize revenue.
