Answer
a) $\$14471.4$ million
b) $1998$
c) $1989$
Work Step by Step
Given \begin{equation}
N(t)=376.5 t^2+548.1 t+2318.4,\\
a= 376.5, b= 548.1, c= 2318.4
\end{equation} a) Set $t= 5$ to estimate the annual net sales for Home Depot in $1995$.
\begin{equation}
\begin{aligned}
N(5) & =376.5 \cdot 5^2+ 548.1\cdot 5+2318.4 \\
& =14471.4
\end{aligned}
\end{equation} b) Set $N(t)= 30000$ to find the values of $t$, which gives the year(s) that the sale will reach $\$30,000$ million. \begin{equation}
\begin{aligned}
376.5 t^2+548.1 t+2318.4& =30000 \\
3765 t^2+5481 t+23184&=300000\\
3765 t^2+5481 t-276816&=0.
\end{aligned}
\end{equation} Solve the equation: $$
\begin{aligned}
t& =\frac{-5481 \pm \sqrt{5481^2-4 \cdot 3765(-276816)}}{2 \cdot 3765}\\
& =\frac{-5481 \pm \sqrt{4198890321}}{7530}\\
& \approx -0.728\pm 8.605.
\end{aligned}
$$ The solutions are: $$
\begin{aligned}
t_1& =-0.728- 8.605 \\
&\approx -9.33 \\
t_2 & =-0.728+8.605 \\
& \approx7.88\approx 8.
\end{aligned}
$$ It was in about $1998$ that the sale was $\$30,000$.
c) The vertex of the revenue function will give us the maximum revenue that we are looking to maximize.Use $a= -1.5$ and $b= 30$ into the following formula.
$$
\begin{aligned}
& t=\frac{-b}{2 a}=\frac{-548.1}{2(376.5)}=-0.728 \\
& N_{\text {max }}=N(-0.728 ) \\
&=376.5(-0.728 )^2+548.1 \cdot(10)+ 2318.4\\
&=2118.92
\end{aligned}
$$ The vertex is $(-0.728,2118.92)$. This means that the company had the lowest sales of $\$2118.92$ million in $1989$.
