College Algebra, Chapter 1, 1.6, Section 1.6, Problem 94

When a manufacturer sells $x$ units of a certain product, then revenue $R$ and cost $C$ in dollars are given by

$
\begin{equation}
\begin{aligned}
R &= 20x\\
\\
C &= 2000 + 8x + 0.0025 x^2
\end{aligned}
\end{equation}
$



How many units the manufacturer should sell to have a profit at least \$ 2400?
Recall that, profit = revenue - cost

$
\begin{equation}
\begin{aligned}
20x - \left(2000 + 8x + 0.0025 x^2\right) & \leq 2400 && \text{Model}\\
\\
20x - 2000 - 8x - 0.0025 x^2 & \leq 2400 && \text{Subtract } 2400\\
\\
12x - 4400 - 0.0025x^2 & \leq 0 && \text{Divide both sides by } -0.0025\\
\\
-4800x + 1760000 + x^2 & \geq 0 && \text{Factor}\\
\\
(x - 4400)(x-400) & \geq 0
\end{aligned}
\end{equation}
$


We have, $x \geq 400$ and $x \geq 4400$
It shows that if the number of units sold is in the range of (the intersection of these inequalities) is $x \geq 4400$, then the profit will be at least \$2400.