A company’s profit $P(x)$ in thousands of dollars is modeled by $P(x) = -2x^2 + 12x - 16$, where $x$ is the number of units sold in hundreds. Find the number of units sold to maximize profit.

February 22, 2026

$A company’s profit $P(x)$ in thousands of dollars is modeled by $P(x) = -2x^2 + 12x - 16$, where $x$ is the number of units sold in hundreds. Find the number of units sold to maximize profit.$

Probability of second red card: $\frac{25}{51}$.
Combined probability: $\frac{1}{2} \times \frac{25}{51} = \frac{25}{102}$.
#### $\frac{25}{102}$
The maximum occurs at $x = -\frac{b}{2a}$.
Here, $a = -2$, $b = 12$.
$x = -\frac{12}{2(-2)} = 3$.

A company’s profit \(P(x)\) in thousands of dollars is modeled by \(P(x) = -2x^2 + 12x - 16\), where \(x\) is the number of units sold in hundreds. Find the number of units sold to maximize profit.

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