Maximizing a quadratic profit function

Differentiate the profit function: $P'(x) = -0.04x + 60$. Setting $P'(x) = 0$ gives $-0.04x + 60 = 0$, so $x = 1500$. Since $P''(x) = -0.04 < 0$ for all $x$, the function is concave down everywhere, confirming that $x = 1500$ is indeed a maximum.

Substitute $x = 1500$ into the original profit function: $P(1500) = -0.02(1500)^2 + 60(1500) - 3000 = -45000 + 90000 - 3000 = 42000$. Therefore producing 1500 units maximizes profit, yielding a maximum profit of $42,000.