Maximizing profit from revenue and cost functions

The profit function is $P(x) = R(x) - C(x) = (60x - 0.5x^2) - (10x + 1000) = -0.5x^2 + 50x - 1000$. This is a function of the single variable $x$, defined for $x \geq 0$, so we can proceed to optimize it directly using calculus.

Differentiate to get $P'(x) = -x + 50$. Setting $P'(x) = 0$ gives $x = 50$. Since $P''(x) = -1 < 0$ for all $x$, this critical point is a maximum. Evaluating, $P(50) = -0.5(2500) + 50(50) - 1000 = -1250 + 2500 - 1000 = 250$.

Therefore producing and selling $x = 50$ units gives the maximum profit, and this maximum profit is \250$.