Calculations for profit maximization:

• Revenue function: $R(x) = 200x$
• Cost function: $C(x) = 500,000 + 80x + 0.003x^2$
• Profit function $P(x) = R(x) - C(x)$:
  $$P(x) = 200x - (500,000 + 80x + 0.003x^2) = -500,000 + 120x - 0.003x^2$$
• Differentiate with respect to $x$ and set $P'(x) = 0$:
  $$P'(x) = 120 - 0.006x = 0$$
  $$0.006x = 120 \implies x = \frac{120}{0.006} = 20,000$$
• Second derivative test: $P''(x) = -0.006 \lt 0$, confirming that $x = 20,000$ yields a maximum.
  Since $20,000 \le 30,000$ (within the production capacity), 20,000 units maximize profit. Option 2 is correct.