Step 1: Write price in terms of quantity.
From the demand equation \(Q=100-2P\), solve for \(P\):
\[ P=\frac{100-Q}{2}=50-\frac{Q}{2} \]
Step 2: Write revenue as a function of Q.
Revenue is price times quantity, \(R=P\times Q\):
\[ R=\left(50-\frac{Q}{2}\right)Q=50Q-\frac{Q^2}{2} \]
Step 3: Write profit (before tax) as a function of Q.
Profit is revenue minus cost, \(\pi=R-C\):
\[ \pi=\left(50Q-\frac{Q^2}{2}\right)-\left(Q^2-16Q+2000\right) \]
\[ \pi=50Q-\frac{Q^2}{2}-Q^2+16Q-2000=66Q-\frac{3}{2}Q^2-2000 \]
Step 4: Maximize profit using calculus.
A firm's profit is maximized where marginal revenue equals marginal cost, which is the same as setting the derivative of profit with respect to Q to zero.
\[ \frac{d\pi}{dQ}=66-3Q=0 \implies Q=22 \]
Since the coefficient of \(Q^2\) in \(\pi\) is negative (\(-\tfrac{3}{2}\)), the profit curve is a downward-opening parabola, so this point is indeed a maximum, not a minimum.
Step 5: Confirm by comparing profit at nearby values.
\[ \pi(22)=66(22)-1.5(22)^2-2000=1452-726-2000=-1274 \]
\[ \pi(21.5)=66(21.5)-1.5(21.5)^2-2000=1419-693.375-2000=-1274.375 \]
\[ \pi(20)=66(20)-1.5(20)^2-2000=1320-600-2000=-1280 \]
\[ \pi(19)=66(19)-1.5(19)^2-2000=1254-541.5-2000=-1287.5 \]
Profit is highest (least negative) at \(Q=22\), confirming it beats 21.5, 20 and 19.
Step 6: Rule out the other options.
21.5, 20 and 19 all give a lower profit than 22, as shown above, so none of them can be the profit-maximizing output.
Final Answer:
The profit-maximizing output, before any tax, is 22 units.
\[ \boxed{22} \]